The logarithm of equilibrium constant for the reaction Pd 2 + + 4 Cl - ? PdCl 4 2 - is _________ (Nearest Integer). [2023]

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Solution

Q.
The logarithm of equilibrium constant for the reaction ${Pd}^{2 +} + 4 {Cl}^{-} ⇌ {PdCl}_{4}^{2 -}$ is _________ (Nearest Integer).

Given: $\frac{2.303 R T}{F} = 0.06 V$

${Pd}_{(a q)}^{2 +} + 2 e^{-} ⇌ {Pd}_{(s)} E^{θ} = 0.83 V$

${PdCl}_{4}^{2 -} (a q) + 2 e^{-} ⇌ Pd (s) + 4 {Cl}^{-} (a q) E^{θ} = 0.65 V$ [2023]

Ans.
(6)

$Given reaction: {Pd}^{2 +} + 4 {Cl}^{-} ⇌ {[{PdCl}_{4}]}^{2 -}$

$E_{cell}^{\circ} = E_{{Pd}^{2 +} | Pd}^{\circ} - E_{{({PdCl}_{4})}^{2 -} | Pd}^{\circ} = 0.83 - 0.65 = 0.18$

$E_{cell}^{\circ} = \frac{0.06}{2} \log K_{eq}$

$0.18 = \frac{0.06}{2} \log K_{eq}$

$\log K_{eq} = 6$

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