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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2026 · 22 January · Evening Shift · Chemistry · Electrochemistry

Consider the following reduction processes : Al^ 3+ +3 e^ - Al(s),...

Consider the following reduction processes :
Al3++3e−⟶Al(s),E0=−1.66 V\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{s}), \mathrm{E}^{0}=-1.66 \mathrm{~V}
Fe3++e−⟶Fe2+,E0=+0.77 V\mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^{0}=+0.77 \mathrm{~V}
Co3++e−⟶Co2+,E0=+1.81 V\mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^{0}=+1.81 \mathrm{~V}
Cr3++3e−⟶Cr(s),E∘=−0.74 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 :

Options

  1. AAl>Cr>Fe2+>Co2+\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}
  2. BCr>Fe2+>Al>Co2+\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}
  3. CAl>Fe2+>Cr>Co2+\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}
  4. DAl>Cr>Co2+>Fe2+\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}

Correct answer

A. Al>Cr>Fe2+>Co2+\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}

Step-by-step 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⁺.
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