Does the relationship equation between standard cell potential and equilibrium constant violate potential's intensive properties?

Question

The equation:

$$E^{。}_{cell}= \frac{RT}{nF}\ln K_{eq}$$

We all know cell potential is intensive, not affected by the amount, Because: $volt=\frac{joule}{coulomb}$. Both joule and coulomb will be doubled altogether.

But as seen from the equation, cell potential is affected by the number of electrons transferred.

Answer 1

No, there is no violation of "intensivity". The reason is that $K_{eq}$ depends on $n$, and changes in one cancel the other out.

For example, consider the electrolysis of water:

1. $$\ce{2H2O_{(l)} -> 2H2_{(g)} + O2_{(g)}}$$

The equilibrium constant for this reaction is $K_{1}=\frac{[\ce{H2}]^2 [\ce{O2}]}{1}$ and if you wrote out each electrochemical half reaction separately, $n$ would be 4.

Now consider this reaction:

2. $$\ce{4H2O_{(l)} -> 4H2_{(g)} + 2O2_{(g)}}$$

The equilibrium constant is now $K_{2}=\frac{[\ce{H2}]^4 [\ce{O2}]^2}{1}=(K_1)^2$. If you wrote out the half-reactions for this reaction, $n$ would be 8, twice as big. But the $\ln K_{eq}$ term would also be twice as big, since $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.