N in Nernst Equation
The Nernst equation is as follows,
\begin{align*}
    E_\text{red}&amp;=E_\text{red}^{0}-\frac{RT}{nF}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}
where $c$ is the concentration of the species specified (usually the chemical activity, $a$, is used, but the equations follow the same principle). In the first case, $n=2$. So,

\begin{align*}
    E_\text{red}&amp;=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}

Suppose we double the coefficients in the equation. Now, $n=4$. However, the concentrations of all the species have also doubled. This requires the reaction quotient (the stuff inside the natural logarithm) to be squared. Therefore,
\begin{align*}
    E_\text{red}&amp;=E_\text{red}^{0}-\frac{RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)^2}.
\end{align*}

The laws of logarithms means that we can move the power of 2 to be a multiplication of the logarithm,

\begin{align*}
    E_\text{red}&amp;=E_\text{red}^{0}-\frac{2RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},\\\\
    &amp;=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},
\end{align*}
which is the exact equation we had previously, thus, nothing has "changed".

So to answer your equation. Yes, $n=4$, but the equation does not change, as the reaction quotient alters accordingly, also.