In the electrolysis of a CuSO_4 solution,how | vedclass

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(A) The $	ext{moles of } O_2 = \frac{	ext{Given volume}}{	ext{Molar volume}} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \ mol$. For the anode reaction:

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$31.75$

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(A) The $	ext{moles of } O_2 = \frac{	ext{Given volume}}{	ext{Molar volume}} = \frac{5.6 \ L}{22.4 \ L/mol} = 0.25 \ mol$. For the anode reaction: $2H_2O ightarrow O_2 + 4H^+ + 4e^-$,the $n$-factor for $O_2$ is $4$. $	ext{Equivalents of } O_2 = 	ext{moles} 	imes n	ext{-factor} = 0.25 	imes 4 = 1$. According to Faraday's law of electrolysis,$	ext{Equivalents of } O_2 = 	ext{Equivalents of } Cu$. For the cathode reaction: $Cu^{2+} + 2e^- ightarrow Cu$,the $n$-factor for $Cu$ is $2$. $	ext{Equivalents of } Cu = \frac{	ext{mass of } Cu}{	ext{Equivalent mass of } Cu} = \frac{	ext{mass of } Cu}{63.5 / 2}$. Equating the equivalents: $1 = \frac{	ext{mass of } Cu}{31.75}$. $	ext{mass of } Cu = 31.75 \ g$.

In the electrolysis of a $CuSO_4$ solution,how many grams of $Cu$ are plated out on the cathode,in the time that is required to liberate $5.6 \ L$ of $O_{2(g)}$,measured at $1 \ atm$ and $273 \ K$,at the anode (in $g$)?

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