Standard electrode potentials (Mn+/M) of four metals A, B, C and D are given below:
(A) - 1.2 V
(B) + 0.5 V
(C) + 1.2 V
(D) - 3.0 V
The correct increasing order for the reducing power of these metals will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C), (B), (A), (D).
A standard electrode potential E°(Mn+/M) is a reduction potential: it measures how easily Mn+ gains electrons to become M. A metal acts as a reducing agent by doing the reverse, losing electrons:
\( \displaystyle \ce{M -> M^{n+} + ne^-} \)
A more negative E° means Mn+ is hard to reduce, so the metal M loses electrons easily: a stronger reducing agent.
A more positive E° means Mn+ is easily reduced, so the metal holds its electrons: a weaker reducing agent.
So reducing power increases as E° decreases. Taking the given metals from the most positive to the most negative E°:
(C) — first (weakest reducing agent). E° = +1.2 V, the most positive value, so Cn+ is the easiest to reduce and C has the least tendency to lose electrons.
(B) — second. E° = +0.5 V. It is still positive, but lower than that of C, so B gives up electrons a little more easily.
(A) — third. E° = −1.2 V. A negative value means A is a better reducing agent than hydrogen, and a much better one than B and C.
(D) — last (strongest reducing agent). E° = −3.0 V, the most negative value, close to that of lithium (−3.05 V), the most powerful reducing agent in aqueous solution.
NCERT (2026-27 edition):
Measurement of Electrode Potential – “… Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction.”