The equilibrium constant of the reaction A_{( | vedclass

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(D) The given reaction is $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$.Here,the number of electrons transferred,$n = 2$.Give

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$10$

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(D) The given reaction is $A_{(s)} + 2B^{2+}_{(aq)} ightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$.Here,the number of electrons transferred,$n = 2$.Given $E^{\circ}_{	ext{cell}} = 0.0295 \ V$.The relationship between equilibrium constant $K_c$ and $E^{\circ}_{	ext{cell}}$ is given by the Nernst equation at equilibrium:$E^{\circ}_{	ext{cell}} = \frac{0.059}{n} \log K_c$.Substituting the values:$0.0295 = \frac{0.059}{2} \log K_c$.$0.0295 = 0.0295 \log K_c$.$\log K_c = 1$.$K_c = 10^1 = 10$.

The equilibrium constant of the reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ with $E^{\circ}_{\text{cell}} = 0.0295 \ V$ is (Given: $\frac{2.303RT}{F} = 0.059$)

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