How to calculate the pH of a redox reaction between nickel(IV) oxide and silver?
For the following reaction:
\begin{gather}
\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -&gt; Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\\
E^\circ = \pu{2.48 V}
\end{gather}
Calculate the $\mathrm{pH}$ of the solution if $E = \pu{2.23 V}$ and $[\ce{Ag^+}] = [\ce{Ni^2+}] = \pu{0.023 M}$.

The important thing to understand that was not immediately obvious to me and a lot of other people was the fact that $n = 2$, and this changes how you calculate your answer.
$\ce{Ni}$ does not change oxidation state in its transition from $\ce{NiO2}$ to $\ce{Ni^2+}$, as its oxidation state in $\ce{NiO2}$ is $+2$.  The 4 $\ce{H^+}$ ions also do not change oxidation state in their transition from $\ce{H^+}$ to $\ce{H2O}$, oxidation state in $\ce{H^+}$ is $+1$ and oxidation state in $\ce{H2O}$ for $\ce{H}$ is $+1$ for 2 $\ce{H}$ atoms.  That leaves us with the only $\ce{e^-}$ transfer occurring to $\ce{Ag}$.  There are 2 moles of $\ce{Ag}$ atoms going to oxidation state $+1$.  Therefore, 2 moles of $\ce{e^-}$ are transferred.  Hence, $n=2$
The Nernst Equation:
$$E = E^\circ  -  (0.0592/n)\log Q$$
Where $E˚$ is the standard cell potential, $E$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$
For this problem, $n(\ce{e-}) = \pu{2 mol}$, and that
$$Q = \frac{[0.023]^3}{[\ce{H^+}]^4}.$$
Following the above calculation with this difference leads to $\mathrm{pH} = 3.34$, which is the correct solution.
\begin{align}
\tag1  
2.23 &amp;= 2.48 
- \frac{0.0592}{2}\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag2 
0.25 &amp;= \frac{0.0592}{2} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag3 
0.50 &amp;= \frac{0.0592}{1} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag4 
0.50 &amp;= 0.0592\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag5 
\frac{0.50}{0.0592} &amp;= \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag6 
\frac{0.50}{0.0592} 
&amp;= \log\left(0.023^3\right) -\log\left([\ce{H^+}]^4\right)\\
\tag7 
\frac{0.50}{0.0592} - \log\left(0.023^3\right) 
&amp;= -\log\left([\ce{H^+}]^4\right)\\
\tag8 
\frac{0.50}{0.0592} - \log\left(0.023^3\right) 
&amp;= -4\log\left([\ce{H^+}]\right)\\
\tag9 
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} 
&amp;= -\log\left([\ce{H^+}]\right)\\
\tag{10} 
\mathrm{pH} &amp;= -\log\left([\ce{H^+}]\right)\\
\tag{11} 
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} &amp;= 3.34
\end{align}