The logarithm of equilibrium constant ( log_10 K ) for the reaction Pd^2+_( aq ) +4 Cl^-_( aq ) ⇌ PdCl_4^2-_( aq ) is x × 10^-1. The value of x is \…

The logarithm of equilibrium constant $\displaystyle \left(\log _{10} \mathrm{~K}\right)$ for the reaction $$\mathrm{Pd}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{Cl}^{-}{ }_{(\mathrm{aq})} \rightleftharpoons \mathrm{PdCl}_4^{2-}{ }_{(\mathrm{aq})} $$ is $\displaystyle x \times 10^{-1}$. The value of $\displaystyle x$ is \_\_\_\_. (Nearest integer) [Given emf for the half-cell at $\displaystyle 25$°C $$\begin{aligned} & \mathrm{Pd}^{2+}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}_{(\mathrm{s})} \quad \mathrm{E}^{\circ}=0.83 \mathrm{~V} \\ & \left.\mathrm{PdCl}_4{ }^{2-}{ }_{(\mathrm{aq})}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}(\mathrm{~s})+4 \mathrm{Cl}^{-} \quad \mathrm{E}^{\circ}=0.64 \mathrm{~V}\right] \end{aligned} $$