From the redox reaction Zn+ Ni^2+--> Zn^2+ Ni we can say that?

I have to write ionic equations for the following and I am unsure of how to do so. Here is the question and my attempt at each:

a) chromium dipped into silver nitrate
Cr3+ +Ag+(NO3)- --> Cr 3+(NO3)- +Ag+

b) gold immersed in hydrocholoric acid
NO REACTION and therefore NO EQUATION

c) nickel pellets dropped into a bath of calcium acetate
Ni2+ + Ca2+ (C2H3O2)2- --> Ca2+ + Ni 2+ (C2H3O2)2-

d) aluminium dropped into a bath of sulphuric acid
Al3+ + H22+(SO4)2- --> H22+ + Al3+(SO4)2-

e) zinc dipped inot lead(II) nitrate

Zn2+ + Pb2+ (NO3-)2 --> Pb 2+ +Zn 2+ (NO3)2

I have the rules for writing the equations but the examples given are not helping me answer these questions. I think my mistakes are:
1) I am only supposed to ionize (aq) substances and I ionized everything
2) when writing a compound I am supposed to just give the charge of the whole molecule instead of each part.

Other than that I am lost.
) yes, this is the problem. "Chromium" is Chromium metal, Cr NOT the ion, likewise Nickel is Ni, etc.

2) You can give the charge for each part, that's fine, and even write them seperately as this will make it easier to spot the spectator ions.

eg. Lithium dropped into aqueous HBr

HBr(aq) <-----> H+(aq) + Br-(aq)

Li(s) + H+(aq) + Br-(aq) -----> LiBr(aq) + 1/2H2(g)

But,

LiBr(aq) <-----> Li+(aq) + Br-(aq)

So overall,

Li(s) + H+(aq) + Br-(aq) -----> Li+(aq) + Br-(aq) + 1/2H2(g)

So Bromide is a spectator ion, and the net ionic equation is

Li(s) + H+(aq) -----> Li+(aq) + 1/2H2(g)

Okay so a) might look like this:

a) chromium dipped into silver nitrate
Cr(s) +Ag+(NO3)- (aq)--> Cr3+(NO3)- +Ag(s)

Then seperate the aqueous ions right? so:

Cr(s) + Ag+(aq) + NO3-(aq) --> Cr3+(aq) + NO3-(aq) + Ag(s)

Then we can see NO3 is a spectator ion so we would rewrite it as:

Cr(s) + Ag+(aq) --> Cr3+(aq) + Ag(s)

Cr(s) +Ag+(NO3)- (aq)--> Cr3+(NO3)- +Ag(s
Your formula for Chromium(III) Nitrate is wrong and the charge is not balanced. It should be Cr(NO3)3

Quote
Cr(s) + Ag+(aq) --> Cr3+(aq) + Ag(s)
Your previous mistake carries forward itself here. It resulted in your charges are not balanced. It should be:

Cr(s) + 3 Ag+(aq) --> Cr3+(aq) + Ag(s)
Redox reactions - in which charge is transferred between reagents - can be balanced by inspection, although it can be extremely difficult. But lets start with very simple example:

Cu + Fe3+ -> Cu2+ + Fe2+

At first sight equation seems to be already balanced, but if we check charges - it is not. There is +3 charge on the left and +5 charge on the right, so we have to do something about it. Using inspection method we should select most complicated particle first. Let's say it is Fe3+:

Cu + 1Fe3+ -> Cu2+ + Fe2+

If so, we can already add 1 in front of Fe2+:

Cu + 1Fe3+ -> Cu2+ + 1Fe2+

Trying to balance copper will lead us nowhere (atoms are balanced, but charges will be left unbalanced as they were from the beginning), so let's look at the charge - it can be balanced just like atoms. We have +3 on the left and +2 on the right - so we need additional +1 on the right side. Do you remember that we can use fractions at this stage? Half Cu2+ will do:

Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+

What is wrong now is the left side of equation - too much copper at the moment. Once again, half will do:

1/2Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+

Final touch - multiply everything by 2 to remove fractions:

Cu + 2Fe3+ -> Cu2+ + 2Fe2+

Atoms - balanced (one of copper and two of iron on both sides). Charge - balanced (+6 on both sides). We have just balanced redox equation using inspection method. But in the case of more difficult ones - like for example

FeSO4 + KMnO4 +H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O

this approach is a waste of time. Much easier and faster ways of balancing redox reactions are oxidation numbers method and half reaction method.
Oxidation takes place when electrons are removed from an atom or ion and reduction is the process by which an atom or ion gains electrons. Oxidation (loss of electrons) and reduction (gain of electrons) must always accompany one another in a reaction.
An increase in oxidation number is called oxidation; and a reduction in oxidation number is called reduction.
A substance that gains electrons in a reaction is called an oxidizing agent. A substance that transfers or loses electrons in a reaction is called a reducing agent. The stronger the tendency of an oxidizing agent to gain electrons, the greater its strength. The weaker the tendency for a reducing agent to hold electrons, the greater is its strength as a reducing agent. See the table of reduction potentials.

Data and Observations

1. Write the balanced equations for the reactions in Part 1 (only where a reaction occurred).

Cu+2 + Pb --> if nothing happened then write N.R. for no rxn
if a rxn occurs we have to indicate the products. Since this is a redox equation, I know electrons are being transferred. I remember Pb has a +2 charge from nomenclature. Cu has a +1 or +2. How do I know which to use? Well, your lab tells you you start with +2 and the observation says that you have a copper colored metal coating the Pb strip. This, of course, is Cu 0
Cu+2 + Pb --> Cu + Pb+2
The good news is all these have a +2 charge as ions and a 0 charge as an element so they equation will be balanced as written.

Your observations will tell you whether to write an equation or not.
Keep this with your observations. On the facing page WRITE THE CORRECT RESULTS AND EQUATIONS. I will give those to you now.
Cu+2 + Pb --> Cu + Pb+2 You should have seen Cu "growing" on the lead piece
Cu+2 + Zn --> Cu + Zn+2 You should have seen Cu "growing" on the Zinc piece
Pb+2 + Cu --> NR
Pb+2 + Zn --> Pb + Zn+2 You should have seen Zn (gray) "growing" on the lead piece
Zn+2 + Pb --> NR
Zn+2 + Cu --> NR

2.Which metal is the strongest reducing agent? Which metal is the weakest reducing agent?

To answer this we need to look and see which metal had the most rxns. (which test tubes showed products)
in the first two rxns above Cu+2 has been changed to Cu. The ox # has gone down = reduction so the Pb is the reducing agent. Same for the rest of the rxns. What we see is
the Pb can reduce Cu+2
Zn can reduce Cu+2
Zn can reduce Pb+2
Since Zn reduces the other two it is the strongest reducing agent. Cu can’t reduce anything so it is the weakest.
3. Use the reactions in Part 1 to prepare a table of half-reactions with the strongest oxidizing agent at the top of the table. (Example: Cu2+ + 2e- ® Cu.) See attached table for example

Use your results from Question #2. If Zn tubes had 2 rxns happen, and Cu had one then I know Zn is better at giving electrons away than Cu. Giving away electrons is oxidation. So with the example Zn would be a better at giving away electrons than Cu. You have to be able to think backwards too. If Zn is better at giving than Cu then Cu is better at accepting than Zn. Giving away makes it a reducing agent and taking makes it an oxidizing agent. So Cu is a stronger oxidizing agent than Zn
so I would see
Cu2+ + 2e- --> Cu then
Pb+2 + 2e- --> Pb then
Zn+2 + 2e- --> Zn

4. The molecular equation for the reaction in Step 10 is given. Split all soluble substances into ions

Fe+2 + SO4-2 + K+ +MnO4- + H+ +SO4-2 --> Fe+3 + SO4-2 + Mn + SO4-2 + K+ + SO4-2 + H2O

Eliminate substances that are identical and on opposite sides of the arrow
Fe+2 +MnO4- + H+ --> Fe+3 + Mn + H2O this shows an acidic environment so do a normal reodx ½ rxn balance

To label the reducing agent and oxidizing agents remember they must be reactants. If electrons are on the reactant side of the balanced ½ rxn, this is reduction. A substance that is reduced is classified as an oxidizing agent. If electrons are on the right this is oxidation and the reactant Is classified as a reducing agent
Questions and Conclusions

1.Using the Standard Reduction Table, List the half-reactions for the halogens in order with the strongest oxidizing agent on top.

To do this you need to look at the table on the last page of your lab. Find the Halogens (Group VII). The table shows each substance gaining electrons (reduction) so this is called a reduction table. The Eo column gives us the relative values for each rxn. Looking at the table F2 (a halogen) has a value of +2.87, the largest value in the chart. This makes it the best at gaining electrons. (This makes a little sense since you know that Fluorine is the most electronegative element we have. Electronegativity is the tendency to gain electrons) The larger the value, the better it is at gaining electrons. (This is another way to say oxidizing agents because oxidizing agents have to accept the electrons they take from another substance)
2. Write the balanced total reaction for Zn + Cu+2, Zn + Pb2+, and Pb + Cu+2.

Same as the rxns we wrote for the observation
3. Using The Standard Reduction Table, note that a reducing agent (on the right side of the table) has the possibility of reacting with any oxidizing agent (on the left). (The rate of the reaction and the concentration of ions are not taken into account.)
After studying the table, name the substance which can:

a. reduce Pb+2 to Pb but cannot reduce Ni+2 to Ni

Look for something on the right that is above Ni+2 but below Pb+2
b. Oxidize I- to I2 but cannot oxidize Br-1 to Br2

This is reverse of the one above so look for something on the right above I- but below Br-
4. The following equations describe the reaction of copper with nitric acid. Using the first, show both the oxidation and the reduction half-reactions.

Cu + HNO3 --> Cu+2 + NO + H2O (NO is colorless)

Copper is going from Cu which is 0 to Cu+2 (figure out what this is)
So some other reactant must be doing the opposite. Eliminate the spectators in the beginning so HNO3 becomes NO3-1 (what does it change into?), Now just balance like our homework. (O’s by adding water, H+ to balance H’s, etc)
NO + O2 --> NO2 (NO2 is brown and noxious)
ACIDS and BASES
Acids = substances which ionize to form H+ in solution, usually referred to as the HYDRONIUM ION H3O+
Common acids are HCl, HNO3, CH3COOH (acetic acid or vinegar), lemon or lime juice (citric acid), vitamin C (ascorbic acid).
Bases = substances which react with the H+ ions formed by acids, usually by causing an increase in the OH-concentration in aqueous solution. The most common bases are NaOH, KOH, and Ca(OH)2.
another base which does not contain OH- is NH3 (ammonia).
when NH3(g) dissolves in water it forms NH4+ and OH-.
Ammonia is a weak base, and is a weak electrolyte.
Strong acids and bases...see lecture.
Classification of Electrolytes
Compounds can be classified as strong electrolyte, weak electrolyte, and non electrolyte by looking at their solubility.
If a compound is water soluble and ionic, then it is probably a strong electrolyte.
If a compound is water soluble and not ionic, and is a strong acid, then it is a strong electrolyte.
Similarly, if a compound is water soluble and not ionic, but is a strong base, then it is a strong electrolyte.
If a compound is water soluble and not ionic, and is a weak acid or weak base, then it is a weak electrolyte.
Otherwise, the compound is probably a non-electrolytes.

Neutralization and salt formation
Acid + base -----> salt + water
A salt is a compound in which the anion is from an acid and the cation is from a base.

IONIC EQUATIONS.
These are used to highlight reactions between ions
See lecture for details.

REACTIONS IN AQUEOUS SOLUTIONS.
Metathesis Reactions
Metathesis reactions involve swapping ions in solution:

AX + BY -----> AY + BX.
Metathesis reactions will lead to a change in solution if one of three things occurs:
1. an insoluble solid is formed (precipitate),
2. weak or nonelectrolytes are formed, or
3. an insoluble gas is formed.
1. Precipitation Reactions
A precipitate is an insoluble solid that forms when two solutions are mixed.
TO KNOW WHETHER SOMETHING IS SOLUBLE OR NOT...

A solute is soluble in water if more than 0.01 mol of the substance
will dissolve in enough water to make 1 liter of solution.
LEARN SOLUBILITY RULES P. 97!
Consider

2KI(aq) + Pb(NO3)2(aq) ----> PbI2(s) + 2KNO3(aq)
Both KI(aq) + Pb(NO3)2(aq) are colorless solutions.
When mixed, they form a bright yellow precipitate of PbI2 and a solution of KNO3.
The final product of the reaction contains solid PbI2, aqueous K+ and aqueous NO3- ions.

The molecular equation lists all the species as molecules:

2KI(aq) + Pb(NO3)2(aq) -----> PbI2(s) + 2KNO3(aq)
However, we know that certain substances exist as ions in solution.
The full ionic equation lists all ions:
2K+(aq) +2I-(aq) + Pb2+(aq) + 2NO3-(aq) -----> PbI2(s) + 2K+(aq) + 2NO3-(aq)
The net ionic equation cancels spectator ions that are unchanged:
2I-(aq) + Pb2+(aq) ----> PbI2(s)
2. weak or nonelectrolytes are formed

E.g. Dissolution of Mg(OH)2 in Acid
Milk of magnesia is a suspension of Mg(OH)2 in water.
Mg(OH)2 is relatively insoluble in neutral water.
In acidic solutions the Mg(OH)2 dissolves.
In acidic solution, the hydronium ions (H3O+) react with the OH- from the Mg(OH)2 to form water.
In the process Mg2+ ions are free to move about the solution.
If HCl is used as the acid, the overall chemical equation is

Mg(OH)2(s) + 2HCl(aq) -----> MgCl2(aq) + 2H2O(l)
The net ionic equation is
Mg(OH)2(s) + 2H3O+(aq) -----> Mg2+(aq) + 4H2O(l)
Or Iron (III) Oxide with nitric acid.....see lecture

3. if insoluble gases are formed
E.g. FeS + 2HCl ----> H2S + FeCl2
Net ionic equation is
FeS + 2H+ ----> H2S + Fe2+

Introduction to Oxidation-Reduction Reactions
Oxidation and Reduction
Oxidation = Loss of Electrons
e.g. Zn ------> Zn2+ + 2 e-
Here zinc has been oxidized to zinc (II) ions

Reduction = Gain of Electrons
e.g. 2H3O+ + 2 e- ------> H2 + 2H2O
Here protons (represented as the hydronium ion) have been reduced to hydrogen gas.

In all reduction-oxidation (redox) reactions, one species is reduced at the same time another is oxidized.
The species that causes oxidation is called the oxidizing agent.
The species that causes reduction is called the reducing agent.
The oxidizing agent is always reduced and the reducing agent oxidized.
The substance that is oxidized loses electrons to the substance that is reduced.

Oxidation of Metals be Acids and Salts
It is common for metals to produce hydrogen gas when they react with acids. e.g. Mg and HCl (aq)

Mg(s) + 2HCl(aq) ------> MgCl2(aq) + H2(g).

In the process the metal is oxidized and the H+ is reduced.
It is possible for metals to be oxidized in the presence of a salt:

Fe(s) + Ni(NO3)2(aq) ------> Fe(NO3)2(aq) + Ni(s).

The net ionic equation shows the redox chemistry well:

Fe(s) + Ni2+(aq) ------> Fe2+(aq) + Ni(s).

In this reaction iron has been oxidized to Fe2+ while the Ni2+ has been reduced to Ni.

The Activity Series
We can list metals in order of decreasing ease of oxidation.
This list is the activity series.
A metal in the activity series can only be oxidized by a metal ion below it.
The metals at the top of the activity series are called active metals.
The metals at the bottom of the activity series are called noble metals.

Formation of Silver Crystals on Copper Wire :(Demonstration : narrative)
Copper wire is placed in a beaker.
A solution of silver nitrate is added to the beaker.
Copper reduces Ag+ to Ag.
The Cu is oxidized to Cu2+

The full molecular equation

2AgNO3(aq) + Cu(s) ------> Cu(NO3)2(aq) + 2Ag(s)
The net ionic equation is
2Ag+(aq) + Cu(s) -----> Cu2+(aq) + 2Ag(s)
Note that the solution changes from colorless to green indicating the presence of copper(II) nitrate.
Silver crystals form on the Cu wire.

Solution Stoichiometry and Chemical Analysis
Recognize that there are two different types of units:
1. laboratory units (the macroscopic units which we measure in lab) and
2. chemical units (the microscopic units which relate to moles).
Always convert the laboratory units into chemical units first.
Grams are converted to moles using molar mass.
Volume or molarity are converted into moles using M = mol/L.
Use the stoichiometric coefficients from the balanced chemical equation to move between reactants and product.
Convert the laboratory units back into the required units.
Moles are converted to grams using molar mass.
Moles are converted to molarity or volume using M = mol/L.
Example worked in lecture.

Titrations
A titration is an experiment in which the molarity of a substance is measured by knowing the molarity of another substance.
Example: Suppose we know the molarity of an NaOH solution and we want to find the molarity of an HCl solution.
WE know....
molarity of NaOH, volume of HCl.
What do we want?
Molarity of HCl.
What do we do?
Take a known volume of the HCl solution (20.0 mL, say) and measure the number of mL of NaOH solution required to react completely with the HCl solution.
What do we get?
Volume of NaOH. Since we already have the molarity of the NaOH, we can calculate moles of NaOH.
Next step?
We also know HCl + NaOH -----> NaCl + H2O. Therefore, we know moles of HCl.
Can we finish?
Knowing mol (HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.
Example will be worked in lecture.

simple example: NH3 (aq) + ClO- (aq) ® N2H4 (l) + Cl- (aq)

The half-reactions will be : NH3 ® N2H4 and ClO- ® Cl-

So we get 2 NH3 ® N2H4 + 2H+ + 2e-

and 2H+ + 2e- + ClO- ® Cl- + H2O

Another example: SO32- + MnO4- ® SO42- + Mn+2 in acid solution

(sulfurous acid) permanganate (sulfate)

SO3-2 ® SO42- + e- (oxidation)

MnO4- + e- ® Mn+2 (reduction)

To balance first eq. need to add H+ and H2O:

SO3-2 + H2O ® SO42- + 2e- + 2 H+

Need H2O on RHS to balance O: MnO4- + e- ® Mn+2 + 4H2O

Then need H+ on LHS to balance charge and H's: MnO4- + 8H+ + 5e- ® Mn+2 + 4H2O

However, one half-reaction has 5e's the other 2, so 10 is the lowest common denominator:

5SO32- + 5H2O ® 5SO42- + 10e- + 10H+

and 2MnO4- + 16H+ + 10e- ® 2Mn+2 + 8H2O

So adding these two together we get: 2MnO4- + 5SO32- + 6H+ ® 2Mn+2 + 3H2O + 5SO42-

Check that the answer is balanced etc.

Oxidation of zinc by copper: A neat demonstration is to take a strip of zinc metal (its a shiny metal which looks like silver) and stick it in a solution of copper sulfate (which is a beautiful deep blue). After a while the solution becomes colorless and the zinc strip becomes copper-colored, because it has become coated with copper!

The reaction is Zn + CuSO4 (blue) ® Cu + ZnSO4(colorless)

The oxidation half-reaction is:

Zn ® Zn+2 + 2e-

The reduction half-reaction is:

Cu+2 + 2e- ® Cu

Since the number of e's in each half-reaction is the same we just add the two to get the final balanced equation:

Zn + CuSO4 ® Cu + ZnSO4

Suppose we set up the system so that we could transfer the e-'s from one half-reaction to theother, we could then use the system to generate electricity! This is the principle of what's known as an

ELECTROCHEMICAL CELL

and is the basis of many common batteries. If we just stuck a piece of Zn in, say, a solution of zinc sulfate, and a piece of copper in a solution of copper sulfate, and connected the metals by a wire with a voltmeter in the circuit, we would find that nothing happened! Why not? Because the set-up would separate charge in the sense of making more Zn2+ than sulfate in one compartment and less Cu2+ than sulfate in the other. So what we need is a way of transporting ions from one compartment to another. This is usually done with what is known as a salt-bridge or a porous divider.

(see figure).

The purpose of the salt bridge is to allow ions to transfer to maintain electrical balance in each electrode compartment. The cell then works as follows:

Electrons are lost by the zinc (oxidation): Zn ® Zn+2 + 2e-

which generates zinc ions - these go into the solution and are neutralized by sulfate (coming through the salt-bridge from the adjacent compartment). The electrons flow through the wire to the copper electrode (a conductor in the solution - usually either the metal undergoing oxidation or reduction, or a "neutral" conductor such as carbon or platinum). The voltmeter records the electrical potential, which is analogous to gravitational potential (a useful analogy is to consider electrons similar to water - both flow "downhill".)

For 1 M solutions of zinc and copper sulfate (standard conditions) the voltage produced by this cell is 1.1 V. A volt is a measure of the electrical potential, or if you like, the "electron pressure".

In the second part of the cell the copper ions become reduced (Cu2+ + 2e- ® Cu), thus copper ions are lost from the solution and deposit as copper metal on the electrode (it can be made of any conductor). The loss of the copper ions leads to an excess of sulfate which migrates across the salt-bridge to the other compartment. (The salt bridge need not be made of a sulfate salt.)

A few important definitions:

Oxidation occurs at the anode (remember O and A come before C and R) (and is marked negative - because its where the electrons are released).

Reduction occurs at the cathode

Anions flow to the anode

Cations flow to the cathode

An electrolyte is a salt solution that allows the flow of both a current (and potentially ions).

Electrode - a conducting material, may or may not be part of the oxidation or reduction half-reactions. Common examples are platinum, carbon, zinc.

ELECTRICAL POTENTIAL or EMF (electromotive force)

An electron moves from an electrode of higher electrical potential to one of lower potential, i. e. energetically downhill. In so doing the electron can do work. The amount of work done is proportional to the number of electrons (the charge) and the potential energy difference,

Electrical work = charge x potential difference

A Coulomb (C) is the quantity of charge that passes a point in an electrical circuit when a current of 1 amp flows for 1 sec.

The charge on one electron is 1.602 x 10-19 C.

Electrical potential is measured in volts where 1 V is defined as 1 J of work performed when 1 C of charge passes through a potential difference of 1 V. i. e. 1 V = 1 J/C

Since the potential of a cell depends on the concentration of reactants, electrode potentials are given as standard potentials for conditions of 1 M or 1 atm.

They are denoted as Eo and are normally for 25° C.

For half-reactions the Eo potentials are called standard reduction potentials.

They are measured using the hydrogen electrode cell (2 H+ (aqueous, 1M) + 2 e- ® H2 (g)(1 atm) - with a Pt electrode) as one half-reaction, with its standard potential chosen as 0 V.

Any electrode at which a reduction half-reaction shows a greater tendency to occur than does 2 H+ + 2 e- ® H2 has a positive Eo. (And vice versa)

By convention, standard electrode potentials are given for the half-cell reaction written as a reduction e. g.

Zn+2 + 2e- ® Zn Eo = -0.76 V

Cu2+ + 2e- ® Cu Eo = +0.337 V

Eocell = Eoox + Eored = +0.76 + 0.34 = 1.1 V (as we will see shortly, if the cell potential is positive it means the reaction is spontaneous.)

NOTE for the oxidation reaction Zn ® Zn2+ + 2 e- we have to change the sign, since we are now considering the reverse reaction i. e. oxidation.

There are tables of standard electrode potentials e. g. Table 17-1 and in the appendix of your text.

Knowing the standard electrode potentials for each half-cell reaction in a cell allows us to calculate the cell potential. The more easily an element is reduced, the higher the standard electrode potential.

In the analogy with water, the more positive the Eo the higher the "elevation" of the electron, i. e. the more potential energy it has.

Here's an example: we could replace the Cu in the Cu/Zn cell with nickel. The measured cell Eo at 25° C is +0.51 V. What is the Eo for the Ni half-cell reaction?

We can tell that the Ni2+ gets reduced, because we see Ni plating out on the Ni electrode.

First we write the half-reactions:

Zn ® Zn2+ + 2e- Eo = +0.76 V

Ni2+ + 2 e- ® Ni Eo = ?

Net

Reaction Zn + Ni 2+ ® Zn2+ + Ni Eonet = +0.51 V (remember Eocell will always be +)

Therefore Eo for Ni2+ + 2 e- ® Ni must be 0.51 - 0.76 = -0.25 V.

Remember in Tables of standard reduction potentials:

The Eo values are for the oxidized form + electrons ® reduced form

If writing the half reaction in the reduced form ® oxidized form + electrons you must change the sign of Eo

All half-reactions are reversible

The more positive the value for Eo the better the oxidizing ability of the species on the left side of the reaction - thus F2 + 2e- ® 2 F- with Eo = +2.8 V is the best oxidizing agent in table 17-1, and the Li+ ion is the weakest oxidizing agent.

The more negative Eo is the more likely the species will occur in the opposite direction, i. e. Li metal is a very strong reducing agent.

The reaction of any species on the left in the Table with one below it on the right, will lead to a product-forming reaction i. e. a cell which works.

Electrochemical potentials depend on the nature of the reactants and products and their concentrations, NOT on the quantities of material used. I. e. the coefficients used to multiply half-reactions to balance the number of electrons do not affect Eo. For example

Fe3+ + e- ® Fe2+ Eo = +0.77 V

and 2 Fe3+ + 2 e- ® 2 Fe2+ Eo = +0.77 V
Electrolytes and Non-electrolytes

In solid NaCl the ions in the lattice are held in place by strong ionic bonds.
This means the ions cannot move about in an electric field.
Therefore, solid NaCl does not conduct electricity.
When NaCl is added to water, the salt dissociates into Na+(aq) and Cl-(aq).
In the presence of an electric field, the solution can conduct electricity.
Strong electrolytes exist in solution entirely (or almost entirely) as ions.
Non-electrolytes (e.g. sugar) do not conduct electricity at all. (ie. there are no ions in solution
Strong and Weak Electrolytes

Water has a low conductivity so the bulb does not burn in pure water.
Hydrogen chloride is soluble in water.
In water, HCl ionizes into H+ and Cl-.
Since HCl is a strong electrolyte, in water there are no HCl molecules, only H+ and Cl- ions.
When the HCl(g) is bubbled through the solution, the bulb glows brightly because of the presence of the ions.
When acetic acid replaces the HCl solution the bulb glows less brightly.
Acetic acid exists as a mixture of acetic acid molecules and H+ and acetate ions in solution.
Because acetic acid is a mixture of ions and parent molecules in solution, acetic acid is a weak electrolyte.
Therefore, the acetic acid solution does not cause the bulb to glow brightly.

--------------------------------------…

Dissolution of an Ionic Compound

Fig. 4.3 pg. 124

Properties of Solutes in Aqueous Solution

Ionic Compounds in Water

When an ionic compound dissolves in water, the ions dissociate.
This means that in solution, the solid no longer exists as a well ordered arrangement of ions in contact with each other.
Instead, each ion is surrounded by water molecules.
The positive ions have the oxygen atoms of water pointing towards the ion, negative ions have the hydrogen atoms of water pointing towards the ion.
The transport of ions through the solution causes electric current to flow through the solution.
When a molecular compound dissolves in water (e.g., CH3OH), there are no ions formed. Therefore, there is nothing in the solution to transport electric charge and the solution does not conduct electricity.

--------------------------------------…

Dissolution of NaCl

http://hogan.chem.lsu.edu/matter/chap26/…

Dissolution of NaCl

When an ionic solid is placed in water, the ions dissociate.
If the attractions between the ions and water molecules overcome the ionic attractions in the lattice, the salt is soluble.
Cations are attracted to the O atoms in water (through lone pairs).
Anions are attracted to the H atoms in water.
In solution the ions are separated.
Potassium Iodide and Lead Nitrate

Fig. 4.4 pg. 126

Metathesis Reactions

Metathesis reactions involve swapping ions in solution:

AX + BY AY + BX.
Ion swapping will lead to a chemical reaction in solution if one of three things occurs:
an insoluble solid is formed (precipitate),
weak or nonelectrolytes are formed, or
an insoluble gas is formed.
Precipitation Reactions

A solute is soluble in water if more than 0.01 mol of the substance will dissolve in enough water to make 1 liter of solution.
A precipitate is an insoluble solid that forms when two solutions are mixed.
Consider 2KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2KNO3(aq)
Both KI(aq) + Pb(NO3)2(aq) are colorless solutions. When mixed, they form a bright yellow precipitate of PbI2 and a solution of KNO3.
The final product of the reaction contains solid PbI2, aqueous K+ and aqueous NO3- ions.
The molecular equation lists all the species as molecules:

2KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2KNO3(aq)
However, we know that certain substances exist as ions in solution.
The full ionic equation lists all ions:

2K+(aq) +2I-(aq) + Pb2+(aq) + 2NO3-(aq) PbI2(s) + 2K+(aq) + 2NO3-(aq)
The net ionic equation cancels those ions that are unchanged:

2I-(aq) + Pb2+(aq) PbI2(s)
Solubility of Ionic Compounds

Table 4.1 pg. 127

Metathesis Reactions

In order to determine whether or not a substance will dissolve or form a precipitate, certain rules apply.
These rules often have important exceptions.
Net Ionic Equations

Text slide.

Net Ionic Equations

Write down balanced chemical reaction described in problem.
Rewrite this reaction in ionic form (ionic equation).
Cancel out equal quantities of all ions which are identical on both sides of reaction arrow.
Convert the reactions below into net ionic equations:

Reaction between Na2CO3(aq) and MgSO4(aq)

Reaction between Pb(NO3)2 (aq) and Na2S(aq)

Reaction between (NH4)3PO4(aq) and CaCl2(aq)

--------------------------------------…

Space Filling Models of Acids

Top of Fig. 4.6 pg. 131

Properties of Solutes in Aqueous Solution

Acids

Definitions:

Dissociation = pre-formed ions in a solid move apart in solution.
Ionization = neutral substance forms ions in solution.
Acid = substances which ionizes to form H+ in solution.
Common acids are HCl, HNO3, CH3CO2H (acetic acid or vinegar), lemon, lime, vitamin C.
Shown here are HCl (top), nitric acid (middle), and acetic acid (bottom). O atoms are shown in red, H in white, Cl in green, N in blue and C in black.
Bases = substances which react with the H+ ions formed by acids.
Common bases are NH3 (ammonia), drano, milk of magnesia.

--------------------------------------…

Aqueous Acids

http://hogan.chem.lsu.edu/matter/chap26/…

Introduction to Aqueous Acids

Acids increase the concentration of [H+] when dissolved in water.
When HCl is added to water it completely dissociates into H+ and Cl-.
HCl is a strong electrolyte.
There is no undissociated HCl at the end of the reaction, so HCl is a strong acid.
HNO3 is also a strong acid. Therefore, in water it forms H+ and NO3- with no undissociated HNO3 left.
Acetic acid is a weak acid. Therefore, it exists as a mixture of undissociated acetic acid, H+, and acetate ions in solution.

--------------------------------------…

Reaction Between Ammonia and Water

Bottom of Fig. 4.6 pg. 131

Properties of Solutes in Aqueous Solution

Strong and Weak Acids and Bases

Strong acids and bases are strong electrolytes.
Therefore, they are completely ionized in solution.
We write this as HCl H+ + Cl-.
Weak acids and bases are weak electrolytes.
Therefore, they are partially ionized in solution.
Since H+ is a naked proton, we refer to acids as proton donors and bases as proton acceptors.
In this figure, we see the proton transfer between NH3 (a weak base) and water (a weak acid). Since there is a mixture of NH3, H2O, NH4+, and OH- in solution, we write:

--------------------------------------…

Aqueous Bases

http://hogan.chem.lsu.edu/matter/chap26/…

Introduction to Aqueous Bases

Bases increase the [OH-] concentration in aqueous solution.
The most common bases are NaOH, KOH, and Ca(OH)2.
When solid NaOH is added to water it completely dissociates into Na+ and OH- in solution.
Since there is no undissociated NaOH at the end of the reaction, NaOH is a strong base.
Strong bases are strong electrolytes.
NH3 is an example of a base which does not contain OH-.
However, when NH3(g) dissolves in water it forms NH4+ and OH-.
Ammonia is a weak base, so there is an equilibrium mixture of NH3, NH4+, and OH- in solution.
Most of the ammonia exists as NH3 in solution, it is a weak electrolyte.

--------------------------------------…

Strong Acids and Bases

Table 4.2 pg. 132

Strong Acids and Bases

Table 4.2 Common Strong Acids and Bases

Strong Acids Strong Bases
Hydrochloric, HCl Group 1A metal hydroxides (LiOH,
Hydrobromic, HBr NaOH, KOH, RbOH, CsOH)
Hydroiodic, HI .
Chloric, HClO3 Heavy group 2A metal hydroxides
Perchloric, HClO4 (Ca(OH)2, Sr(OH)2, Ba(OH)2)
Nitric, HNO3 .
Sulfuric, H2SO4 .

--------------------------------------…

Weak and Strong Electrolytes

Text slide.

Properties of Solutes in Aqueous Solution

Compounds can be classified as strong electrolytes, weak electrolytes, and nonelectrolytes by looking at their solubilities and classifications.

Water-soluble and ionic = strong electrolyte (NaCl: ionic salt, NH4Cl: ionic weak acid, NaC2H3O2: ionic weak base).
Water-soluble strong acid or strong base = strong electrolyte (HCl: ionic strong acid, NaOH: ionic strong base).
Water-soluble but not ionic weak acid or base = weak electrolyte (NH3: molecular weak base, HC2H3O2: molecular weak acid).
Not water-soluble but ionic = weak electrolyte (CaSO4: "insoluble" ionic).
Water-soluble but not ionic = nonelectrolyte (CH3OH: soluble molecular).
Not water-soluble and not ionic: = nonelectrolyte (C6H6: insoluble molecular).

--------------------------------------…

Natural Indicators

http://hogan.chem.lsu.edu/matter/chap27/…

Natural Indicators

An indicator is a substance that changes color as a function of pH.
Most indicators are dyes.
Some indicators can be extracted from natural products.
Example: the dye which is characteristic of the red color in red cabbage leaves is an indicator.
The colorless dye is extracted with methanol.
The dye turns yellow/green in base and red in acid.
Tea is another natural indicator. It changes from brown to yellow/orange in acid. In this experiment, lemon juice is used as the acid.

--------------------------------------…

Dissolution of Mg(OH)2 by Acid

Fig. 4.8 pg. 135.

Dissolution of Mg(OH)2 with HCl

Phillip's Milk of Magnesia bottle in Fig. 4.8 a contains milky mixture of Mg(OH)2 and water.
Concentrated HCl solution is added from clear bottle in Fig. 4.8 b.
All Mg(OH)2 has dissolved in Fig. 4.8 c.
Mg(OH)2(s) + 2 HCl(aq) MgCl2 (aq) + 2 HOH(l)
Net ionic equation: Mg(OH)2(s) + 2 H+(aq) Mg2+(aq) + 2 HOH(l)

--------------------------------------…

Magnesium Hydroxide

http://hogan.chem.lsu.edu/matter/chap26/…

Dissolution of Mg(OH)2 in Acid

Milk of magnesia is a suspension of Mg(OH)2 in water.
Mg(OH)2 is relatively insoluble in neutral water.
In acidic solutions the Mg(OH)2 dissolves.
The solid Mg(OH)2 consists of layers of Mg2+ ions sandwiched between OH- ions.
In water there is an equilibrium between the solid Mg(OH)2 and the ions.
In acidic solution, the hydronium ions (H3O+) react with the OH- from the Mg(OH)2 to form water.
In the process Mg2+ ions are free to move about the solution.
If HCl is used as the acid, the overall chemical equation is:

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
The net ionic equation is:

Mg(OH)2(s) + 2H3O+(aq) Mg2+(aq) + 4H2O(l)

--------------------------------------…

Electron Transfer Reactions

Right margin pg. 139

Introduction to Oxidation-Reduction Reactions

Oxidation and Reduction

In all reduction-oxidation (redox) reactions, one species is reduced at the same time another is oxidized.
The species that causes oxidation is called the oxidizing agent.
The species that causes reduction is called the reducing agent.
The oxidizing agent is always reduced and the reducing agent oxidized.
The substance that is oxidized loses electrons to the substance that is reduced.

--------------------------------------…

Redox I

http://hogan.chem.lsu.edu/matter/chap26/…

Oxidation-Reduction Reactions I

Consider zinc metal in the presence of oxygen at high temperature.
The zinc reacts to form ZnO.
Electrons are transferred from zinc to oxygen.
So, the zinc becomes Zn2+ while the oxygen becomes O2-.
Zinc has been oxidized (i.e. its oxidation state has increased) by oxygen (the species causing the oxidation).
Oxygen has been reduced (i.e. its oxidation state has decreased) by zinc (the reducing agent).
In general, the loss of electrons is oxidation: Zn Zn2+ + 2e-.
The gain of electrons is reduction: O + 2e- O2-.

--------------------------------------…

Redox II

http://hogan.chem.lsu.edu/matter/chap26/…

Oxidation-Reduction Reactions II

When Zn is added to dilute acid, H2(g) and Zn2+ ions are formed:

Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
H3O+ attacks the zinc atoms and an electron is transferred from the hydronium ion (H3O+) to the zinc metal.
After two such transfers, zinc is oxidized to Zn2+ while H+ is reduced to hydrogen gas.
When zinc is placed in aqueous copper sulfate, a similar reaction occurs: Zn is oxidized by Cu2+ to Zn2+ while Cu2+ is reduced to Cu0 by Zn.
When zinc is oxidized it causes reduction.
Therefore, zinc is the reducing agent.
The substances that are reduced, cause oxidation and are the oxidizing agents.

--------------------------------------…

Oxidation of Mg by Acid

Fig. 4.13 pg. 141

Introduction to Oxidation-Reduction Reactions

Oxidation of Metals be Acids and Salts

It is common for metal to produce hydrogen gas when they react with acids. Shown here is the reaction between Mg and HCl:

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g).
In the process the metal is oxidized and the H+ is reduced.
It is possible for metals to be oxidized in the presence of a salt:

Fe(s) + Ni(NO3)2(aq) Fe(NO3)2(aq) + Ni(s).
The net ionic equation shows the redox chemistry well:

Fe(s) + Ni2+(aq) Fe2+(aq) + Ni(s).
In this reaction iron has been oxidized to Fe2+ while the Ni2+ has been reduced to Ni.

--------------------------------------…

Formation of Silver Crystals

http://hogan.chem.lsu.edu/matter/chap27/…

Formation of Silver Crystals on Copper Wire

See Fig. 4.15 pg. 125
Copper wire is placed in a beaker (or test tube).
A solution of silver nitrate is added to the beaker (test tube).
Copper reduces Ag+ to Ag.
The Cu is oxidized to Cu2+.
The full molecular equation:

2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s)
The net ionic equation is:

2Ag+(aq) + Cu(s) Cu2+(aq) + 2Ag(s)
Note that the solution changes from colorless to blue indicating the presence of copper(II) nitrate.
Silver crystals form on the Cu wire.

--------------------------------------…

Solution Formation From Solid

Fig. 4.16 pg. 146

Solution Composition

Molarity

A solution is made when one substance (the solute) is dissolved in another (the solvent).
The solute is the substance that is present in smallest amount.
Solutions in which water is the solvent are called aqueous solutions.
Solutions can be prepared with different concentrations by adding different amounts of solute to solvent.
The amount (moles) of solution per liter of solution is the molarity of the solution.
By knowing the molarity of a quantity in liters of solution, we can easily calculate the number of moles (and, by using molar mass, the mass) of solute.
In this figure, blue copper(II) sulfate pentahydrate, CuSO4·5H2O, is weighed (62.4 g, 0.250 mol) and placed in a 250 mL volumetric flask. A little water is added and the flask swirled to ensure the copper sulfate dissolves. When all the copper sulfate has dissolved, the flask is filled to the mark with water. The molarity of the solution is 0.250 mol / 0.250 L = 1.00 M.

--------------------------------------…

Solution Formation

http://hogan.chem.lsu.edu/matter/chap26/…

Solution Formation from a Solid

To form a solution of known concentration (standard solution) a certain amount of solid must first be accurately weighed.
To prepare 250 mL of 1.00 M CuSO4 solution, 0.250 moles of copper sulfate are required.
Copper sulfate occurs as the pentahydrate, CuSO4·5H2O (FW = 249.7 g/mol). Therefore, 62.40 g of CuSO4·5H2O are required.
The copper sulfate is added to a 250 mL volumetric flask.
Some water is added and the flask is swirled to dissolve the salt.
Once all the salt is dissolved, enough water is added to bring the solution to the mark on the volumetric flask.

--------------------------------------…

Dilution

http://hogan.chem.lsu.edu/matter/chap26/…

Solution Formation By Dilution

See Fig. 4.19 pg.131
Suppose we want to prepare 250 mL of a 0.100 M solution of copper sulfate starting with a 1.00 M stock solution.
We carefully pipet 25.00 mL of the stock solution and add it to a 250 mL volumetric flask.
Some water is added to the volumetric flask and the flask is swirled to ensure good mixing.
Then enough water is added to make the total volume of the solution 250 mL.

--------------------------------------…

Volumetric Calculations

Text slide.

Volumetric Calculations

To work problems involving molarity, volume, and grams first calculate number of moles of solute you are working with.

(vol of solution) x (concentration) = (moles of solute)
(grams of solute) _ (MM of solute) = (moles of solute)
(vol of solute) x (density of solute) _ (MM of solute) = (moles of solute)
Next use dimensional analysis to calculate the final quantity you need.

How many g of K2Cr2O7 in 50.0 mL of 0.850 M solution?
What is molarity of 2.50 g of (NH4)2SO4 in a 250 mL solution?
How many mL of 0.387 M CuSO4 contains 1.00 g of solute?
How many g of AgNO3 needed for 100.0 mL of 0.200 M solution?
How much 6.0 M HNO3 needed to create 250 mL of 1.0 M HNO3 ?
Calculate molarity of 250 mL glycerol (C3H8O3) solution made using 50.000 mL of pure glycerol. Density = 1.2656 g/mL.

--------------------------------------…

Titration

Fig. 4.21 pg. 153

Solution Stoichiometry and Chemical Analysis

Titrations

A titration is an experiment in which the molarity of a substance is measured by knowing the molarity of another substance.
Example: Suppose we know the molarity of an NaOH solution and we want to find the molarity of an HCl solution.
First let us examine what we know: molarity of NaOH, volume of HCl.
What do we want? Molarity of HCl.
What do we do? Take a known volume of the HCl solution (20.0 mL, say) and measure the number of mL of NaOH solution required to react completely with the HCl solution.
What do we get? Volume of NaOH. Since we already have the molarity of the NaOH, we can calculate moles of NaOH.
Next step? We also know HCl + NaOH NaCl + H2O. Therefore, we know moles of HCl.
Can we finish? Knowing mol(HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.

--------------------------------------…

Titration

http://hogan.chem.lsu.edu/matter/chap26/…

Acid-Base Titration

A titration is an experiment in which the concentration of an unknown acid or base is deduced from the concentration of a standard solution of base or acid.
Consider the addition of a standard NaOH solution added to a solution of HCl.
A pH meter is used to measure the acidity of the solution.
When 0.100 M NaOH solution is added to a 0.100 M solution of HCl, the pH increases from 1.0.
As more base is added, the pH gradually increases.
Near the equivalence point (the point defined by stoichiometry as the point at which enough base has been added to neutralize the acid), the pH begins to change dramatically with small additions of base.
The last few drops of base change the pH from around 3 to 7.
At the equivalence point the pH is 7.0 because the solution is neutral.
Addition of one drop of NaOH after equivalence causes the pH to increase to about 10.0.
Note that the phenolphthalein indicator changes color from colorless to red after the equivalence point.
The end point of the titration is the point at which we observe a color change.
The end point is experimentally determined, the equivalence point is determined by stoichiometry
Claims:
We claim:

1. A method for preparing a chromate-treated zinc-plated steel strip, comprising

effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to 100 C/dm.sup.2.

2. A method for preparing a chromate-treated zinc-plated steel strip, comprising

effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, 0.05 to 25 grams per liter, calculated as Al.sub.2 O.sub.3, of alumina sol, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to 100 C/dm.sup.2.

3. A method for preparing a chromate-treated zinc-plated steel strip, comprising

effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to substantially less than 30 C/dm.sup.2.

4. A method for preparing a chromate-treated zinc-plated steel strip, comprising

effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, 0.05 to 25 grams per liter, calculated as Al.sub.2 O.sub.3, of alumina sol, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, and substantially free of sulfate and chloride, at a current quantity density of 1 to 50 A/dm.sup.2 and to an electric of 5 to substantially less than 30 C/dm.sup.2.

Description:
BACKGROUND OF THE INVENTION

This invention relates to chromate-treated zinc-plated steel strips having high corrosion resistance without coating, good coating adherence, and firm adhesive bond to vinyl chloride and similar resins, as well as a method for making the same.

Most of currently available zinc-plated steel strips are zinc electroplated steel strips and zinc hot dipped or galvanized steel strips. Since they are not necessarily sufficient in corrosion resistance, various zinc alloy plated steel strips including Zn-Ni, Zn-Fe, and Zn-Al alloy plated ones have been developed and marketed. These advanced products may be used as such, but are often used after a chromate treatment which serves for white rust prevention and as a primary treatment for subsequent coating.

Most currently used chromate treatments are reactive chromate treatments which are applied to those products which require a white rust generating time of 24 to 100 hours in the standard salt spray test. In the reactive chromate treatments, the quantity and nature of the resulting chromate film are largely affected by the reactivity of the underlying metal. More particularly, because of their relatively high reactivity, zinc-plated steel strips can be coated with a chromate film only by dipping the strips in conventional chromate solutions having a relatively low etching power. Since zinc alloy-plated steel strips, however, are low reactive, a chromate film can not fully grow thereon in the conventional chromate solutions. Although corrosion resistance is improved by increasing the quantity of a chromate film deposited, an excessively built-up chromate film turns to be yellow due to hexavalent chromium and thus exhibits an undesirable appearance. When such thickly chromated strips are coated with paint, the adherence between the chromate film and the paint is poor.

As a high speed plating line becomes widespread, post-treatment procedures also want speeding up. In order for the reactive chromate treatment to produce a competent quantity of a uniform chromate film, continuous dipping or spraying for a certain period of tim, typically 4 to 10 seconds is necessary. A common approach for accommodating with the high speed line is to increase the number of tanks to extend the reaction time.

Another class of chromate treatment including coating and electrolytic chromate treatments becomes recently available because these treatments are little affected by the reactivity of steel strips and take a short time to completion. The coating chromate treatment is applied to those products which require a corrosion resisting time of 200 hours or more in the standard salt spray test. The electrolytic chromate treatment results in more improved adherence to a coating as compared with the reactive and coating chromate treatments because the resulting chromate film consists essentially of trivalent chromium.

The coating chromate treatment is generally practiced by a method of adding colloidal silica as a film forming agent as disclosed in Japanese Patent Publication No. 42-14050. Another method for conducting the coating chromate treatment involves applying a chromate solution containing an organic polymer by roll coating or dipping and roll squeezing, followed by drying with or without water rinsing. The coating chromate treatment, however, has the disadvantages that it is difficult to control the quantity of a chromate film deposited and that a high speed treatment frequently invites inconsistencies because the chromate film tends to be nonuniform in a transverse direction to the feed direction. It is needed to develop a technique enabling uniform film formation. Another disadvantage is that the resulting chromate film has poor adherence to a coating because the film is thick and retains hexavalent chromium unchanged throughout the film. Also, the chromate film provides a poor adhesive bond to vinyl chloride and similar resins.

The electrolytic chromate treatment is applied by subjecting a steel strip to cathodic electrolysis whereby hexavalent chromium is electrically reduced to trivalent chromium to form a hydrated oxide film at the strip surface. The electrolytic chromate treatment can not only readily accommodate with speeding-up because the quantity of a chromate film can be controlled by a quantity of electricity, but also be applied to various types of steel strips because hexavalent chromium ions in the chromate solution are reduced electrically rather than by redox reaction. The chromate film resulting from the electrolytic chromate treatment consists essentially of trivalent chromium and has higher coating adherence as compared with the reactive and coating chromate treatments, but is less corrosion resistant as compared with the reactive chromate treatment.

One prior art method for carrying out an electrolytic chromate treatment is disclosed in Japanese Patent Publication No. 47-44417 which is incorporated herein by reference. This method is successful in forming a good, but thin chromate film only at a relatively low current density. The chromate layer cannot be further grown even by increasing electricity quantity. Differently stated, the method fails to form a thick chromate film on a zinc alloy plated steel strip. As previously indicated, in general, the electrolytic chromate film is less corrosion resistant as compared with the reactive and coating type chromate films having the same amount of chromium deposited. This is probably because the electrolytic chromate film tends to be porous due to evolution of hydrogen gas during film formation and because the chromate film composed mainly of trivalent chromium contains an insufficient amount of hexavalent chromium to seal such pores or defects, that is, lacks a self-healing ability.

Another method for carrying out an electrolytic chromate treatment is disclosed in Japanese Patent Application Kokai No. 60-110896 which is incorporated herein by reference. A chromate film is formed in a bath containing hexavalent chromium (Cr.sup.6+)+cationic colloidal silica+H.sub.2 SO.sub.4 +optional NaOH for pH adjustment. Due to the inclusion of sulfate residues in the bath, metallic Cr tends to deposit in a chromate film particularly at a high current density and thus, the chromate film often becomes black colored. The cationic colloidal silica and sulfate residues serve as film forming agents while processing inconsistencies often occur. An observation of chromate films under a scanning electron microscope has indicated that chromate films resulting from a bath containing a fluoride additive are more uniform and dense than those from a bath containing sulfuric acid.

SUMMARY OF THE INVENTION

Therefore, an object of the present invention is to provide a novel and improved chromate-treated zinc alloy-plated steel strip having a chromate film exhibiting high corrosion resistance, good adherence to a coating, and a firm adhesive bond to vinyl chloride and similar resins.

Another object of the present invention is to provide a method for making the same wherein an electrolytic chromate treatment can be carried out on any type of zinc alloy plating within a short time to a sufficient thickness of chromate film to meet the intended application.

In the initial of developing a zinc or zinc alloy electroplated steel strip having a chromate film exhibiting satisfactory corrosion resistance, coating adherence, and adhesive bond, we attempted to carry out a coating adherence improving treatment on a reactive chromate film. This attempt, however, requires two treatments. It also requires a choice between thick and thin films. A thick film must be formed to insure corrosion resistance when it is intended to use the final product without coating. A thin film will suffice when the final product is coated on use. A compromise is to form a chromate film of moderate thickness having a minimized content of hexavalent chromium in the outermost surface layer.

Intending to produce a chromate film fulfilling the requirements of corrosion resistance, coating adherence, and adhesive bond by only an electrolytic chromate treatment, we have discovered that the object can be attained by controlling the composition of a chromate film.

More particularly, it is desired that the outermost surface region or layer of a chromate film have an effective composition to provide corrosion resistance and coating adherence.

We have discovered it effective in enhancing corrosion resistance that (1) an appropriate amount of hexavalent chromium is contained in the chromate film predominantly comprising trivalent chromium to impart a self-sealing or self-healing ability, (2) the film thickness is increased to form a reinforced barrier by adding a film forming agent such as silicon dioxide, and (3) the film is rendered uniform by adding an etching agent.

We have also discovered it effective in enhancing coating adherence that (4) the outermost surface layer is a thin region composed predominantly of trivalent chromate. (5) SiO.sub.2 is effective in enhancing coating adherence, but tends to cause delamination in the chromate film as the film becomes thick. It will be advantageous that the chromate film be bonded to a resin laminated board with an adhesive. We have discovered that (6) the adhesive bond can be improved by adding Al.sub.2 O.sub.3 to the chromate bath along with SiO.sub.2. The present invention is predicated on these findings.

According to a first aspect of the present invention, there is provided a chromate-treated zinc-plated steel strip comprising

a steel substrate,

a zinc base plating on at least one surface of the substrate,

a metallic chromium layer on the zinc base plating,

a chromium oxide layer on the metallic chromium layer, consisting essentially of the oxide of trivalent chromium, and

an outermost surface layer on the chromium oxide layer, consisting essentially of silicon dioxide and oxides of a major proportion of trivalent chromium and an effective proportion of hexavalent chromium and hydrates thereof.

According to a second aspect of the present invention, there is provided a method for preparing a chromate-treated zinc-plated steel strip, comprising

effecting cathodic electrolysis on a zinc-plated steel strip in a bath containing 2.6 to 78 grams per liter of hexavalent chromium, 0.5 to 50 grams per liter, calculated as SiO.sub.2, of colloidal silica, and 0.05 to 5.0 grams per liter, calculated as F, of a fluoride, at a current density of 1 to 50 A/dm.sup.2 and to an electricity quantity of 5 to 100 C/dm.sup.2.

In one preferred embodiment of the present invention, the outermost surface layer further contains aluminum oxide. In this case, the electrolytic chromate bath used in preparing a corresponding chromate-treated zinc-plated steel strip further contains 0.05 to 25 grams per liter, calculated as Al.sub.2 O.sub.3, of alumina sol in addition to the above-defined ingredients.

BRIEF DESCRIPTION OF THE DRAWINGS

The above and other objects, features, and advantages of the present invention will be readily understood by reading the following description when taken in conjunction with the accompanying drawings, in which:

FIG. 1 is a diagram showing the proportions of metallic Cr, Cr.sup.3+, and Cr.sup.6+ in the chromate film analyzed by ESCA;

FIG. 2 is a diagram showing the relative proportions of Si and Cr in the chromate film analyzed by GDS;

FIG. 3 is a diagram showing the weight of chromium deposited as a function of electricity quantity in the chromate treatment of Example 1;

FIG. 4 is a diagram showing the percent white rust of chromate treated steel strips produced in Example 2 and Comparative Examples 2 and 3 as a function of salt spray test time;

FIGS. 5 and 6 graphically show the weight of chromium deposited as a function of electricity quantity in the chromate treatment of various zinc-plated steel strips in different baths in Example 3; and

FIG. 7 is a diagram showing the percent white rust of chromate treated steel strips produced in Example 4 and Comparative Example 4 as a function of salt spray test time.

DETAILED DESCRIPTION OF THE INVENTION

In the present disclosure, the term zinc plated steel strips is used to encompass steel strips plated with zinc and zinc based alloys. Typical examples of the zinc plated steel strips include zinc electroplated (or electrogalvanized), zinc hot dipped (or galvanized), galvannealed, Zn-Ni alloy plated, Zn-Fe alloy plated, and Zn-Al alloy plated steel strips. These plating surfaces are different in metal or alloy phase and particularly in reactivity during a subsequent treatment, for example, a heat treatment to form an oxide coating.

According to the present invention, electrolysis is effected on various zinc-plated steel strips in a chromate bath with the strips made cathode, by supplying constant current. Hexavalent chromium ions typically present in the form of Cr.sub.2 O.sub.7.sup.2- and CrO.sub.4.sup.2+ in the bath are electrochemically reduced to trivalent chromium ions to form a chromate film predominantly comprising Cr.sup.3 +. Thus the formation of chromate film is little affected by the underlying layer, that is, zinc plating. The amount of chromate film formed is proportional to a quantity of electricity supplied so that the thickness of chromate film may be controlled over a wide range from thin to thick films depending on the intended application of the chromated strip.

The chromate-treated, zinc-plated steel strip according to the present invention has a chromate film consisting of

(1) a layer most adjacent to the zinc base plating which consists of metallic chromium,

(2) an intermediate layer which consists essentially of the oxide of trivalent chromium, and

(3) an outermost surface layer which consists essentially of silicon dioxide (SiO.sub.2), optional aluminum oxide (Al.sub.2 O.sub.3), and oxides of a major proportion of trivalent chromium and an effective proportion of hexavalent chromium and hydrates thereof.

The metallic chromium layer disposed in direct contact with the zinc base plating is not critical in the practice of the present invention, but is naturally deposited in a small amount from the chromate bath operated under the electrolytic conditions according to the present invention. The metallic chromium layer may be discontinuous. Excess deposition of metallic chromium is undesirable because the amount of subsequently formed hydrated oxides is reduced. The weight of metallic chromium deposited is preferably limited to the maximum of 20 mg/m.sup.2

In the outermost surface layer, trivalent chromium and an effective proportion of hexavalent chromium coexist. The effective proportion of hexavalent chromium means a sufficient amount of hexavalent chromium to exert a full self-healing effect. The proportion of hexavalent chromium preferably ranges from 1/100 to 1/5 of the total weight of chromium in the chromate film. The lower limit of hexavalent chromium is set to 1/100 or 1% below which hexavalent chromium is too less to provide a self-healing effect, failing to improve corrosion resistance. The presence of hexavalent chromium in excess of 1/5 or 20% of the total weight of chromium will result in a colored film and detract from coating adherence.

FIG. 1 shows the proportions of metallic, trivalent and hexavalent chromiums based on the total weight of chromium in the chromate film according to the present invention. The proportions of Cr(0), Cr(III) and Cr(VI) are determined in a thickness direction of the film by electron spectroscopy for chemical analysis (ESCA) and expressed as their ratio to the total chromium.

The chromate film is preferably deposited to a weight of 20 to 200 mg/m.sup.2 calculated as Cr. A chromate film having less than 20 mg/m.sup.2 of Cr exhibits poor corrosion resistance without coating as demonstrated by the white rust generating time of about 24 hours in the standard salt spray test (SST). Conversely, a chromate film having more than 200 mg/m.sup.2 of Cr exhibits yellow color in appearance and poor coating adherence.

According to the present invention, the chromate film contains silicon dioxide (SiO.sub.2) The present invention is characterized in that silicon dioxide is preferentially present in the outermost surface layer. FIG. 2 illustrates the results of measurement of the chromate film by glow discharge spectrometry (GDS). The proportion of SiO.sub.2 preferably ranges from 1/40 to 1/2 of the total weight of chromium in the chromate film. Less than 1/40 of the total chromium weight of SiO.sub.2 is insufficient to exert its essential effect of film formation. The content of SiO.sub.2 is limited by such processing factors as transfer to rolls during manufacturing process. The presence of more than 1/2 of the total chromium weight of SiO.sub.2 results in a rather thick film and adversely affects coating adherence.

In one preferred embodiment of the present invention, the chromate film further contains aluminum oxide (Al.sub.2 O.sub.3) in its outermost surface layer. Aluminum oxide is introduced to enhance coating adherence and particularly, adhesive bond characteristics. The amount of Al.sub.2 O.sub.3 preferably ranges from 10 to 1/2 of the weight of SiO.sub.2. Inclusion of Al.sub.2 O.sub.3 in amounts of less than 1/10 of the SiO.sub.2 content could not attain its own purpose of enhancing coating adherence whereas more than 1/2 of the SiO.sub.2 content of Al.sub.2 O.sub.3 renders the adsorption of SiO.sub.2 to the plating surface less uniform.

It has been found that when aluminum oxide is contained in the chromate film along with silicon dioxide, the aluminum oxide contributes to significant improvements in corrosion resistance, coating adherence, and adhesive bond. Although the reason is not fully understood and the present invention is not bound to any theory, we suppose the following mechanisms. In general, alumina sol is positively charged in an acidic bath. Thus alumina is uniformly deposited on the cathode to form a rigid film during cathodic electrolysis of a steel strip. Furthermore, active hydroxyl groups on alumina colloid surface will form a firm hydrogen bond with functional groups of a subsequently applied coating or adhesive.

In summary, the chromate film of the chromate-treated zinc-plated steel strip according to the present invention has the following composition:

______________________________________
Cr 20-200 mg/m.sup.2,
precisely,
Cr.sup.0 0-20 mg/m.sup.2,
Cr.sup.6+ 0.2-40 mg/m.sup.2,
Cr.sup.3+ balance,
SiO.sub.2 0.5-100 mg/m.sup.2, and
optionally,
Al.sub.2 O.sub.3
0.05-50 mg/m.sup.2.
______________________________________

The chromate-treated, zinc-plated steel strips organized as above according to the present invention may be manufactured as follows.

In the chromate bath, first of all, hexavalent chromium is necessary as a main component for forming a chromate film. A source of Cr.sup.6+ may be selected from CrO.sub.3, chromate salts, and bichromate salts although the most common source is CrO.sub.3. In a bath containing CrO.sub.3 alone, electrolysis will grow little hydrated oxide, resulting in an extremely thin chromate film. This is because in a very initial stage of electrolysis, a hydrated oxide film covers the surface to retard electrolysis. In order to break thin portions of the hydrated oxide film to enable further growth of the film, an amount of etching agent is needed. A common practice is to use sulfate ion (see Japanese Patent Publication No. 47-44417) and fluoride ion. Sulfate ion tends to help metallic chromium to deposit to blacken the film when the CrO.sub.3 concentration or the current density is high.

Therefore, the present method favors the use of a fluoride as the etching agent. Typical examples of the fluorides include sodium (Na) and potassium (K) salts of AlF.sub.6.sup.3-, SiF.sub.6.sup.2-, BF.sub.4.sup.- and F.sup.-. They may be added alone or in admixture as long as a necessary level of fluoride ion is reached.

The concentration of Cr.sup.6+ is limited to the range of 2.6 to 78 grams per liter of the solution. Concentrations of less than 2.6 g/l furnish insufficient hexavalent chromium to the plating interface to form a sound film. Concentrations of more than 78 g/l not only tend to help metallic chromium to deposit so that the hydrated oxide film becomes thin, but also invite zinc dissolution reaction at the same time so that the film becomes yellowish brown and unacceptable in appearance.

The fluoride is added to provide a concentration in the range of 0.05 to 5.0 grams of fluoride (F) per liter of the solution. Less than 0.05 g/l of F is less aggressive and fails to grow the film. More than 5.0 g/l of F has a too high etching ability and thus causes to dissolve the hydrated oxide film itself or etch the surface of the plating to give rise to zinc dissolution, resulting in complicated reaction.

The chromate bath containing only Cr.sup.6+ and a fluoride yields a chromate film which is still thin and less resistant against corrosion. The film cannot be further grown simply by increasing the electricity quantity.

According to the present invention, colloidal silica is added as the third component to the chromate bath. Colloidal silica or SiO.sub.2 sol is added as a film forming agent at a concentration of 0.5 to 50 grams of SiO.sub.2 per liter of the solution. Because of its adsorption power and steric structure, colloidal silica is effective in producing a thick chromate film. Examples of the colloidal silica include anionic colloidal silica commercially available as Snowtex O add C (trademarks) and cationic colloidal silica commercially available as Snowtex AK and BK (trademarks), all manufactured by Nissan Chemical K.K. Particularly, cationic colloidal silica is preferred because the transfer of colloidal silica to the plating surface which is made cathode is promoted. In addition, cationic colloidal silica having adsorbed on its surface such anions as Cr.sub.2 O.sub.7.sup.2- and CrO.sub.4.sup.2- in the chromate solution is adsorbed to the cathode so that the resulting chromate film is a fully corrosionresistant film containing a self-healing amount of the hexavalent chromium component.

The amount of colloidal silica added is limited to the range of 0.5 to 50 grams of SiO.sub.2 per liter of the solution. Less than 0.5 g/l is little effective. Inclusion of colloidal silica in excess of 50 g/l of SiO.sub.2 results in a chromate bath having a low electric conductivity and a too thick chromate film which is unacceptably colored or nonuniform in thickness.

According to the preferred aspect of the present invention, alumina sol is added to the electrolytic chromate solution along with colloidal silica for the purpose of improving the bond of the chromated steel strip to a vinyl chloride or similar resin sheet with the aid of an adhesive. Alumina or Al.sub.2 O.sub.3 sol is added in a proportion of 1/10 to 1/2 of the weight of SiO.sub.2, that is, in a concentration of 0.05 to 25 grams of Al.sub.2 O.sub.3 per liter of the solution. Less than 1/10 of the SiO.sub.2 content of Al.sub.2 O.sub.3 cannot attain the purpose of enhancing the adhesive bondability whereas more than 1/2 of the SiO.sub.2 content of Al.sub.2 O.sub.3 will disturb the adsorption of SiO.sub.2 to the plating surface.

In the practice of the present invention, silica and alumina may be added to the electrolytic chromate solution in the following two ways.

(1) SiO.sub.2 sol and Al.sub.2 O.sub.3 sol are separately added in appropriate amounts.

(2) SiO.sub.2 having Al.sub.2 O.sub.3 sol adsorbed thereon is added in an appropriate amount.

In either of (1) and (2), the electrolytic chromate treatment can be carried out in an acceptable manner. The addition of SiO.sub.2 having Al.sub.2 O.sub.3 sol adsorbed thereon (2) is more advantageous in controlling the colloid sol content of the chromate film.

The above-formulated chromate bath is preferably operated at a temperature of 30.degree. to 60.degree. C. using an insoluble anode such as a Pb-Sn (Sn 5%) electrode as the anode. The bath is operated by supplying electricity at a current density of 1 to 50 A/dm.sup.2 (ampere per square decimeter) although the exact density depends on the processing time required. Within this current density range, the amount of chromate film deposited is increased with the quantity of electricity supplied. By controlling current density and electricity quantity in accordance with the line speed associated with the chromate treatment, any desired amount of chromate film can be deposited.

The electricity quantity preferably ranges from 5 to 100 C/dm.sup.2 (coulomb per square decimeter). An electricity quantity of less than 5 C/dm.sup.2 is insufficient to form a chromate film beyond 20 mg/m.sup.2 whereas an electricity quantity of more than 100 C/dm.sup.2 will result in a chromate film beyond 200 mg/m.sup.2.

After the electrolytic chromate treatment, the steel strip is roll squeezed for film thickness control and then dried, or washed with flowing water, roll squeezed for film thickness control and then dried. The former procedure is employed when corrosion resistance is important. Generally, the latter procedure involving washing is useful to present a film having a uniform appearance free of processing variations.

The present invention is distinguishable over the prior art method disclosed in Japanese Patent Application Kokai No. 60-110896 using a bath containing hexavalent chromium, cationic colloidal silica, sulfuric acid, and optional sodium hydroxide. As demonstrated in Example 4 and FIG. 7, samples treated in a bath containing CrO.sub.3 + colloidal silica+fluoride according to the present invention exhibit evidently superior corrosion resistance to those treated in a bath containing CrO.sub.3 +cationic colloidal silica+H.sub.2 SO.sub.4 according to the prior art, provided that the amount of chromate film deposited is equal. It is supposed that while colloidal silica acts as a film forming agent, the fluoride removes an oxide coating on the plating surface to allow hydrated chromium oxides to uniformly adhere thereto and at the same time, etches away thin weak portions or readily dissolvable portions of the chromate film itself to allow a new film to grow in these sites. In the chromate bath according to the present invention, the double actions of film formation and etching occur in a well-balanced harmony so as to produce a uniform corrosion resistant film.

Although it will occur to add other anions to the bath, they have some problems. More particularly, chloride ion will color the chromate film in yellowish brown. Phosphate ion will react with the zinc plating so that a substantial amount of phosphate residue is introduced in the chromate film. Thus, corrosion resistance is less improved irrespective of the amount of chromate film deposited.

As previously indicated, among the anionic and cationic colloidal silicas, the latter is more readily adsorbed to the zinc plating surface because the zinc plated strip is made cathode during electrolytic chromate treatment. Cationic colloidal silica is thus effective even in a relatively low concentration, say 0.5 to 10 g/l of SiO.sub.2 Conversely, anionic colloidal silica is used in a relatively high concentration, say 10 to 30 g/l of SiO.sub.2 to obtain a satisfactory result.

As described above, the chromate film obtained from the prior art bath of hexavalent chromium, cationic colloidal silica, and sulfuric acid is rather irregular and exhibits poor corrosion resistance unless its thickness is increased to a level corresponding to an electricity quantity of more than 30 C/dm.sup.2. By virtue of the fluoride, the chromate bath of the present invention can produce a dense chromate film having an aesthetic uniform appearance and high corrosion resistance even with a reduced thickness corresponding to an electricity quantity of less than 30 C/dm.sup.2 and irrespective of whether the bath uses either cationic or anionic colloidal silica.

EXAMPLES

In order that those skilled in the art will readily understand the practice of the present invention, examples are given below by way of illustration and not by way of limitation. In the examples, g/l is gram per liter of solution, g/m.sup.2 or mg/m.sup.2 is gram or milligram per square meter of surface, A/dm.sup.2 is ampere per square decimeter, and C/dm.sup.2 is coulomb per square decimeter.

EXAMPLE 1

The zinc plated steel strip used in this example was a zinc electroplated steel strip having a zinc coating weight of 20 g/m.sup.2 It was subjected to a chromate treatment in a bath containing 50 g/l of CrO.sub.3, 0.27 g/l calculated as F of Na.sub.3 AlF.sub.6, and 3 g/l calculated as SiO.sub.2 of Snowtex AK (trademark, manufactured by Nissan Chemical K.K.) in water while the quantity of electricity supplied across the strip was varied. The bath temperature was 50.degree. C. and the current density was set to 5 A/dm.sup.2 and 10 A/dm.sup.2. In Comparative Example 1, a chromate treatment was effected in a bath containing 50 g/l of CrO.sub.3 and 0.27 g/l calculated as F of Na.sub.3 AlF.sub.6 in water under the same conditions as described above. The results are shown in FIG. 3.

In the conventional bath free of colloidal silica (Comparative Example 1), the amount of chromium deposited is only slightly increased by increasing the electricity quantity. In the bath according to the present invention (Example 1), the amount of chromium deposited is increased in approximate direct proportion to the electricity quantity. If it is desired to form a thick chromate film having a chromium weight of approximately 100 mg/m.sup.2, the chromate treatment according to the present invention can produce the film by supplying electricity at a current density of 5 A/dm.sup.2 to a quantity of 15 C/dm.sup.2, that is, within 3 seconds. To match with a high speed plating line, approximately the same weight of chromium can be deposited by supplying electricity at 10 A/dm.sup.2 to the same quantity of 15 C/dm.sup.2, that is, within 1.5 seconds.

EXAMPLE 2

A sample was prepared by effecting a chromate treatment on a zinc plated steel strip in a bath containing 30 g/l of CrO.sub.3, 1.0 g/l calculated as F of K.sub.2 SiF.sub.6, and 10 g/l calculated as SiO.sub.2 of Snowtex O (trademark, manufactured by Nissan Chemical K.K.) in water by supplying electricity at a current density of 10 A/dm.sup.2 to a quantity of 10 C/dm.sup.2. The sample was subjected to a salt spray test (SST) according to JIS Z 2371 to determine the variation of percent white rust area with time. In Comparative Example 2, a chromate treatment was effected in a bath containing 30 g/l of CrO.sub.3 and 10 g/l calculated as SiO.sub.2 of Snowtex O in water under the same conditions as described above. In Comparative Example 3, a chromate treatment was effected in a bath containing 30 g/l of CrO.sub.3 and 1.0 g/l calculated as F of K.sub.2 SiF.sub.6 in water under the same conditions as described above. The comparative samples were also examined for corrosion resistance. The results are shown in FIG. 4 in which the percent white rust area is plotted as a function of the time of SST.

The present sample treated in the three-component bath had a satisfactory chromate film which experienced no white rust even after 90 hours of SST. The treating time of the present sample was 1 second, indicating the possible matching with a high speed line.

EXAMPLE 3

Different types of zinc plated steel strips including galvanized, electrogalvanized, and Zn-Ni plated ones were chromate treated according to the present method. The results are shown in FIGS. 5 and 6 in which the weight of chromium deposited is plotted as a function of electricity quantity. In the graphs, EG corresponds to an electrogalvanized (or zinc electroplated) steel strip having a coating weight of 20 g/m.sup.2, Zn-Ni corresponds to a Zn-Ni alloy plated steel strip having a coating weight of 20 g/m.sup.2 and a nickel content of 13% by weight, and GI corresponds to a galvanized (or zinc hot dipped) steel strip having a coating weight of 60 g/m.sup.2. It is evident that an equal amount of chromate film is formed on different zinc plated steel strips regardless of their zinc plating type.

In FIG. 5, the strips were treated in a bath containing 50 g/l of CrO.sub.3, 0.30 g/l calculated as F of Na.sub.2 SiF.sub.6, and 10 g/l calculated as SiO.sub.2 of Snowtex O in water by supplying electricity at a current density of 10 A/dm.sup.2. In FIG. 6, the strips were treated in a bath containing 50 g/l of CrO.sub.3, 0.69 g/l calculated as F of NaBF.sub.4, and 2 g/l calculated as SiO.sub.2 of Snowtex O in water by supplying electricity at a current density of 10 A/dm.sup.2.

EXAMPLE 4

A zinc plated steel strip was subjected to electrolysis in a bath containing 50 g/l of CrO.sub.3, 1.29 g/l calculated as F of Na.sub.2 SiF.sub.6, and 6 g/l calculated SiO.sub.2 of Snowtex AK in water by supplying electricity at a current density of 10 a/dm.sup.2 to a quantity of 10 C/dm.sup.2. The resulting sample was subjected to a salt spray test (SST) according to JIS Z 2371 to determine the variation of percent white rust area with time.

In Comparative Example 4, a similar electrolytic chromate treatment was effected in a bath containing 50 g/l of CrO.sub.3, 0.2 g/l of H.sub.2 SO.sub.4, and 6 g/l calculated as SiO.sub.2 of Snowtex AK in water by supplying electricity at a current density of 10 A/dm.sup.2 to a quantity of 10 C/dm.sup.2. The comparative sample was also examined for corrosion resistance by SST. Both the samples had a chromium coating weight of 100 mg/m.sup.2.

The results are shown in FIG. 7 in which the percent white rust area is plotted as a function of the time of SST. It is evident that the chromate film (Example 4) obtained by the present method has improved corrosion resistance over that (Comparative Example 4) obtained from the bath containing Cr.sup.6+ plus cationic colloidal silica plus H.sub.2 SO.sub.4 by the prior art method described in Japanese Application Kokai No. 60-110896.

EXAMPLE 5

An electrogalvanized steel strip having a zinc coating weight of 20 g/m.sup.2 was subjected to cathodic electrolysis in a bath containing 5 to 150 g/l of Cr.sub.3, 0.05 to 5 g/l calculated as F of Na.sub.2 SiF.sub.6, 0.5 to 50 g/l calculated as SiO.sub.2 of colloidal silica, and 0 to 25 g/l calculated as Al.sub.2 O.sub.3 of colloidal alumina in water by supplying electricity at a current density of 1 to 50 A/dm.sup.2.

For comparison purposes, a reactive chromate treatment was carried out. In this comparative run designated Comparative Example R, the same electrogalvanized steel strip was treated in a commonly used reactive chromate bath containing 20 g/l of CrO.sub.3 and 1 g/l of F, yielding a sample having a chromium coating weight of 40 mg/m.sup.2.

Additionally, a coating chromate treatment was carried out. In this comparative run designated Comparative Example C, the same electrogalvanized steel strip was treated by applying an aqueous solution containing 30 g/l of CrO.sub.3 and 80 g/l of colloidal silica and squeezing the coated strip between rolls to control the coating weight to 80 mg/m.sup.2 of Cr.

Then an acrylic resin coating composition was applied to the thus obtained samples of this Example and Comparative Examples and baked at 160.degree. C. for 20 minutes. The coated samples were subjected to several tests as described below. The results are shown in Table 1.

TEST PROCEDURES AND EVALUATION

(1) Corrosion resistance

A salt spray test (SST) was carried out according JIS Z 2371, one cycle including salt water spraying for 8 hours and allowing to stand for 16 hours (total 24 hours). The sample was examined every cycle (24 hours) to determine the time taken until white rust appeared.

(2) Coating adherence

(2-1) Erichsen scribed adhesion test

The coated sample was scribed to define 100 square sections of 1 mm by 1 mm in the coating, cup drawn to a depth of 7 mm by means of an Erichsen drawing machine, and then examined for separation of coating sections by applying and removing an adhesive tape.

(2-2) duPont adhesion test

An impact was applied to the coated sample by dropping a 1/2 inch diameter weight of 500 grams from a height of 500 mm according to the duPont impact test. The sample was then examined for separation of coating pieces by applying and removing an adhesive tape.

(2-3) Immersion scribed adhesion test

The coated sample was immersed in boiling water for 3 24 hours, scribed to hours, allowed to stand in air for define 100 square sections of 1 mm by 1 mm in the coating, and then examined for separation of coating sections by applying and removing an adhesive tape.

Evaluation was made according to the following criterion.

______________________________________
Symbol Observation
______________________________________
O no separation
.DELTA. faintly separated
X apparently separated
______________________________________

(3) Adhesive bond

A polyvinyl chloride sheet was bonded to each of the samples of Example 5 and Comparative Examples R, C, using a thermosetting acrylic adhesive, SC-457 manufactured by Sony Chemical K.K. The sample was scribed to define 25 square sections of 2 mm by 2 mm down into the coating, cup drawn to a depth of 8 mm by means of an Erichsen drawing machine, and then visually examined for separation of coating.

Evaluation was made according to the following criterion.

______________________________________
Symbol Observation
______________________________________
O no separation
.DELTA. faintly separated
X apparently separated
______________________________________
TABLE 1
________________________________________…
Chromate film composition Corrosion
Metallic Resistance
Coating Adherence
Sample
Cr Cr.sup.3+
Cr.sup.6+
SiO.sub.2
Al.sub.2 O.sub.3
White rust
Erichsen
duPont
Immersion
Adhesive
No. mg/m.sup.2
mg/m.sup.2
mg/m.sup.2
mg/m.sup.2
mg/m.sup.2
time (hr.)
adhesion
adhesion
adhesion
bond Remarks
________________________________________…
1 10 60 10 15 -- 144 O O O .DELTA.
2 20 40 8 10 -- 96 O O O O
3* 30 15 5 5 -- 48 O O O O
4* 5 10 1 1 -- 24 O O O O
5 5 15 2 2 -- 48 O O O O
6 5 45 5 10 -- 96 O O O O
7 5 90 20 15 -- 192 O O O .DELTA.
8 5 155 40 20 -- 240 O .DELTA.
O .DELTA.
9* 5 200 50 30 -- 240 .DELTA.
.DELTA.
.DELTA.
X
10* 5 120 0.5 5 -- 72 O O O .DELTA.
11 5 120 1.5 5 -- 120 O O O .DELTA.
12 5 80 10 10 -- 144 O O O .DELTA.
13 5 40 3 15 -- 96 O O O O
14* 5 60 20 10 -- 192 .DELTA.
.DELTA.
.DELTA.
.DELTA.
irregularities
observed
15* 5 70 5 1 -- 72 O .DELTA.
O .DELTA.
16 5 70 5 2 -- 96 O O O .DELTA.
17 5 70 5 15 -- 144 O O O .DELTA.
18 5 150 15 85 -- 240 O .DELTA.
O .DELTA.
19* 5 150 15 100 -- 240 .DELTA.
X X X irregularities
observed
20 5 70 5 10 1 120 O O O O
21 5 70 5 10 2 144 O O O O
22 5 70 5 10 5 144 O O O O
23* 5 70 5 10 10 168 O O O O irregularities
observed
24 5 80 5 20 2 144 O O O O
25 5 80 5 20 4 168 O O O O
26 5 80 5 20 10 192 O O O O
27* 5 80 5 20 20 216 O O O O irregularities
observed
28 5 150 20 50 5 210 O O O O
29 5 150 20 50 10 210 O O O O
30 5 150 20 50 20 240 O O O O
31* 5 150 20 50 40 240 O O O O irregularities
observed
CE-R 40 -- -- 24 O O O .DELTA.
reactive type
CE-C 80 240 -- 216 X X X X coating
________________________________________…
type
*samples falling outside the scope of the invention

According to the present invention, any desired amount of chromate film can be deposited on a variety of zinc-plated steel strips within a short time by subjecting the strips to a cathodic electrolytic treatment in a bath containing hexavalent chromium, an etching agent in the form of fluoride, and a film forming agent in the form of silicon dioxide. The resulting chromate-treated zinc-plated steel strip has a chromate film possessing excellent corrosion resistance and coating adherence. Such products cannot be produced by the conventional reactive, immersing or coating type chromate treatment methods. The present method can carry out a necessary electrolytic chromate treatment at a high speed and is convenient in controlling the amount of chromate film. Inclusion of aluminum oxide in the chromate film along with silicon dioxide further improves the adhesive bond of the chromate treated steel strip to a vinyl chloride or similar resin.

Yahoo! Answers

Breaking News Visit Yahoo! News for the latest.

Ask

Answer

Discover

Resolved Question

From the redox reaction Zn+ Ni^2+--> Zn^2+ Ni we can say that?

Best Answer - Chosen by Voters

Other Answers (1 - 30 of 83)

Source(s):

Source(s):

Source(s):

Source(s):

Source(s):

Source(s):

Source(s):

Discover Questions in Chemistry

Answers International

Yahoo! Answers

Breaking News Visit Yahoo! News for the latest.

Ask

Answer

Discover

Resolved Question

From the redox reaction Zn+ Ni^2+--> Zn^2+ Ni we can say that?

Best Answer - Chosen by Voters

Other Answers (1 - 30 of 83)

Source(s):

Source(s):

Source(s):

Source(s):

Source(s):

Source(s):

Source(s):

Go to Page

Discover Questions in Chemistry

Trending Now

Related Questions

Categories

Who found this interesting?

Answers International