38.6 A of current is passed for 100 s throu | vedclass

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(C) The reduction potential of $Cu^{2+}$ is higher than $H^+$,hence $Cu^{2+}$ is reduced at the cathode.Reaction at the cathode: $Cu^{2+} + 2e^- \righ

Vedclass · Accepted Answer

$1.27 \ g, 0.224 \ L$

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(C) The reduction potential of $Cu^{2+}$ is higher than $H^+$,hence $Cu^{2+}$ is reduced at the cathode.Reaction at the cathode: $Cu^{2+} + 2e^- ightarrow Cu(s)$Reaction at the anode: $2H_2O(l) ightarrow O_2(g) + 4H^+(aq) + 4e^-$Total charge passed: $Q = I 	imes t = 38.6 \ A 	imes 100 \ s = 3860 \ C$.Mass of $Cu$ deposited: $W = \frac{M 	imes Q}{n 	imes F} = \frac{63.54 	imes 3860}{2 	imes 96500} = 1.27 \ g$.For $O_2$ gas at the anode,$4 \ mol$ of electrons produce $1 \ mol$ ($22.4 \ L$ at $STP$) of $O_2$.$n_{e^-} = \frac{3860}{96500} = 0.04 \ mol$.$n_{O_2} = \frac{0.04}{4} = 0.01 \ mol$.Volume of $O_2 = 0.01 	imes 22.4 \ L = 0.224 \ L$.

$38.6 \ A$ of current is passed for $100 \ s$ through an aqueous $CuSO_4$ solution using platinum electrodes. The mass of copper deposited from the solution and the volume of gas liberated at $STP$ are respectively (molar mass of $Cu = 63.54 \ g \ mol^{-1}$)

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