JEE Main 2025 Jan 24 Shift 2, Chemistry Q1: Oxidation and Reduction

JEE Main2025Jan 24, Shift 2Chemistry

Based on the data given below:

$E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} = 1.33 V$ $E_{C l_{2} / C l^{-}}^{0} = 1.36 V$

$E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V$ $E_{C r^{3 +} / C r}^{0} = - 0.74 V$

the strongest reducing agent is :

A
$M n^{2 +}$
B
$C r$
C
$M n O_{4}^{-}$
D
$C l^{-}$

Solution

The strongest reducing agent is the species that is most easily oxidised, which corresponds to the half-cell with the most negative (lowest) standard reduction potential.

Comparing the four given potentials:

$E_{C r^{3 +} / C r}^{0} = - 0.74 V$ is the only negative value, and it is the lowest of the four.

$∴$ The species on the reduced side of this couple, namely $C r$ (metallic chromium), acts as the strongest reducing agent.

Correct option: (2)

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JEE Main 2025 Jan 24 Shift 2, Chemistry Q1: Oxidation and Reduction

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