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(D) At the cathode,the reduction reaction is: $Ag^{+} + e^{-} \rightarrow Ag$. The equivalent weight of $Ag$ is $E_{Ag} = \frac{108}{1} = 108 \ g/eq$.

Vedclass · Accepted Answer

$21.6$

Vedclass · Answer

(D) At the cathode,the reduction reaction is: $Ag^{+} + e^{-} ightarrow Ag$. The equivalent weight of $Ag$ is $E_{Ag} = \frac{108}{1} = 108 \ g/eq$.At the anode,the oxidation reaction is: $2H_{2}O ightarrow O_{2} + 4H^{+} + 4e^{-}$. The equivalent weight of $O_{2}$ is $E_{O_{2}} = \frac{32}{4} = 8 \ g/eq$.According to Faraday's law of electrolysis,the number of equivalents of substances deposited or liberated at the electrodes are equal:$\frac{W_{Ag}}{E_{Ag}} = \frac{W_{O_{2}}}{E_{O_{2}}}$Substituting the given 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 $Ag_{2}SO_{4}$ and has platinum electrodes. $A$ current is passed until $1.6 \ g$ of $O_{2}$ is liberated at the anode. The amount of silver deposited at the cathode would be ............ $g$.

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