Q4 / 75 · JEE Main 2026 (22 January Shift 2) · Chemistry

Consider the following reduction processes :
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{s}), \mathrm{E}^{0}=-1.66 \mathrm{~V}$
$\mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^{0}=+0.77 \mathrm{~V}$
$\mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^{0}=+1.81 \mathrm{~V}$
$\mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{s}), \mathrm{E}^{\circ}=-0.74 \mathrm{~V}$
The tendency to act as reducing agent decreases in the order :

  • Correct answer
    A
    $\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}$
  • B
    $\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}$
  • C
    $\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}$
  • D
    $\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}$

Solution

Reducing agent strength is determined by oxidation potential (opposite sign of reduction potential). Stronger reducing agents have more negative oxidation potentials (easier to oxidize).
Oxidation potentials:
Al(s) → Al³⁺ has E° = +1.66 V (strongest reducing agent);
Cr(s) → Cr³⁺ has E° = +0.74 V;
Fe²⁺ → Fe³⁺ has E° = -0.77 V;
Co²⁺ → Co³⁺ has E° = -1.81 V (weakest reducing agent).
Order of decreasing reducing agent tendency: Al > Cr > Fe²⁺ > Co²⁺.

Hence, the correct answer is option: 'A'.

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