Why do Cu⁺ ions spontaneously form copper metal and Cu²⁺ ions in solution?

Question

The standard electrode potentials for three reactions involving copper and copper ions are:

$$ \begin{align} \ce{Cu^2+(aq) + e- &-> Cu+(aq)} &\quad E^\circ &= \pu{+0.15 V} \\ \ce{Cu^2+(aq) + 2e- &-> Cu(s)} &\quad E^\circ &= \pu{+0.34 V} \\ \ce{Cu+(aq) + e- &-> Cu(s)} &\quad E^\circ &= \pu{+0.52 V} \\ \end{align} $$

Which statement is correct?

A. $\ce{Cu^2+}$ ions are a better oxidizing agent than $\ce{Cu+}$ ions.
B. Copper metal is a better reducing agent than $\ce{Cu+}$ ions.
C. $\ce{Cu+}$ ions will spontaneously form copper metal and $\ce{Cu^2+}$ ions in solution.
D. Copper metal can be spontaneously oxidized by $\ce{Cu^2+}$ ions to form $\ce{Cu+}$ ions.

I thought that because copper is more likely to get oxidised than $\ce{Cu+},$ it is a better reducing agent and thus B is the answer. But the given answer is C. Why?

Answer 1

Both reactions involving $\ce{Cu+}$ have the lowest and highest standard electrode potentials, respectively. Thus, $\ce{Cu+}$ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation.

Because the formation of $\ce{Cu}$ has a higher SEP than $\ce{Cu+},$ it is more likely to get reduced and therefore is a better oxidizing agent — not a better reducing agent — than $\ce{Cu+}$ (which is more likely to get oxidized as it has a higher SEP).