JEE Main · 2019 · Shift-ImediumRDX-033

Given that, EO2/H2O = +1.23\ V; ES2O82-/SO42- = +2.05\ V EBr2/Br- = +1.09\ V; EAu3+/Au = +1.4\ V The strongest…

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

Given that,

$E^\circ_{\mathrm{O_2/H_2O}} = +1.23\ \mathrm{V};\quad E^\circ_{\mathrm{S_2O_8^{2-}/SO_4^{2-}}} = +2.05\ \mathrm{V}$ $E^\circ_{\mathrm{Br_2/Br^-}} = +1.09\ \mathrm{V};\quad E^\circ_{\mathrm{Au^{3+}/Au}} = +1.4\ \mathrm{V}$

The strongest oxidizing agent is

Options

a
$\mathrm{O_2}$
b
$\mathrm{S_2O_8^{2-}}$
✓
c
$\mathrm{Br_2}$
d
$\mathrm{Au^{3+}}$

Correct Answerb

$\mathrm{S_2O_8^{2-}}$

Detailed Solution

Principle: The species with the highest standard reduction potential ( $E^\circ$ ) is the strongest oxidising agent (most easily reduced).

Step 1 — Compare $E^\circ$ values:

| Species | $E^\circ$ (V) | |---|---| | $\mathrm{O_2}$ | $+1.23$ | | $\mathrm{S_2O_8^{2-}}$ | $+2.05$ | | $\mathrm{Br_2}$ | $+1.09$ | | $\mathrm{Au^{3+}}$ | $+1.40$ |

Step 2 — Identify the highest $E^\circ$ : $\mathrm{S_2O_8^{2-}}$ has the highest $E^\circ = +2.05\ \mathrm{V}$ → strongest tendency to be reduced → strongest oxidising agent

Answer: Option (2) — $\mathrm{S_2O_8^{2-}}$

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