Question

Given below some standard reduction potentials:
$S{n}^{2+}+2e^{-}\to Sn,E^{\circ }=-0.14V$
$H{g}_2^{2+}+2e^{-}\to 2Hg$, $E^{\circ }=0.79V$
Which of the following statement is correct?

• A. $Hg$ can reduce $S{n}^{2+}$ to $Sn$
• B. $S{n}^{2+}$ can oxidize $Hg$ to $H{g}_2^{2+}$
• C. $H{g}_2^{2+}$ can oxidze $Sn$ to $S{n}^{2+}$
• D. Between $H{g}_2^{2+}$ and $S{n}^{2+}$, $S{n}^{2+}$ is stronger oxidising agent than $H{g}_2^{2+}$

Answer 1

The correct option is C $H{g}_2^{2+}$ can oxidze $Sn$ to $S{n}^{2+}$

If $E_{cell}^{\circ }$ is positive for a reaction, the process would take place otherwise not.
For the cell reaction,
$H{g}_2^{2+}+Sn⟶2Hg+S{n}^{2+}$
$E_{cell}^{\circ }=E_{H{g}^{2+}/Hg}^0-E_{S{n}^{2+}/Sn}^0=0.79V-(-0.14V)=0.93V$. As, $E_{cell}^{\circ }$ is positive for this reaction, it follows that $H{g}_2^{2+}$ can oxidize $Sn$ to $S{n}^{2+}$
Hence, (C) is the correct option.