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andselisk ♦
andselisk ♦
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Answer: BothBoth reactions involving Cu+$\ce{Cu+}$ have the lowest and highest Standard Electrode Potentialsstandard electrode potentials, respectively. Thus, Cu+$\ce{Cu+}$ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation.

Because the formation of Cu$\ce{Cu}$ has a higher SEP than Cu+,$\ce{Cu+},$ it is more likely to get reduced and therefore is a better oxidizing agent — - NOTnot a better reducing agent -— than Cu+$\ce{Cu+}$ (which is more likely to get oxidized as it has a higher SEP).

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).

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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Mayuri Vaish
Mayuri Vaish
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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).