Question

Given: $E^0\left(S{n}^{2+}/S{n}^{4+}\right)=0.15V,E^0\left(H{g}_2^{2+}/H{g}^{2+}\right)=0.92V$ and $E^0\left(P{b}^{2+}/Pb{O}_2\right)=1.45V.$
Based on this data, which of the following statements is incorrect?

• A. $S{n}^{4+}$ is a stronger oxidizing 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 oxidizing 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 $H{g}_2^{2+}$

The oxidising agent is reduced and reducing agent is oxidised in the process. So a better oxidising agent is the one having more tendency to get reduced i.e. its reduction potential is high.
So, higher reduction potential means better oxidising agent.
Similalry, lower reduction potential means better reducing agent. The given data is of oxidation potential but reduction potential = - oxidation potential. So, reduction potential follows the order:
$P{b}^{4+}<H{g}^{2+}<S{n}^{4+}$
As $S{n}^{4+}\text{to}S{n}^{2+}$ has a reduction potential of $-0.15V$ whereas $P{b}^{4+}\text{to}P{b}^{2+}$ has a reduction potential of $-1.45V$ which is lesser than $-0.15V$. So, $S{n}^{4+}$ is stronger oxidizing agent than $P{b}^{2+}$.

Similarly we can check the statements in options (c) and (d) and verify that these statements are correct.
For option (b), $S{n}^{2+}$ is not a better reducing agent than
$H{g}_2^{2+}$ because a better reducing agent needs higher value of oxidation potential but $S{n}^{2+}$ has smaller oxidation potential($0.15V$) than $H{g}_2^{2+}$($0.92V$).
Hence, (b) is the incorrect statement.