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I was hoping for some help with choosing half-reactions for an electrolytic cell. This was the problem that sparked the question (Question #6 from the 2020 USNCO Part II):

Question #6 2020 USNCO Part II

The correct answer is $$2\ce{Cu^2+ + 2H2O (l) -> 2 Cu (s) + O2 (g) + 4H+},$$ which fits in with the details about gas evolution. However, my question if we were not given these details, and rather just the solution and table of half-reactions, how would you determine the correct combination of half-reactions? Is it always the reaction that has the greatest value of $E^{\circ}$? And does this change when a current is being passed through the solution?

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    $\begingroup$ It would be better to cite the text of the question and not including it as an image. Images cannot be searched and often enough have not the best readability. (Also it is sad to see that USNCO has such a high regard for correct typesetting.) $\endgroup$ – Martin - マーチン Apr 22 at 22:24
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Well I'm not completely sure but I'll try to answer.

To answer your first question, yes, it is almost always certain that the half reaction with the greater $E^\circ$ will determine the correct combination. But, as always there are instances when the above is not always true due to reaction factors.

For instance, in the electrolysis of a $\ce{NaCl}$ solution, we have the following possibilities at the anode: \begin{align} \ce{Cl^- (aq) &-> 1/2Cl (g) + e-}, & E^\circ &= \pu{-1.36 V}\\ \ce{2H2O(l) &-> O2 (g) + 4H+ (aq) + 4e-},& E^\circ &= \pu{-1.23 V} \end{align}

It would suggest that the correct half reaction is the latter, but practically speaking, the $\ce{O2 (g)}$ formed forms a layer of bubbles over the metal surface of the electrode which slows down the reaction quite a lot, and so to increase the rate, an overpotential[1] or bubble potential is applied, but that makes the former reaction kinetically favorable, leading to the oxidation of $\ce{Cl-}$ and not $\ce{H2O}$ as expected.
Besides some exceptions like the above, it is always the half reaction with the greater $E^\circ$.

To answer your second question, if we look at the expression for $E^\circ$, $$ \Delta G^∘=-nFE^\circ $$ and $\Delta G^∘$ only depends on temperature, so unless the current causes the solution to heat thereby changing its temperature, the value of $E^\circ$ will not change.

References

  1. Wikipedia: overpotential
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