If the $E_{\text{cell}}^{\circ}$ of the given hypothetical cell reaction is 0.046 V, the value of $\log K_{c}$ for this reaction at 298 K would be:

$ \text{A(s)}+2\text{B}^{+}(aq)\rightarrow\text{A}^{2+}(aq)+2\text{B(s)} $

1.556

The correct answer is Option (1) → 1.556

Relation between standard cell potential and equilibrium constant:

$E^\circ_{\text{cell}} = \left( \frac{0.0591}{n} \right) \log K_c$

Stepwise Calculation

Given:

$E^\circ_{\text{cell}} = 0.046 \text{ V}$

From reaction:

$A \rightarrow A^{2+} + 2e^-$

Number of electrons transferred ($n$):

$n = 2$

Substitute in formula:

$0.046 = \left( \frac{0.0591}{2} \right) \log K_c$

$\log K_c = \frac{0.046 \times 2}{0.0591}$

$\log K_c = \frac{0.092}{0.0591}$

$\log K_c = 1.556$