electrochemistry - 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:

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

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.