What does a negative \(E^o\) (standard reduction potential) value of a redox couple indicate?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Redox couple is a stronger reducing agent than the \(H^+ / H_2\) couple
Standard electrode potentials are measured against the standard hydrogen electrode, whose potential is taken as zero, so the sign of E° compares a couple with the \( \ce{H^+} \)/\( \ce{H2} \) couple. For example, \( \ce{Zn^2+} \)/Zn has E° = −0.76 V, and zinc displaces hydrogen from dilute acids; \( \ce{Cu^2+} \)/Cu has E° = +0.34 V, and copper does not.
Why the other options are incorrect
Option (2) Redox couple is a weaker reducing agent than the H+/H2 couple — Incorrect. A weaker reducing agent than \( \ce{H^+} \)/\( \ce{H2} \) is what a positive E° shows.
Option (3) Redox couple is a stronger oxidizing agent than the H+/H2 couple — Incorrect. A stronger oxidising agent than \( \ce{H^+} \)/\( \ce{H2} \) also corresponds to a positive E°, as for \( \ce{F2} \)/F– (+2.87 V).
Option (4) Redox couple carries equivalent oxidizing power as the H+/H2 couple — Incorrect. Equal oxidising power to \( \ce{H^+} \)/\( \ce{H2} \) would mean E° = 0, not a negative value.
NCERT (2026-27 edition):
Measurement of Electrode Potential – …
