# 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.

• – Poutnik May 21 at 10:56
• @Poutnik - thanks, yes I did. But that does not explain why in terms of the standard electrode potentials involved. I'm having a hard time understanding how you could infer that from the SEPs given. Do you have any idea? Thanks in advance.! – Mayuri Vaish May 21 at 11:02
• Did you compare standard potentials of all 3 half reactions ? – Poutnik May 21 at 11:04
• as they are directly related to delta G via nF factor. – Poutnik May 21 at 11:13
• @Poutnik I got it!!! Thank you!!! Essentially, both reactions involving Cu+ have the lowest and highest SEPs respectively - so they have to be the strongest oxidizing/reducing agents - thus undergoing disproportionation. :))) – Mayuri Vaish May 21 at 12:09