Question

Given:
$Hg _{2}^{2+}+2 e^{-} \longrightarrow 2 Hg ; \,\,\, E^{\circ}=0.789\, V$
$Hg ^{2+}+2 e^{-} \longrightarrow Hg ; \,\,\, E^{\circ}=+0.854\, V$
Then equilibrium constant for
$Hg _{2}^{2+} \longrightarrow Hg + Hg ^{2+} $ is:

• A. $3.13 \times 10^{-3}$
• B. $6.26 \times 10^{-3}$
• C. $3.13 \times 10^{-4}$
• D. $6.26 \times 10^{-4}$

Answer 1

Answer: (B)

$Hg _{2}^{2+}+2 e^{-} \longrightarrow 2 Hg ; \,\,\,\, E^{\circ}=0.789 V$
$Hg ^{2+}+2 e^{-} \longrightarrow Hg ; \,\,\,\, E^{\circ}=+0.854 V$
$Hg _{2}^{2+} \longrightarrow Hg + Hg ^{2+} \,\,\,\, E^{\circ}=+0.789-0.854=-0.065 V$
$K=$ antilog $\left[\frac{n E^{\circ}}{0.059}\right]=$ antilog $\left[\frac{2 \times(-0.065)}{0.059}\right]=6.26 \times 10^{-3}$