For a $\mathrm{Ag}-\mathrm{Zn}$ button cell, net reaction is

$\begin{gathered} \mathrm{Zn}(s)+\mathrm{Ag}_2 \mathrm{O}(s) \longrightarrow \mathrm{ZnO}(s)+2 \mathrm{Ag}(s) \\ \Delta G_f^{\circ}\left(\mathrm{Ag}_2 \mathrm{O}\right)=-11.21 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta G_f^{\circ}(\mathrm{ZnO})=-318.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{gathered}$

Then, $E^{\circ}$ cell of the button cell is

At $25^{\circ} \mathrm{C}$, the molar conductance at infinite dilution for the strong electrolytes $\mathrm{NaOH}, \mathrm{NaCl}$ and $\mathrm{BaCl}_2$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \mathrm{~Sm}^2 \mathrm{~mol}^{-1}$ respectively. $\lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}(\mathrm{OH})_2$ in $\mathrm{Sm}^2 \mathrm{~mol}^{-1}$ is

Assertion (A) For a Daniell cell $\mathrm{Zn} / \mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+} \mid \mathrm{Cu}$ with $E_{\text {cell }}=1.1 \mathrm{~V}$, the application of opposite potential greater than $1.1 \mathrm{~V}$ results into the flow of electrons from cathode to anode.

Reason (R) Zn is deposited at zinc electrode and $\mathrm{Cu}$ is dissolved at copper electrode.