[NeoTrion]

Having a problem solving a question here.  This one has two equations.  The second one I believe I have solved, but the first one is giving me some trouble.  I can get what I think is the half reactions on the first one, but nothing as far as the acidic and basic solutions go...

Balance the following redox reaction by the ion-electron method for both acidic and basic solutions. Be sure to identify and write the half reactions for oxidation and reduction.

a. S2O3- (aq) + I2(aq) → S4O62- (aq) + I- (aq)
b. MnO4- (aq) + Fe2+ (aq) → Fe3+ (aq) + Mn2+ (aq)

a. S2O3- (aq) + I2(aq) → S4O62- (aq) + I- (aq)

2e- + I2 → 2I-               Reduction
2S2O3- → S4O62-          Oxidation

But if I am right there, then I am stuck and have nothing to add water to to make an acidic solution.  Here is what I did on B.

b. MnO4- (aq) + Fe2+ (aq) → Fe3+ (aq) + Mn2+ (aq)

Fe2+ → Fe3+ + e-              Oxidation
5e- + MnO4- → Mn+2          Reduction

5Fe2+ → 5Fe3+ + 5e-
8H+ + MnO4- + 5e- → Mn+2 + 4H20

8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O  Acidic

8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
8OH- + 8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O + 8OH-

4H20 + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 8OH-

Hopefully someone can point me in the direction on A and check B to make sure I know what I am doing.

[Borek]

Both problems are tricky.

a. You are right - there is no need to balance separately for acidic and alkalic solutions.

b. Products in this reaction are highly pH dependent. See for example http://www.titrations.info/permanganate-titration. Also think what happens to Fe³⁺ in alkalic solution.

[NeoTrion]

In A it just seemed weird to include it.  I was curious if there was a way, but I don't see the reason to do it either.  I thought that was as far as the instructions wanted us to take the reaction on B.  I just wanted to make sure I wasn't going crazy on Part A.  I have been banging my head against the wall to figure out how to get 2 e- to cancel so I can start adding some water to get the acidic and then the basic solution.  But if another set of eyes sees it as I do, then I think I can go on to a bigger and better question. LOL

[NeoTrion]

Ok just had a talk with the professor and he did say the oxidation state was wrong.  It should be...

a. S2O3-2 (aq) + I2(aq) → S4O62-2 (aq) + I- (aq)

He said it was very tricky.  It had to do with the oxidation state of S.  I am still clueless, so maybe someone out there can help.

[Borek]

a. S2O3-2 (aq) + I2(aq) → S4O62-2 (aq) + I- (aq)

Actually S4O6-2 (S₄O₆^2-). But there is really nothing tricky here - unless your prof wants to force you to use oxidation numbers method (no idea what for). Using half reactions it is a breeze.