Under what condition is $E_{\text {cell }}=0$ or $\Delta_r G=0$ ?
At the stage of chemical equilibrium in the cell.
$$\begin{aligned} E_{\text {cell }} & =0 \\ \Delta_r G^{\circ} & =-n F E_{\text {cell }}^{\circ} \\ & =-n \times F \times 0=0 \end{aligned}$$
What does the negative sign in the expression $E^{\mathrm{s}} \mathrm{Zn}^{2+} / \mathrm{Zn}=-0.76 \mathrm{~V}$ mean?
Greater the negative reactivity of standard reduction potential of metal greater is its reactivity. It means that Zn is more reactive than hydrogen. When zinc electrode will be connected to standard hydrogen electrode, Zn will get oxidised and $\mathrm{H}^{+}$will get reduced.
Thus, zinc electrode will be the anode of the cell and hydrogen electrode will be the cathode of the cell.
Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.
Different masses of Cu and Ag will be deposited at cathode. According to Faraday's second law of electrolysis amount of different substances liberated by same quantity of electricity passes through electrolyte solution is directly proportional to their chemical equivalent weight.
$$\frac{W_1}{W_2}=\frac{E_1}{E_2}$$
where, $E_1$ and $E_2$ have different values depending upon number of electrons required to reduce the metal ion. Thus, masses of Cu and Ag deposited will be different.
Depict the galvanic cell in which the cell reaction is
$$\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow 2 \mathrm{Ag}+\mathrm{Cu}^{2+}$$
In a galvanic cell, oxidation half reaction is written on left hand side and reduction half reaction is on right hand side. Salt bridge is represented by parallel lines $\mathrm{Cu}\left|\mathrm{Cu}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$.
Value of standard electrode potential for the oxidation of $\mathrm{Cl}^{-}$ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is $\mathrm{Cl}^{-}$oxidised at anode instead of water?
Under the condition of electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential. $\mathrm{So}, \mathrm{Cl}^{-}$is oxidized at anode instead of water.
Possible oxidation half cell reactions occurring at anode are
$$\begin{array}{ll} \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \frac{1}{2} \mathrm{Cl}_2(g)+\mathrm{e}^{-} ; & E_{\mathrm{cell}}^{\mathrm{s}}=1.36 \mathrm{~V} \\ 2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{O}_2(g)+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} ; & E_{\mathrm{cell}}^{\circ}=1.23 \mathrm{~V} \end{array}$$
Species having lower $E_{\text {cell }}^{\circ}$ cell undergo oxidation first than the higher value but oxidation of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ is kinetically so slow that it needs some overvoltage.