Which of the following will have a highest Eo value?

The correct answer is Option (3) → Mn³⁺/Mn²⁺

The standard electrode potential E^°(M³⁺/M²⁺) measures how easily M³⁺ is reduced to M²⁺. A large positive value means M³⁺ readily gains an electron, so M²⁺ is the more stable ion. A negative value means M²⁺ readily loses an electron, so M³⁺ is the more stable ion. The value depends largely on the stability of the d-electron configuration of the two ions.

Mn (Z = 25) is [Ar]3d⁵4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Mn^{3+}} \): 3d⁴. On gaining one electron it becomes \( \ce{Mn^{2+}} \), 3d⁵, which is a half-filled d subshell with extra stability. So \( \ce{Mn^{3+}} \) is strongly oxidising, and among the given couples Mn³⁺/Mn²⁺ has the highest E^° value.

Why the other options are incorrect

• Option (1) Cr³⁺/Cr²⁺ — Incorrect. Cr (Z = 24) is [Ar]3d⁵4s¹; losing three electrons gives \( \ce{Cr^{3+}} \): 3d³, which has a stable half-filled t_2g level, so \( \ce{Cr^{2+}} \) is a reducing agent and E^° is negative (−0.41 V).

• Option (2) Sc³⁺/Sc²⁺ — Incorrect. Sc (Z = 21) is [Ar]3d¹4s²; losing all three electrons gives \( \ce{Sc^{3+}} \) the stable noble gas configuration [Ar], and \( \ce{Sc^{2+}} \) is practically unknown. \( \ce{Sc^{3+}} \) has no tendency to be reduced, so the value is strongly negative.

• Option (4) Ti³⁺/Ti²⁺ — Incorrect. Ti (Z = 22) is [Ar]3d²4s², so \( \ce{Ti^{2+}} \) is 3d² and \( \ce{Ti^{3+}} \) is 3d¹. \( \ce{Ti^{2+}} \) is easily oxidised to \( \ce{Ti^{3+}} \), so E^° is negative (−0.37 V).

NCERT (2026-27 edition):
Trends in the M³⁺/M²⁺ Standard Electrode Potentials