Statement I: Copper is less reactive than hydrogen.
Statement II: The standard electrode potential E°Cu2+/Cu is negative.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Statement I is correct but Statement II is incorrect
The standard electrode potential of a half-cell is measured against the standard hydrogen electrode (E° = 0). A positive value means the ion is reduced more easily than \( \ce{H^+} \); a negative value means the metal is oxidised by \( \ce{H^+} \).
Statement I: Copper is less reactive than hydrogen. — True. \( \ce{H^+} \) ions cannot oxidise copper, so copper does not displace hydrogen from acids; it does not dissolve in HCl.
Statement II: The standard electrode potential E°Cu2+/Cu is negative. — False. For \( \ce{Cu^2+ + 2e^- -> Cu} \), E° = +0.34 V, a positive value.
NCERT (2026-27 edition):
2.2.1 Measurement of Electrode Potential – “… The measured emf of the cell: Pt(s) | H2(g, 1 bar) | H+(aq, 1 M) || Cu2+(aq, 1 M) | Cu is 0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction: Cu2+(aq, 1M) + 2e– → Cu(s) …”
“The positive value of the standard electrode potential in the first case indicates that Cu2+ ions get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions cannot oxidise Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under the standard conditions described above. Thus, Cu does not dissolve in HCl.”