[sinthreck]

Calculate the pH of this cell when the potential is 0. (At 25 Degrees C).

Standard Reduction Potential:
MnO₄^- + 8H⁺ + 5e^-  Mn²⁺ + 4H₂O    E^o = +1.51 V
Au³⁺ + 3e^-  Au(s)            E^o = +1.42 V

The contents of the half cells are:
0.36 M MnO₄^-
0.004 M Au³⁺
0.001 M Mn²⁺

I am assuming the above concentrations are equilibirum concentrations, otherwise I don't know how to solve the problem. Is this assumption ok?

Balanced Equation:
3MnO₄^- + 24H⁺ + 5Au(s)  3Mn²⁺ + 5Au³⁺ + 12H₂O

We know n = 15
E^o = 0.09

.09 = (.0592/15) * log ([Mn²⁺]³ * [Au³⁺]⁵  /  [MnO₄^-]³ * [H+]²⁴]   <--- Are these coefficients ok?

Solving for the math:

[H+] = 0.01701 M
p(H) = 1.769

[Borek]

Approach and exponents are correct. However, I am never sure about signs and I have not checked the result thoroughly.

[sinthreck]

What do you mean by signs?

As in, I stuffed up a calculation?

I'm ok with the math, so as long as my approach is ok. I'm satisfied.

[Borek]

Signs like whether the potential should be 0.09 or -0.09.

[Abstractineum]

Because MnO₄^- has higher reduction potential than Au³⁺, and it is being reduced, I believe the standard cell potential should be positive.

[sinthreck]

The answer is correct. I just submitted my quiz and got full marks )