[iseeu]

2MnO4- (aq) +16H+(aq) +5 Pb(s) + 5SO2-4 (aq) --> 5PbSO4 (s) 2 Mn2+ (aq) + 8H2O (l)

I determined the cell potential to be E= 1.63 Volts.

So if sodium sulfite was added to the cell to precipitate Mn2+ ions, would the cell potential increase or decrease? please describe to me why

I think more solid would give the cell less volts? I have no idea though.


Another question..what is the value of the equilibrium constant K @ 25º C?

I tried using the equations: ΔGº = -nFEº and ΔGº= -RTln(Keq)

but they didnt work out. (my calculator couldn't even handle the answer)

however, in an earlier part of the problem, he also gave us concentrations for all of the ions. So should I just calculate K that way? (I already did) It just seems too easy. Let me know what you think!

(I calculated K to be 8.11 x 10 ^32, so it's really big!! It just seems you should be able to get the same answer through the other equations though, so something's wrong)

[Borek]

So if sodium sulfite was added to the cell to precipitate Mn2+ ions, would the cell potential increase or decrease?

Sulfite will get oxidized by MnO₄^-, in this context I don't see how it could precipitate Mn²⁺.

[iseeu]

Well that was a question, and it's a take home test. Maybe it's just a more conceptual or general question, like what happens to ε when you have more solid in a system.

Oh yea he gave us ion concentrations: [MnO^-4] =.4M [SO^2-4] =.6M [Mn²⁺] = 2x10^-4M  pH=2.4 so [H+]=.00398 M

I was thinking maybe the precipitation will affect Q because Mn2+ would just drop out of the equation. Is this right?

[Borek]

Could be it was the general idea, but if so, you would need K_sp to calculate [Mn²⁺]. Unless it is just a matter of Yes/No answer.