# How do I go about balancing these equations?

I have to use oxidization number approach.

$\ce{Co^2+ (aq) + H2SO3(aq) -> Co^3+(aq) + S2O3^2- (aq)}$

This is my approach so far:

The change in Co is 1 electron, the change is S is 2 electrons, so CO must be multiplied by two

$\ce{2 Co^2 + H2SO3(aq) -> 2 Co^3+ + S2O3^2-}$

I balance the O's, then balance the H's and I get

$\ce{2 Co^2+ + 2H2SO3(aq) + 2H^+ -> 2 Co^3+ + S2O3^2- + 3H2O(l)}$

I did it the half reaction way as well, I got the same answer but the coefficients on the Co's were 4. I suspect my problem is my assumptions about the oxidization number change at the beginning.

• Also it is in acidic conditions – Blobd Mar 17 '15 at 0:28
• You've got two sulphurs... – Gerhard Mar 17 '15 at 0:31
• You need to balance charges as well. – Curt F. Mar 17 '15 at 5:45

$$\ce{4 Co^2+(aq) + 2H2SO3(aq) + 2H^+ (aq)-> 4 Co^3+(aq) + S2O3^2- (aq) + 3H2O(l)}$$
The problem is in your assumptions about the oxidization number change at the beginning, and more precisely about sulfur. The oxidation number of sulfur in $\ce{H2SO3}$ is four. But in thiosulfate ion, you have two inequivalent sulfur atoms: Please see: Oxidation states of the sulfur atoms in the thiosulfate ion The oxidation number is $\ce{+IV}$ for the central one, while it's zero for the terminal one.
In sum, the variation in the oxidation number for sulfure is four, not two. (As if you have two sulfur atoms with oxidation number of four in $\ce{2H2SO3}$. One of them changes its oxidation number to zero. while the other one stays intact).