From a past USNCO test:
40. What is the standard reduction potential of $\ce{Hg^2+(aq)}$ to $\ce{Hg(l)}$?
\begin{align} \ce{2 Hg^2+(aq) + 2 e- &-> Hg2^2+(aq)} &\quad E^\circ &= \pu{+0.90 V} \\ \ce{Hg2^2+(aq) + 2 e- &-> 2 Hg(l)} &\quad E^\circ &= \pu{+0.80 V} \end{align}
(A) $\pu{+1.70 V}$
(B) $\pu{+0.85 V}$
(C) $\pu{+0.10 V}$
(D) $\pu{-0.10 V}$
While I understand the method behind it (converting potentials to free energy first, adding, then converting back to $E),$ I don't understand why we have to do that in the first place, since when calculating a typical $E_\text{cell}$, one just adds the potentials.
I've heard people say it's because the number of electrons transferred affects the potential, but it doesn't really make sense to me, as a typical potential is intensive. Why is this case the exception?
The answer is $\pu{0.85 V},$ as calculated from the $\Delta G$ method and $\pu{1.70 V}$ by simply adding the reduction potentials.