Redox Reactions and Electrochemistry: JEE Main 2023 Chemistry question

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