The logarithm of equilibrium constant for the reaction Pd 2 + + 4 Cl - ⇌ PdCl 4 2 -    is…

Given, $\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{V}$

Using gibbs free energy equation,

$\mathrm{\Delta G}°=-\mathrm{RT} \mathrm{lnK}$

$-{\mathrm{nFE}}_{\text{cell }}^{\mathrm{o}}=-\mathrm{RT}\times 2.303\left({\log }_{10} \mathrm{K}\right)$

$\frac{{\mathrm{E}}_{\text{Cell}}^0}{0.06}\times \mathrm{n}=\log \mathrm{K}---------\left(\mathrm{i}\right)$

${\mathrm{Pd}}^{2+} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pd}\left(\mathrm{s}\right), {\mathrm{E}}_{\text{cathode, reduction }}^0 = 0.83 \mathrm{V}$

$\mathrm{Pd} \left(\mathrm{s}\right)+4{\mathrm{Cl}}^{-} (\mathrm{aq}.) \rightleftharpoons {\mathrm{PdCl}}_4^{2-} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-}, {\mathrm{E}}_{\text{anode, oxidation}}^{\mathrm{o}} = 0.65\mathrm{V}$

Net Reaction $\to {\mathrm{Pd}}^{2+} (\mathrm{aq}.) + 4{\mathrm{Cl}}^{-} (\mathrm{aq}.) \rightleftharpoons {{\mathrm{PdCl}}_4}^{2-} (\mathrm{aq}.)$

${\mathrm{E}}_{\text{cell }}^{\mathrm{o}} = {\mathrm{E}}_{\text{cathode,reduction}}^{\mathrm{o}} - {\mathrm{E}}_{\text{anode, oxidation }}^{\mathrm{o}}$

$$\begin{gathered} \Rightarrow {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}} = 0.83 - 0.65 \\ \Rightarrow {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}} = 0.18--------\left(\mathrm{ii}\right) \end{gathered}$$

Also $\mathrm{n}=2--------\left(\mathrm{iii}\right)$

Using equation $\left(\mathrm{i}\right), \left(\mathrm{ii}\right)$& $\left(\mathrm{iii}\right)$, we get

$\frac{{\mathrm{E}}_{\text{Cell}}^0}{0.06}\times \mathrm{n}=\log \mathrm{K}$

$\Rightarrow \frac{0.18}{0.06}\times 2 = \log \mathrm{K}$

$\Rightarrow \mathrm{logK} = 6$