For the cell: $\mathrm{Zn}(\mathrm{s})\left|\mathrm{\: Zn}^{2+}(a \mathrm{\: M}) \| \mathrm{Ag}^{+}(b \mathrm{\: M})\right| \mathrm{Ag}(\mathrm{s})$ Write Nernst equation to show how Ecell vary with concentration of Zn2+ and Ag+ ions. Given E- (Zn2+| Zn) = -0.76 V,  E-(Ag+| Ag) = 0.80 V.Option: 1 Option: 2 Option: 3 Option: 4 None of above

Anode reaction:
$\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Zn}^{2+}(a \mathrm{\: M})+2\mathrm e^{-}$
Cathode reaction:
$2 \mathrm{\: Ag}^{+}(b \mathrm{\: M})+2 \mathrm e^{-} \longrightarrow 2 \mathrm{Ag}(\mathrm{s})$
Cell reaction:
$\mathrm{Z n(s)}+2 \mathrm{A g^{+}(b \: M)} \longrightarrow$ $\mathrm{Zn}^{2+}(\mathrm{a\: M})+2\: \mathrm{Ag}(\mathrm{s})$
$\\E^{-}_{\text {cell }}=\left(E^{-} \text { reduction }\right)_{c}-\left(E^{-} \text { reduction }\right)_{\mathbf{a}}\\\\E^{-}_{\text {cell }} =0.80-(-0.76)=1.56 \mathrm{\: V}$
Using the Nernst equation,
$\mathrm E_{\text {cell }}=\mathrm E^-_{\text {eell }}-\frac{0.059}{\mathrm n_{\text {cell }}} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$
$\mathrm E_{\text {cell }}=\mathrm E^{-}_{\text {cell }}-\frac{0.059}{2} \log \left[\mathrm{Zn}^{2+}\right]+\frac{0.059}{2} \log \left[\mathrm{Ag}^{+}\right]^{2}$

$\begin{array}{l}{\text { Therefore, } \mathrm E_{\text {cell}} \textup{ wil decrease when }\left[\mathrm{Zn}^{2+}\right] \text { increases }} \\ {\left[\mathrm{Ag}^{+}\right] \text { decreases. }}\end{array}$

Therefore, option(1) is correct