How to determine the cell potential of chromium(III) — permanganate cell?

Question

Determine the cell potential of the following cell:

$$\ce{Pt | Cr^3+ (\pu{0.125 M}), Cr2O7^2- (\pu{0.200 M}), H+ (\pu{0.600 M}) || MnO4- (\pu{0.150 M}), Mn^2+ (\pu{0.400 M}), H+ (\pu{0.250 M}) | Pt}$$

As far as I can tell, I've balanced the reactions properly:

$$ \begin{align} \ce{MnO4- + 8 H+ 5 e- &-> Mn^2+ + 4 H2O} &\quad E^\circ &= \pu{+1.507 V} \tag{R1} \\ \ce{Cr2O7^2- + 14 H+ + 6 e- &-> 2 Cr^3+ + 7 H2O} &\quad E^\circ &= \pu{+1.360 V} \tag{R2} \end{align} $$

I don't know how to proceed. I know that when I have my voltage, I can use the Nernst equation to use the concentrations here, but the issue is that I'm not sure what comes before that.

Do I balance them by multiplying one by $6$ and the other one by $5$? which one is being reduced and which oxidized, and which one needs to be flipped around as consequence?

Answer 1

You have asked two questions. Regarding "which one is being reduced" and "which one is being oxidized", you can clarify that by using the Nernst equation

$$\Delta E=\Delta E^\circ-\dfrac{RT}{nF}\ln Q$$

In your example, you know that the reduction of $\ce{MnO4-}$ to $\ce{Mn^2+}$ (oxidation state of +7 to +2) and the oxidation of $\ce{Cr^3+}$ to $\ce{Cr2O7^2-}$ (oxidation state of +3 to +6) will take place at standard state, because $\Delta E^\circ=0.147\ \mathrm V$, and therefore $\Delta G_\mathrm r^\circ=-nF\Delta E^\circ\lt0$. Therefore, this is the redox couple that will spontaneously function as a galvanic cell at standard state.

In general, this will also be true for most exercises, since the logarithm part will not deviate much of $\Delta E^\circ$. But, in the general case you can't know in advance, so just do the following:

1. Arbitrarily choose the reduction and oxidation reactions.
2. Use the Nernst equation and calculate $\Delta E$.
3. If $\Delta E\gt0$, then you have guessed them right, and those electrochemical reactions will spontaneously take place in a galvanic cell. If $\Delta E\lt0$, then the answer is still correct, but you have to clarify that those reactions will not spontaneously take place. You would be right if you state that this is the external voltage that you, as surroundings, have to apply to the terminals of your cell in order for the reactions in that direction to take place.

As you said at the end, you multiply those electrochemical equations by 5 and 6 to eliminate the electrons, so you know the powers of your reactants to correctly calculate $Q$.