The logarithm of equilibrium constant for the reaction $\mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_{4}^{2-}$ is ___________ (Nearest integer)

Given : $\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V}$

$$ \mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V} $$

$$ \begin{aligned} & \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned} $$

Sol. $\Delta G^{\circ}=-R T \ell n K$ <br/><br/>$$ \begin{aligned} & -\mathrm{nFE}_{\text {cell }}^{\mathrm{o}}=-\mathrm{RT} \times 2.303\left(\log _{10} \mathrm{~K}\right) \\\\ & \frac{\mathrm{E}_{\text {Cell }}^{\mathrm{o}}}{0.06} \times \mathrm{n}=\log \mathrm{K} .......(1) \\\\ & \mathrm{Pd}^{+2} \text { (aq.) }+ \mathrm{2e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}), \mathrm{E}_{\text {cat, }{ reduction}}^{\mathrm{o}}=0.83 \\\\ & \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-} \text {(aq.) } \rightleftharpoons \mathrm{PdCl}_4^{2-}, \text { (aq) }+2 \mathrm{e}^{-}, \mathrm{E}_{\text {Anode, Oxidation }}^0=0.65 \end{aligned} $$ <br/><br/>Net Reaction $\rightarrow \mathrm{Pd}^{2+}$ (aq.) $+4 \mathrm{Cl}^{-}$(aq.) $\rightleftharpoons \mathrm{PdCl}_4^{2-}$ (aq.) <br/><br/>$$ \begin{aligned} & \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cat,red }}^{\mathrm{o}}-\mathrm{E}_{\text {Anded, oxidid }}^0 \\\\ & \mathrm{E}_{\text {cell }}^{\mathrm{o}}=0.83-0.65 \\\\ & \mathrm{E}_{\text {cell }}^0=0.18 .........(2) \end{aligned} $$ <br/><br/>Also $\mathrm{n}=2$ .......(3) <br/><br/>Using equation (1), (2) and (3) <br/><br/>$\log K=6$