Relationship between Yield and Applied Potentials

This question revolves around this question from Atkins' *Chemical Principles* 5th ed., Chapter 13, Question 13.116:



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> Consider the electroplating of a metal $+1$ cation from a solution of unknown concentration according to the half-reaction $\ce{M+ (aq) + e- -> M (s)}$, with a standard potential $E^\circ$. When the half-cell is connected to an appropriate oxidation half-cell and current is passed through it, the $\ce{M+}$ cation begins plating out at $E\_1$. To what value ($E\_2$) must the applied potential be adjusted, relative to $E\_1$, if $\pu{99.99 \%}$ of the metal is to be removed from the solution?
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How is this problem solvable? I know that at applied potential $E\_1$ the cell potential is $0$ ($\Delta G = 0$), so that means at $E\_2$, the equilibrium constant must demand a yield of $\pu{99.99 \%}$. However, doesn't this yield depend on what the oxidation half-cell is?


For example, the oxidation reaction could be one from cerium(III) to cerium(IV) or from chloride ions to chlorine gas, both of which have different effects on the equilibrium constant. I've gotten as far as $-nFE = RT\ln{K}$, where I do believe $E = E\_2 - E\_1$ and $E$ is positive to ensure negative Gibbs free energy change.


This is an even number question so there is no given answer. I would really appreciate a solution or at least explanations clearing up any misconceptions that I may have / clues to how to solve this problem. This is not a homework problem.


