Question

$E^{\circ }$ for some half cell reactions are given below
$S{n}^{+4}+2e^{-}\to S{n}^{2+};E^{\circ }=0.151V$
$2H{g}^{2+}+2e\to H{g}_2^{2+};E^{\circ }=0.92V$
$Pb{O}_2+4H^{+}2e^{-}\to P{b}^{2+}+2H_{2}O;E^{\circ }=1.45V$
Based on the given data which statement is correct.

• A. $S{n}^{+4}$ is a stronger oxidising agent than $P{b}^{+4}$
• B. $S{n}^{2+}$ is a stronger reducing agent than $H{g}_2^{2+}$
• C. $H{g}^{2+}$ is a stronger oxidising agent than $P{b}^{+4}$
• D. $P{b}^{+2}$ is a stronger reducing agent than $S{n}^{2+}$

Answer 1

The correct option is B

$S{n}^{2+}$ is a stronger reducing agent than $H{g}_2^{2+}$

A stronger oxidising agent gets reduced itself and would have a more +ve reduction potential.

A stronger reducing agent gets oxidised and would have a more +ve oxidation potential or a more -ve reduction potential.

$S{n}^{+2}\to S{n}^{+4}+2e^{-}{E}^{\circ }=-0.15VH{g}_2^{2+}\to 2H{g}^{2+}+2e^{-}{E}^{\circ }=-0.92VS{n}^{+2},H{g}_2^{2+}$