[antoinetta]

I'm having a lot of trouble with this question, so I would appreciate any input.

Calculate the reduction potential (at 25°C) of the half-cell MnO₄^- (1.30×10^-1 M)/ Mn²⁺ (1.40×10-² M) at pH = 4.00.  The half-reaction is given as MnO₄^- + 8H⁺ + 5e^- --> Mn²⁺ + 4H₂O E° = 1.51 V)

So I found the [H⁺] = 1 * 10^-4, but for some reason when I sub the numbers into the equation, I'm not getting the right answer ...

[Borek]

Show the equation you are using.

[antoinetta]

E=Ecell - (0.0257/n)lnQ

Q = [products] / [reactants]= 0.014 / (0.13 * 1 *10^-32)
n = 5
Ecell = 1.51

[Borek]

Looks OK to me, although I prefer version +log([Ox]/) (instead of products/reagents).

What result do you get and what you should get? Looks tome like 1.14 is the right answer (as long as we are omitting ionic strength of the solution).

[antoinetta]

1.14 is the right answer ... but for some reason I've been getting 1.04    thanks for checking the calculations.