The logarithm of equilibrium constant for the reaction $\text{Pd}^{2+} + 4\text{Cl}^- \rightleftharpoons \text{PdCl}_4^{2-}$ is $\underline{\hspace{1cm}}$ (Nearest integer).
Given: $\frac{2.303\text{RT}}{\text{F}} = 0.06\text{V}$
$\text{Pd}^{2+}_{(\text{aq})} + 2\text{e}^- \rightleftharpoons \text{Pd(s)} \, \text{E}^\theta = 0.83\text{V}$
$\text{PdCl}_4^{2-}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Pd(s)} + 4\text{Cl}^-(\text{aq}) \, \text{E}^\theta = 0.65\text{V}$