On electrolysis of aqueous CuSO_4 solution us | vedclass

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(C) According to Faraday's law of electrolysis,the mass of substance deposited or dissolved is proportional to the charge passed: $W = Z 	imes Q = (E

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$2.5 \ g$ of $Cu$ will be dissolved

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(C) According to Faraday's law of electrolysis,the mass of substance deposited or dissolved is proportional to the charge passed: $W = Z 	imes Q = (E / 96500) 	imes Q$. In the electrolysis of $CuSO_4$ using $Cu$ electrodes,the cathode reaction is $Cu^{2+} (aq) + 2e^- ightarrow Cu (s)$ and the anode reaction is $Cu (s) ightarrow Cu^{2+} (aq) + 2e^-$. Since the same amount of charge passes through both electrodes,the mass of $Cu$ deposited at the cathode is equal to the mass of $Cu$ dissolved at the anode. Therefore,if $2.5 \ g$ of $Cu$ is deposited at the cathode,$2.5 \ g$ of $Cu$ will be dissolved from the anode.

On electrolysis of aqueous $CuSO_4$ solution using $Cu$ electrodes,if $2.5 \ g$ of $Cu$ is deposited at the cathode,then at the anode:

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