# Why does ΔG for a reaction scale with reaction quantity but E does not? For example, ΔG0rxn for the combustion of 1 mol of hydrogen is 1 × –237 kJ∕mol = –237 kJ,while ΔG0rxn for the combustion of 2 mol of hydrogen is 2 × –237 kJ∕mol =–474 kJ. In both case

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Feb 9, 2018

${E}_{c e l l}^{\circ}$ scales with the mols of electrons, as well as the mols of reactants... But they cancel out.

$\Delta {G}_{r x n}$ is related to $R T \ln Q$, and we know already that $Q \to {Q}^{c}$ if the reaction is scaled, so $\Delta {G}_{r x n}$ is scaled by $c$ since $R T \ln {Q}^{c} = c R T \ln Q$.

In the Nernst equation we have

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$,

and the units work out as

$\left(\left({\text{V" cdot "C/mol reactant" cdot "K")("K"))/(("mol e"^(-)"/mol reactant")("C/mol e}}^{-}\right)\right)$

to give units of $\text{V}$. It must, or else ${E}_{c e l l}^{\circ}$ can't add to $- \frac{R T}{n F} \ln Q$...

If you scale the reaction by a factor of $c$, you raise $Q$ to the $c$ and multiply the number of electrons $n$ by $c$.

For some general redox reaction, we would get something like:

${M}^{+} \left(a q\right) + A \left(s\right) r i g h t \le f t h a r p \infty n s {A}^{+} \left(a q\right) + M \left(s\right)$

$\text{ "" "" } \downarrow$

$c {M}^{+} \left(a q\right) + c A \left(s\right) r i g h t \le f t h a r p \infty n s c {A}^{+} \left(a q\right) + c M \left(s\right)$

The charge contribution on each side clearly changes. Thus, you would get that the number of electrons changes as $n \to c n$ and $Q \to {Q}^{c}$. As a result:

$- \frac{R T}{n F} \ln Q \to \left[- \frac{R T}{c n F} \ln {Q}^{c} = - \frac{R T}{\cancel{c} n F} \cdot \cancel{c} \ln Q\right]$

$= - \frac{R T}{n F} \ln Q$

which is the same expression as before. If we do not affect $- \frac{R T}{n F} \ln Q$, then by inference, ${E}_{c e l l}$ and ${E}_{c e l l}^{\circ}$ do not change due to scaling the reaction by a constant.

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