JEE Main202622 January 2026Evening ShiftChemistryElectrochemistryActual
Consider the following reduction processes : Al ³⁺+3 e ⁻ Al ( s ), E ⁰=-1.66 ~V Fe ³⁺+ e ⁻ Fe ²⁺, E ⁰=+0.77 ~V Co ³⁺+ e ⁻ Co ²⁺, E ⁰=+1.81 ~V Cr ³⁺+3 e ⁻ Cr ( s ), E ^ =-0.74 ~V The tendency to act as reducing agent decreases in the order :
Options
- AAl > Cr > Fe ²⁺> Co ²⁺
- BCr > Fe ²⁺> Al > Co ²⁺
- CAl > Fe ²⁺> Cr > Co ²⁺
- DAl > Cr > Co ²⁺> Fe ²⁺
Correct answer
A. Al > Cr > Fe ²⁺> 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⁺.