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(D) At the cathode: $Ag^{+} + e^{-} 	o Ag$ (Reduction)At the anode: $2OH^{-} 	o H_2O + \frac{1}{2} O_2 + 2e^{-}$ (Oxidation)Equivalent weight of $Ag

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$21.60$

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(D) At the cathode: $Ag^{+} + e^{-} 	o Ag$ (Reduction)At the anode: $2OH^{-} 	o H_2O + \frac{1}{2} O_2 + 2e^{-}$ (Oxidation)Equivalent weight of $Ag$ $(E_{Ag})$ = $\frac{108}{1} = 108$.Equivalent weight of $O_2$ $(E_{O_2})$ = $\frac{	ext{Molar mass}}{	ext{n-factor}} = \frac{32}{4} = 8$.According to Faraday's law of electrolysis: $\frac{W_{Ag}}{E_{Ag}} = \frac{W_{O_2}}{E_{O_2}}$.Substituting the values: $\frac{W_{Ag}}{108} = \frac{1.6}{8}$.$W_{Ag} = \frac{1.6 	imes 108}{8} = 0.2 	imes 108 = 21.6 \ g$.

An electrolytic cell contains a solution of $Ag_2SO_4$ and has platinum electrodes. $A$ current is passed until $1.6 \ g$ of $O_2$ has been liberated at the anode. The amount of silver deposited at the cathode would be ............. $g$.

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