Question

Find the value of equilibrium constant ( $Kc$ ) for the chemical equation: $Hg_{2}^{2 +} \rightarrow Hg+Hg^{2 +}$ .
The following data has been given: $Hg_{2}^{2 +}+2e \rightarrow 2Hg,E^{0}=0.789V$ $\&Hg^{2 +}+2e \rightarrow Hg,E^{0}=0.854V$

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

Answer 1

Answer: (C)

$Hg_{2}^{2 +}+2e^{-} \rightarrow 2Hg,0.789$ Volt
$Hg \rightarrow Hg^{2 +}+2e^{-},-0.854$ Volt
$Hg_{2}^{\star} \rightarrow Hg+Hg^{2 +},-0.065$ Volt
$\Delta G=-2\times \left(\right.-0.065\left.\right)\times 96500=-8.314\times 298lnK_{e q}$
$K_{e q .}=6.3\times 10^{- 3}$