If 1 A of current is passed through CuSO_4 s | vedclass

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(B) The total charge $Q$ passed is given by $Q = I 	imes t = 1 \ A 	imes 10 \ s = 10 \ C$. The deposition of $Cu^{2+}$ ions at the cathode follows t

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$3.1 	imes 10^{19}$

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(B) The total charge $Q$ passed is given by $Q = I 	imes t = 1 \ A 	imes 10 \ s = 10 \ C$. The deposition of $Cu^{2+}$ ions at the cathode follows the reaction: $Cu^{2+} + 2e^- ightarrow Cu$. This shows that $2$ moles of electrons are required to deposit $1$ mole of $Cu$. The charge required to deposit $1$ ion of $Cu^{2+}$ is $2 	imes e^- = 2 	imes 1.6 	imes 10^{-19} \ C = 3.2 	imes 10^{-19} \ C$. The number of $Cu^{2+}$ ions deposited by $10 \ C$ charge is $\frac{10 \ C}{3.2 	imes 10^{-19} \ C/ion} \approx 3.125 	imes 10^{19} \ ions$. Thus,the number of copper ions is about $3.1 	imes 10^{19}$.

If $1 \ A$ of current is passed through $CuSO_4$ solution for $10 \ s$,then the number of copper ions deposited at the cathode will be about

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