Redox - concentration cell

I'm studying galvanic cells and I've found a demonstration that lead me to an impossible result, so there must be a mistake, but I can't find what it is.
My problem is about concentration cells made by two standard hydrogen electrodes (SHE; $\pu{1 bar}$, $\approx\pu{1 atm}$, $\pu{298 K}$). I want to show that the reduction always happens in the most acid solution.


### Demonstration


Consider the half-cell with $x > \pu{1 M}$:


$$\ce{Pt | H+ (aq, $x$\,\pu{M}) | H2 (\pu{1 atm})}$$


$$
\begin{align} 
\Delta G &= \Delta G^\circ + RT\ln{Q}\\
E &= E^\circ - \frac{RT}{nF}\ln{Q}\\
&= -\frac{0.0592}{2}\log{Q}
\end{align}
$$


The reaction is


$$\ce{2 H^+ + 2 e- -> H2}$$


$$
\begin{align} 
Q &= \frac{P(\ce{H2})}{[\ce{H+}]^2}\\
E &= \frac{0.0592}{2}\log{[\ce{H+}]^2}\\
&= 0.0592\cdot\log{[\ce{H+}]}\\
&= -0.0592\cdot\mathrm{pH}
\end{align}
$$


where $E$ is the reduction potential, and therefore the reaction is shown as a reduction.


So, this means that since $E<0$, $\Delta G>0$, this reaction is not spontaneous, consequently it never reduces when linked to a SHE. Can you find the mistake?




The 'standard' in 'Standard Hydrogen Electrode' also indicates the concentration of the electrolyte must be 1mol/l (pH 0).
Therefore, if you link two Standard Hydrogen Electrodes, you won't be able to measure a voltage.


If you don't intend to have standard conditions however, the reduction will indeed take place in the more acedic half-cell (simply spoken). This is also indicated by the electrode potential you calculated: -0,059V \* pH. The reduction takes place in the half-cell with the higher potential and the potential you calculated increases with lower pH (more acedic solution). 


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