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.