The logarithm of equilibrium constant for the reaction $$Pd^{2+} + 4Cl^- \rightleftharpoons PdCl_4^{2-}$$ is (Nearest integer)
Given: $$\dfrac{2.303RT}{F} = 0.06$$ V
$$Pd^{2+}_{(aq)} + 2e^- \rightleftharpoons Pd(s)$$ $$E^o = 0.83$$ V
$$PdCl_4^{2-}(aq) + 2e^- \rightleftharpoons Pd(s) + 4Cl^-(aq)$$ $$
$$E^o = 0.65$$ V

We need to find $$\log K$$ for the reaction $$Pd^{2+} + 4Cl^- \rightleftharpoons PdCl_4^{2-}$$.

To begin,

(1) $$Pd^{2+}(aq) + 2e^- \rightleftharpoons Pd(s)$$, $$E^\circ_1 = 0.83$$ V

(2) $$PdCl_4^{2-}(aq) + 2e^- \rightleftharpoons Pd(s) + 4Cl^-(aq)$$, $$E^\circ_2 = 0.65$$ V

Next,

Subtracting reaction (2) from reaction (1): we reverse reaction (2) and add to reaction (1).

Reversed (2): $$Pd(s) + 4Cl^- \rightarrow PdCl_4^{2-} + 2e^-$$, $$E^\circ = -0.65$$ V

Adding (1) + reversed (2):

$$ Pd^{2+} + 2e^- + Pd(s) + 4Cl^- \rightarrow Pd(s) + PdCl_4^{2-} + 2e^- $$

Cancelling $$Pd(s)$$ and $$2e^-$$ from both sides:

$$ Pd^{2+} + 4Cl^- \rightarrow PdCl_4^{2-} $$

$$ E^\circ_{\text{cell}} = E^\circ_1 - E^\circ_2 = 0.83 - 0.65 = 0.18\;\text{V} $$

From this,

At equilibrium, the Nernst equation gives:

$$ E^\circ_{\text{cell}} = \frac{2.303RT}{nF} \log K $$

We are given $$\frac{2.303RT}{F} = 0.06$$ V, and $$n = 2$$ electrons are transferred.

$$ 0.18 = \frac{0.06}{2} \times \log K $$

Wait -- let us be careful. The correct relation is:

$$ E^\circ = \frac{2.303RT}{nF} \log K = \frac{0.06}{n} \log K $$

$$ \log K = \frac{n \times E^\circ}{0.06} = \frac{2 \times 0.18}{0.06} = \frac{0.36}{0.06} = 6 $$

The logarithm of the equilibrium constant is 6.