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

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Redox Reactions and Electrochemistry

The logarithm of equilibrium constant for the reaction

Pd^{2 +} + 4 Cl^{-} ⇌ PdCl_{4}^{2 -}

is ______ (Nearest integer)

Given:

\frac{2.303 R T}{F} = 0.06 V

Pd_{(aq)}^{2 +} + 2 e^{-} ⇌ Pd (s)

E^{o} = 0.83 V

PdCl_{4}^{2 -} (aq) + 2 e^{-} ⇌ Pd (s) + 4 Cl^{-} (aq)

E^{\circ} = 0.65 V

Answer: 6

Treating the reaction as a cell, reduction of

Pd X^{2 +}

is the cathode (

E^{\circ} = 0.83 V

) and oxidation of palladium to

PdCl X_{4} X^{2 -}

is the anode (

E^{\circ} = 0.65 V

), so

E_{cell}^{\circ} = 0.83 - 0.65 = 0.18 V

with

n = 2

. Since

lo g K = \frac{n E _{cell}^{\circ}}{0.06} = \frac{2 \times 0.18}{0.06}

, the logarithm of the equilibrium constant is

6

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer