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


$$\ce{2H2O_{(l)} -&gt; 2H2_{(g)} + O2_{(g)}}$$


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:


$$\ce{4H2O_{(l)} -&gt; 4H2_{(g)} + 2O2_{(g)}}$$ 


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 $\ln K_2 = \ln{(K_1)^2} = 2 \ln K_1$.