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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?

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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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  • $\begingroup$ Yes you're right, they are not two SHE. Anyway thanks for the answer because i didn't notice that if x > 1 then pH < $0$ so E > $0$ and G < $0$ $\endgroup$ – Mirko Jan 14 '18 at 22:42

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