Question

write and label the half-reactions for the reduction of I₃^- (aq) by S₂O₃^2- (aq)

Answer 1

Write half-reactions and balance them individually:

Oxidation half-reaction:

2 (S2O3)2- ----> (S4O6)2-

All I did above was to balance out the non-hydrogen/non-oxygen atoms. Now, let's figure out the number of electrons being lost. The oxidation state of sulfur in (S2O3)2- is +2 while in (S4O6)2- is mixed, where you have 2 S(V) and 2 S(0). This means that you lose 6 electrons due to formation of 2 S(V) but you gain 4 electrons due to formation of 2 S(0). The net result is that you lose 2 electrons, thus the reaction is:

2 (S2O3)2- ----> (S4O6)2- + 2 e-

The oxygens are balanced on both sides, so this half-reaction is completely balanced.

Now for the reduction half-reaction:

I3- + 2 e- ---> 3 I-

half reactions for the reduction

No hydrogen or oxygen atoms to balance here so its done too. Since the number of electrons in both equations is the same, you can add them both together to get the overall balanced equation:

2 (S2O3)2- + I3- ----> (S4O6)2- + 3 I-

Total cell reaction

Notice that the charge on both sides is -4, which implies that the oxidation-reduction equation has been properly balanced.