The standard electrode potentials, E°(M3+/M2+), of three metals of the first transition series are: Cr = −0.4 V, Mn = +1.5 V and Fe = +0.8 V. Arrange the following ions in increasing order of their stability in acid solution.
(A) Fe3+
(B) Cr3+
(C) Mn3+
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) < (A) < (B)
E°(M3+/M2+) measures how readily the M3+ ion gains an electron and changes to M2+.
The more positive the value, the more readily M3+ is reduced, that is, the stronger it is as an oxidising agent and the less stable it is in solution. The most negative value means the most stable M3+ ion.
(C) Mn3+ (+1.5 V) — first, the least stable. Its very positive value comes from the particular stability of Mn2+ (d5), so Mn3+ takes up an electron readily; Mn3+ is one of the strongest oxidising agents in aqueous solution.
(A) Fe3+ (+0.8 V) — second. Its value is comparatively low because Fe3+ (d5) has extra stability, so it is more stable than Mn3+.
(B) Cr3+ (−0.4 V) — third, the most stable. The negative value means the reverse change, Cr2+ → Cr3+, takes place readily: Cr2+ is a strong reducing agent and even liberates hydrogen from a dilute acid.
\( \displaystyle \ce{2Cr^2+(aq) + 2H+(aq) -> 2Cr^3+(aq) + H2(g)} \)
NCERT (2026-27 edition):
4.3.6 Trends in the M3+/M2+ Standard Electrode Potentials – “…The comparatively high value for Mn shows that Mn2+(d5) is particularly stable, whereas comparatively low value for Fe shows the extra stability of Fe3+ (d5).”
4.3.8 Chemical Reactivity and E° Values – “An examination of the Eo values for the redox couple M3+/M2+ … shows that Mn3+ and Co3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti2+, V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid, e.g., 2 Cr2+(aq) + 2 H+(aq) → 2 Cr3+(aq) + H2(g)”