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




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).


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