Anonymous
Anonymous asked in Science & MathematicsChemistry · 8 years ago

# Reaction: Potassium iodide(aq) and potassium permanganate(aq) reaction forms MnO2(s) and potassium iodate(aq)?

In a basic solution. I need the molecular equation. Here's my work so far (assume aqueous unless otherwise indicated)

Balanced equation:

KI + KMnO4 ---> KMnO2(s) + KIO3

Use half-reaction method:

Oxidation: KI ---> KIO3

= KI + 3H2O ---> KIO3 + 6H+ + 6e-

Reduction: KMnO4(aq) ---> KMnO2(s)

= KMnO4(aq) + 4H+ + 4e- ---> KMnO2(s) + 2H2O

I multiplied the oxidation equation by 2 and the reduction equation by 3 so I could cancel out the electrons. Now I have:

2KI + 3MnO4 --> + KMnO2(s) + 2KIO3

What's wrong?

Update:

Whoops, I don't think my balanced equation is correct.

Update 2:

So, could someone figure out the equation?

Update 3:

KMnO2 is insoluble, so even though everything else ionizes, it won't. That would mean

I + MnO4 --- KMnO2(s) + IO3

That wouldn't make sense because there's a random K.

Relevance

You've mistaken regards the products of this redox reaction.

That makes any chance of a correct balanced equation zero.

One of the products is manganese dioxide. It is not an ion. It is a substance which precipitates.

I find the best approach to these types of problem is to work with the ionic half equations

So:

The iodide is oxidized to the iodate ion

I- + 6OH- --> IO3- + 3H2O + 6e- ..................1)

The permanganate ion is reduced to manganese dioxide

MnO4- + 2H2O + 3e- --> MnO2 + 4OH- ...... 2)

balancing electrons : (1) and 2 x (2)

I- + 6 OH- --> IO3- + 3 H2O + 6e-

2 MnO4- + 4 H2O + 6e- --> 2 MnO2 + 8 OH-

and combining

I- + 2 MnO4- + H20 --> IO3- + 2 MnO2 + 2 OH-

the above is the net ionic eq for the redox reaction. The K+ ions spectates!

To create the non -ionic form we simply add the K+ ions that partner the anions?/cations? {i forget which is which!}

KI(aq) + 2KMnO4(aq) + H2O(l) ----> KIO3(aq) + 2 KOH(aq) + MnO2(s)

• Ferdie5 years agoReport

Your 'balanced' molecular equation is wrong: check the number of oxygen atoms you have on both sides of your equation!

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• KI + KMnO4 --> KMnO2 + KIO3

I in KI = 1- and becomes 5+ in KIO3, oxidized, loses 6 e-

Mn in MnO4 = 7+ and becomes 3+ in MnO2, reduced, gains 4 e-

even though you balanced in an acidic solution, the answer is the same

balance in basic solution

I- + 6OH- --> IO3- + 3H2O + 6e-

MnO4- + 2H2O + 4e- --> MnO2- + 4OH-

2I- + 12OH- --> 2IO3- + 6H2O + 12e-

3MnO4- + 6H2O + 12e- --> 3MnO2 + 12OH-

2I- + 3MnO4- --> 2IO3- + 3MnO2

2KI + 3KMnO4 --> 2KIO3- + 3KMnO2

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