# 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?

• 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 < $0$ Jan 14 '18 at 22:42