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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.

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    $\begingroup$ See related chemistry.stackexchange.com/questions/115660/… $\endgroup$ – Poutnik May 21 at 10:56
  • $\begingroup$ @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.! $\endgroup$ – Mayuri Vaish May 21 at 11:02
  • $\begingroup$ Did you compare standard potentials of all 3 half reactions ? $\endgroup$ – Poutnik May 21 at 11:04
  • $\begingroup$ as they are directly related to delta G via nF factor. $\endgroup$ – Poutnik May 21 at 11:13
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    $\begingroup$ @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. :))) $\endgroup$ – Mayuri Vaish May 21 at 12:09
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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).

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