The logarithm of equilibrium constant for the reaction Pd^2++4 Cl^- leftharpoons PdCl_4^2- is \underline\hspace2cm (Nearest integer) Given : \frac2.303 R ~T~F=0.06 ~V Pd_(aq)^2++2 e^- leftharpoons Pd(s) \quad E^\ominus=0.83 ~V \beginaligned & PdCl_4^2-(aq)+2 e^- leftharpoons Pd(s)+4 Cl^-(aq) E^\ominus=0.65 ~V \endaligned

Electrochemistry PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q17 with Solution

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Correct answer: 6

\mathrm{Pd}^{2+}+2e^- \rightleftharpoons \mathrm{Pd}(s), \qquad E_1^\circ=0.83\text{ V}

\mathrm{PdCl}_4^{2-}+2e^- \rightleftharpoons \mathrm{Pd}(s)+4\mathrm{Cl}^-, \qquad E_2^\circ=0.65\text{ V}

We need the equilibrium constant for:

\mathrm{Pd}^{2+}+4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}

Reverse the first half-reaction’s reverse? More directly:

From

\mathrm{Pd}^{2+}+2e^- \to \mathrm{Pd}(s)

and reverse the second half-reaction:

\mathrm{Pd}(s)+4\mathrm{Cl}^- \to \mathrm{PdCl}_4^{2-}+2e^-

Adding them, electrons and $\mathrm{Pd}(s)$ cancel:

\mathrm{Pd}^{2+}+4\mathrm{Cl}^- \to \mathrm{PdCl}_4^{2-}

which is exactly the required reaction.

For the overall reaction,

reduction: $\mathrm{Pd}^{2+}+2e^- \to \mathrm{Pd}(s), \ E^\circ=0.83\text{ V}$
oxidation: reverse of second half-reaction, so its oxidation potential is $-0.65\text{ V}$ in reduction-potential convention.

Thus,

E^\circ_{\text{cell}}=0.83-0.65=0.18\text{ V}

For a reaction involving $n=2$ electrons,

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

Given:

\frac{2.303RT}{F}=0.06\text{ V}

So,

E^\circ = \frac{0.06}{2}\log K = 0.03\log K

Now substitute $E^\circ=0.18$ V:

0.18 = 0.03\log K

\log K = \frac{0.18}{0.03}=6

\boxed{6}