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(C) The reactions occurring are:$Ag^{+} + e^{-} \rightarrow Ag$$2H_{2}O \rightarrow O_{2} + 4H^{+} + 4e^{-}$From Faraday's laws,the number of equivale

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$5.6$

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(C) The reactions occurring are:$Ag^{+} + e^{-} ightarrow Ag$$2H_{2}O ightarrow O_{2} + 4H^{+} + 4e^{-}$From Faraday's laws,the number of equivalents of $Ag$ deposited equals the number of equivalents of $O_{2}$ evolved.Equivalents of $Ag = \frac{	ext{mass}}{	ext{equivalent mass}} = \frac{0.108}{108} = 0.001 \ 	ext{eq}$.Equivalents of $O_{2} = 0.001 \ 	ext{eq}$.Since the n-factor for $O_{2}$ in water electrolysis is $4$,moles of $O_{2} = \frac{0.001}{4} = 0.00025 \ 	ext{mol}$.Volume of $O_{2}$ at $STP = 0.00025 	imes 22400 \ 	ext{cm}^{3} = 5.6 \ 	ext{cm}^{3}$.

An electric current is passed through a silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed $0.108 \ g$ more at the end of the electrolysis. The volume of oxygen evolved at $STP$ is ............... $cm^{3}$.

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