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I was just wondering if decomposition potentials actually change during electrolysis. For example, if we reversed the Daniell cell reaction so that copper is oxidized, the half cell reactions would be:

$$\ce{Cu <=> Cu^2+ + 2e-} \qquad \pu{+0.34 V}$$ $$\ce{Zn^2+ + 2e- <=> Zn} \qquad \pu{-0.76 V}$$

if we assume both of the solutions are at $\pu{1 M}$ at $\pu{298.15 K}$, the decomposition voltage should be $\pu{1.1 V}$.

If we applied the voltage until half of the zinc cation solution reacted, we would have $\pu{1.5 M}$ $\ce{Cu^2+}$-solution and $\pu{0.5 M}$ $\ce{Zn^2+}$-solution, so that the resulting decomposition voltage should be higher:

$$E_\ce{Cu} = 0.34 + \pu{0.0295 V}\times \log (1.5) = \pu{0.345 V},$$ $$E_\ce{Zn} = -0.76 + \pu{0.0295 V}\times \log (0.5) = \pu{-0.769 V}$$

This should result in a decomposition voltage of $\pu{1.115 V}$ approx. Am I correct on that matter? I haven't found a single article that talks about the actual potential changes during electrolysis, even in my exam I only had to calculate the initial voltage, not what happens afterwards.

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    $\begingroup$ You can simply use Nernst equation for the cell reaction, $\ce{Zn + Cu^2+ <=> Zn^2+ + Cu}$ with $n = 2$. $\endgroup$ Aug 22 at 13:19
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    $\begingroup$ Hah! I was just having the same question a few days ago. Your conclusion is correct. $\endgroup$
    – Buraian
    Aug 22 at 15:10

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