The strongest reducing agent amongst the following is:
(Given: \(E^0_{Ni^{2+}/Ni} = -0.25 V\), \(E^0_{Cu^{2+}/Cu} = 0.34 V\), \(E^0_{Pb^{2+}/Pb} = -0.13 V\), \(E^0_{Cr^{2+}/Cr} = -0.74 V\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Cr2+
A reducing agent must be able to lose electrons. \( \ce{Ni^{2+}} \), \( \ce{Cu^{2+}} \) and \( \ce{Pb^{2+}} \) are in their usual stable oxidation states; they can only gain electrons, so they act as oxidising agents. \( \ce{Cr^{2+}} \) (3d4) is different: it readily loses an electron to become \( \ce{Cr^{3+}} \) (3d3, a half-filled t2g level), so Cr2+ is strongly reducing.
The \( \ce{M^{3+}}/\ce{M^{2+}} \) potentials decide it: E°(\( \ce{Cr^{3+}}/\ce{Cr^{2+}} \)) = −0.41 V, so \( \ce{Cr^{2+}} \) is easily oxidised to \( \ce{Cr^{3+}} \). It is strong enough to liberate hydrogen from a dilute acid:
\( \ce{2Cr^{2+}(aq) + 2H+(aq) -> 2Cr^{3+}(aq) + H2(g)} \)
The given \( \ce{M^{2+}}/\ce{M} \) values show how easily the metals are oxidised to these ions, not how easily the ions themselves are oxidised.
Why the other options are incorrect
Option (1) Ni2+ — Incorrect. \( \ce{Ni^{2+}} \) is the stable state of nickel and is not oxidised further under ordinary conditions; it can only be reduced to Ni.
Option (2) Cu2+ — Incorrect. \( \ce{Cu^{2+}} \) is the stable state of copper in water. It is reduced (to Cu), not oxidised, so it acts as an oxidising agent, not as a reducing agent.
Option (3) Pb2+ — Incorrect. \( \ce{Pb^{2+}} \) is the stable state of lead; it is reduced to Pb (E° = −0.13 V) rather than oxidised.
NCERT (2026-27 edition):
Chemical Reactivity and E° Values – …
