Best Answer:
This reaction is actually an unbalanced equation of the overall reaction: in order to balance it, ignore the H+ and the H2O for now.

CrO4 2- + NO2 - ---> Cr3+ + NO3-

This equation can be separated into two individual equations: an oxidation and reduction

CrO4 2- --->Cr3+

Reduction of Cr from 6+ to 3+

NO2- --->NO3-

Oxidation of N from 3+ to 5+

Separating the reactions leads to an easier way to balance, instead of trying to do everything all at once. Let's start with the easier one to balance.

NO2 - ---> NO3-

The N atom in both equations is balanced (1 on each side), so we can leave it alone. The O, however, is unbalanced (2 on left, 3 on right). In these types of equations, when you want to balance O, add H2O to the side with the lesser number of O. In this case, we need 3 O on each side. There is already 3 on the right, but 2 on the left. Thus, add 1 H2O on the left side.

NO2 - + H2O---> NO3-

We have now balanced the O, but by doing so, have put 2 H on the left side of the equation: to balance the right side, we must add 2 H+.

NO2 - + H2O --->NO3- + 2 H+

All the atoms are balanced now, but the electrons aren't. If you multiply the coefficient of each compound by the charge, you can see, on the reactant side:

NO2- ---->1 mol of compound (1-, the charge)=1-

On the product side:

1(1-) +2(1+)---> -1+2=1+

You need to get the +1 on the product side to become -1 by adding 2 mol electrons.

NO2 - + H2O --->NO3- + 2 H+ + 2 e-

This equation is now balanced: on to the other one.

CrO4 2- ---> Cr3+

This equation is quite similar to the first one: the number of Cr is balanced, but O is not. Add 4 H2O on the right side to get 4 oxygens on each side.

CrO4 2- ---> Cr3+ + 4 H2O

We have just added 8 H to the right, so add 8 H+ to the left.

8 H+ + CrO4 2- ---> Cr3+ + 4 H2O

The atoms are balanced, but the charges are not.

Reactants: 8(1+) + 1(2-)=6+

Products: 1(3+)=3+

To get the 6 to become a 3, add 3 mol electrons to the left side (reduce positivity).

8 H+ + CrO4 2- +3 e- ---> Cr3+ + 4 H2O

We must combine the two individual equations in order to get an overall reaction.

8 H+ + CrO4 2- +3e- ---> Cr3+ + 4 H2O

NO2 - + H2O --->NO3- + 2 H+ + 2 e-

However, in both equations, the number of moles of electrons transferred is different. Both 3 and 2 are factors of 6, so you must multiply each equation by a factor of 6 to get it. Multiply the all coefficients in the first by 2, and in the second, multiply all coefficients by 3: this allows for equal moles of electrons.

16 H+ + 2 CrO4 2- +6e- ---> 2 Cr3+ + 8 H2O

3 NO2 - + 3 H2O --->3 NO3- + 6 H+ + 6 e-

You can cancel out electrons as they appear on different sides of different equations.

16 H+ + 2 CrO4 2- ---> 2 Cr3+ + 8 H2O

3 NO2 - + 3 H2O --->3 NO3- + 6 H+

You can cancel out the H+ as well (for the same reason we canceled out the electrons). However, this is a little different, since the coefficients of the H+ are different: take the lesser coefficient and subtract it from the coefficients of both H+. The H= on the right will cancel in the second equation, but the one on the left of the first will be reduced to 10.

10 H+ + 2 CrO4 2- ---> 2 Cr3+ + 8 H2O

3 NO2 - + 3 H2O --->3 NO3-

Do the same to the water: subtract 3 from the water on the left and right.

10 H+ + 2 CrO4 2- ---> 2 Cr3+ + 5 H2O

3 NO2 - --->3 NO3-

Almost there: combine both equations.

10 H+ + 2 CrO4 2- + 3 NO2- ---> 2 Cr3+ + 5 H2O + 3 NO3-

To check, multiply the coefficients by the compounds charges on both sides of the equation.

Reactants 10(1+) + 2(2-) + 3(1-)=10 -4-3= 3+

Products 2(3+) + 3(1-)=3+

Thus, the equation is:

10 H+ + 2 CrO4 2- + 3 NO2- ---> 2 Cr3+ + 5 H2O + 3 NO3-

Study hard! These types of problems may be intimidating, but are actually quite easy when broken down.

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