Consider the following reduction processes: $\mathrm{Al}^{3+}+3\mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}), \mathrm{E}^{\circ}=-1.66\mathrm{~V}$ $\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=+0.77\mathrm{~V}$ $\mathrm{Co}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Co}^{2+}, \mathrm{E}^{\circ}=+1.81\mathrm{~V}$ $\mathrm{Cr}^{3+}+3\mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}), \mathrm{E}^{\circ}=-0.74\mathrm{~V}$ The tendency to act as reducing agent decreases in the order:

(A) $\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}$

(B) $\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}$

(C) $\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}$

(D) $\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}$

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Redox Reactions and Electrochemistry

Consider the following reduction processes:

Al^{3 +} + 3 e^{-} \to Al (s), E^{\circ} = - 1.66 V

Fe^{3 +} + e^{-} \to Fe^{2 +}, E^{\circ} = + 0.77 V

Co^{3 +} + e^{-} \to Co^{2 +}, E^{\circ} = + 1.81 V

Cr^{3 +} + 3 e^{-} \to Cr (s), E^{\circ} = - 0.74 V

The tendency to act as reducing agent decreases in the order:

(A) $Al > Cr > Fe^{2 +} > Co^{2 +}$ ✓ Correct
(B) $Al > Fe^{2 +} > Cr > Co^{2 +}$
(C) $Al > Cr > Co^{2 +} > Fe^{2 +}$
(D) $Cr > Fe^{2 +} > Al > Co^{2 +}$

Answer: (A)

Solution

A species is a stronger reducing agent when its standard reduction potential is more negative, since the reduced form then gives up electrons more readily. The potentials are

A l^{3 +} /Al = - 1.66

C r^{3 +} /Cr = - 0.74

F e^{3 +} /F e^{2 +} = + 0.77

V and

C o^{3 +} /C o^{2 +} = + 1.81

V. Arranging from the most negative to the most positive potential gives the decreasing reducing tendency as

Al > Cr > F e^{2 +} > C o^{2 +}

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Redox Reactions and Electrochemistry

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