The logarithm of equilibrium constant for the reaction Pd^2++4 Cl^- ⇌ PdCl_4^2- is ____ (Nearest integer) Given: (2.303 RT)/F=0.06 V Pd_( aq )^2++2…

The logarithm of equilibrium constant for the reaction $\displaystyle \mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_4^{2-}$ is $\displaystyle \_\_\_\_$ (Nearest integer) Given : $\displaystyle \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$ $$\begin{aligned} & \mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{~s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V} \\ & \mathrm{PdCl}_4^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{~s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \quad \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned} $$