Why do Cu⁺ ions spontaneously form copper metal and Cu²⁺ ions in solution? The standard electrode potentials for three reactions involving copper and copper ions are:
  
  $$
\begin{align}
\ce{Cu^2+(aq) + e- &amp;-&gt; Cu+(aq)} &amp;\quad  E &amp;= \pu{+0.15 V} \\
\ce{Cu^2+(aq) + 2e- &amp;-&gt; Cu(s)} &amp;\quad   E &amp;= \pu{+0.34 V} \\
\ce{Cu+(aq) + e- &amp;-&gt; Cu(s)} &amp;\quad   E &amp;= \pu{+0.52 V} \\
\end{align}
$$
  
  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.


Why is the answer C, and not B? 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 I cannot explain why C is the correct answer. Why is this?

Any help would be greatly appreciated.
Answer:
Both reactions involving Cu+ have the lowest and highest Standard Electrode Potentials respectively. Thus, Cu+ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation.

Because the formation of Cu has a higher SEP than Cu+, it is more likely to get reduced and therefore is a better oxidizing agent - NOT a better reducing agent - than Cu+ (which is more likely to get oxidized as it has a higher SEP).