A and B are two metals. The standard reductio | vedclass

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(B) The cell reaction is: $A_{(s)} + B^{+}_{(aq)} \rightleftharpoons A^{+}_{(aq)} + B_{(s)}$Here,$A$ is oxidized to $A^{+}$ $(n=1)$ and $B^{+}$ is red

Vedclass · Accepted Answer

$\frac{100}{6}$

Vedclass · Answer

(B) The cell reaction is: $A_{(s)} + B^{+}_{(aq)} \rightleftharpoons A^{+}_{(aq)} + B_{(s)}$Here,$A$ is oxidized to $A^{+}$ $(n=1)$ and $B^{+}$ is reduced to $B$.$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{B^{+}/B} - E^{\circ}_{A^{+}/A} = 0.5 - (-0.5) = 1.0 \ V$Using the relation: $E^{\circ}_{cell} = \frac{0.06}{n} \log K_C$Substituting the values: $1.0 = \frac{0.06}{1} \log K_C$Therefore,$\log K_C = \frac{1.0}{0.06} = \frac{100}{6}$

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