The logarithm of the equilibrium constant for | vedclass

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Question

(D) The cell reaction is the sum of the reduction of $Pd^{2+}$ and the oxidation of $Pd_{(s)}$ to $PdCl_4^{2-}$.
$Pd^{2+}{(aq)} + 2e^{-} \rightleftha

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The cell reaction is the sum of the reduction of $Pd^{2+}$ and the oxidation of $Pd_{(s)}$ to $PdCl_4^{2-}$.
$Pd^{2+}{(aq)} + 2e^{-} ightleftharpoons Pd_{(s)} \quad E^{\circ}_{red} = 0.83 \ V$
$Pd_{(s)} + 4Cl^{-}{(aq)} ightleftharpoons PdCl_4^{2-}{(aq)} + 2e^{-} \quad E^{\circ}_{ox} = -0.65 \ V$
Adding these,the net cell reaction is $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} ightleftharpoons PdCl_4^{2-}{(aq)}$ with $E^{\circ}_{cell} = 0.83 - 0.65 = 0.18 \ V$.
The relationship between equilibrium constant $K$ and standard cell potential is $\log K = \frac{n E^{\circ}_{cell}}{0.0591} \approx \frac{n E^{\circ}_{cell}}{0.06}$.
Here,$n = 2$ (number of electrons transferred).
$\log K = \frac{2 	imes 0.18}{0.06} = \frac{0.36}{0.06} = 6$.
Thus,the logarithm of the equilibrium constant is $6$.

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