Which one represents the correct reduction reaction for the \( \mathrm{Cu^{2+}|Cu} \) half-cell?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}\)
Reduction is the gain of electrons. In a metal ion–metal half-cell such as \( \ce{Cu^2+}|\ce{Cu} \), the reduction half-reaction turns the metal ion in solution into the metal, with the electrons written on the left. A \( \ce{Cu^2+} \) ion has a charge of +2, so it needs two electrons to become a neutral copper atom:
\( \displaystyle \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
This is the reaction at the copper electrode (the cathode) of the Daniell cell. Both mass and charge balance: one Cu on each side, and +2 − 2 = 0 on the left, 0 on the right.
Why the other options are incorrect
Option (2) \(\mathrm{Cu^{2+}(aq)\rightarrow Cu(s)+2e}^{-}\) — Incorrect. The charge does not balance (+2 on the left, −2 on the right); electrons cannot be released when an ion turns into the neutral metal.
Option (3) \(\mathrm{Cu(s)\rightarrow Cu(aq)+2e}^{-}\) — Incorrect. Electrons are given out, so this is an oxidation; the ion formed is also written without its charge (it should be \( \ce{Cu^2+} \)).
Option (4) \(\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu^{+}(s)+e}^{-}\) — Incorrect. Electrons appear on both sides, and \( \ce{Cu+} \) is an ion, not a solid metal; the net change would be \( \ce{Cu^2+ + e^- -> Cu+} \), a different couple.
NCERT (2026-27 edition):
Galvanic Cells
