# N in Nernst Equation [closed]

My teacher told me that in the Nernst equation, N was the number of electrons transferred. For example, $n = 2$ in $\ce{Ni^{2+} + Zn -> Zn^{2+} + Ni}$.

However, say all of the coefficients were doubled: $\ce{2Ni^{2+} + 2Zn -> 2Zn^{2+} + 2Ni}$. Would $n = 2$ or $n = 4$ in that case?

• You transfer four electrons now, hence $n=4$; it is as simple as that. – Ivan Neretin Aug 6 '18 at 22:24
• And not only does n increase, the reaction quotient changes as well. You may want to think about the properties of logarithms and work through both cases to assure yourself that it makes no difference either way. – Michael DM Dryden Aug 7 '18 at 7:27

The Nernst equation is as follows, \begin{align*} E_\text{red}&=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,
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}&=E_\text{red}^{0}-\frac{RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)^2}. \end{align*}
So to answer your equation. Yes, $n=4$, but the equation does not change, as the reaction quotient alters accordingly, also.