Chemistry - balancing redox equations?

Question

Hi. I'm stuck at this balancing equation problem

K2CrO4 + 3Na2SO3 + HCl ® KCl + 3Na2SO4 + CrCl3 + H2O

Can any chemist out there help? Thank you

3Na2SO3 and 3Na2SO4 are already balanced so that electrons gained and lost are equal. Thank you so much.

Answer 1

This is a reaction that occurs in acid conditions.

So if additional "O" is needed, it will come from H2O.  If there is too much, H+ will be needed to produce H2O

The oxidant is Cr+6, which is reduced to Cr+3

The reduced reactant is S+4, which is oxidized to S+6.  

So........

CrO4 (-2) + 3HCl +  5H+   +3e ->  CrCl3 +  4H2O  and

       SO3 (-2) + H2O          ->  SO4(-2) + 2H+ + 2e

You have to account for HCl as both a supply of Cl ions and as an acid source.  If we double the reduction eqtn and triple the oxidation eqtn, we balance  out electrons, so.......

2K2CrO4 +3 Na2SO3 + 10 HCl ->4 KCl + 2CrCl3 + 5H2O +

      3Na2SO4  

Answer 2

K2CrO4 + Na2SO3 + HCl ----->  KCl + Na2SO4 + CrCl3 + H2O

Cr goes from +6 in K2CrO4 to +3 in CrCl3; Cr gains 3e- (reduced)

S goes from +4 in Na2SO3 to +6 in Na2SO4; S loses 2e- (oxidized)

To balance the electrons, take the Cr compounds times 2 and the S compounds times 3 to get 6 electrons gained and lost. This gives:

2K2CrO4 + 3Na2SO3 + HCl ----->  KCl + 3Na2SO4 + 2CrCl3 + H2O

Next, balance the oxygens by inspection (17 oxygens on each side).

2K2CrO4 + 3Na2SO3 + HCl ----->  KCl + 3Na2SO4 + 2CrCl3 + 5H2O

Balance the hydrogens (10 hydrogens on each side)

2K2CrO4 + 3Na2SO3 + 10HCl ----->  KCl + 3Na2SO4 + 2CrCl3 + 5H2O

Balance the chlorines (10 Cl on each side)

2K2CrO4 + 3Na2SO3 + 10HCl ----->  4KCl + 3Na2SO4 + 2CrCl3 + 5H2O

It is now balanced. There are 4 potassiums on each side.