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If 5 ampers of current is passed for 193 seconds through a solution containing Copper salt, $0.32 \mathrm{~g}$ of copper is deposited. What is the oxidation state of the $\mathrm{Cu}$ in the salt?
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$+2$
Charges passing through solution $=5 \times 193=965 \mathrm{C}$
$\mathrm{Cu}^{\mathrm{n}+}+\mathrm{ne} \rightarrow \mathrm{Cu}$
$965 \mathrm{C}$ charge deposit $0.32 \mathrm{~g}$
$\begin{aligned} & \therefore \quad 96500 \mathrm{n} \text { C charge deposit }=\frac{0.32 \times 96500 \mathrm{n}}{965}=63.5 \\ & \therefore \mathrm{n} \simeq \frac{64}{32}=2\end{aligned}$
$\mathrm{Cu}^{\mathrm{n}+}+\mathrm{ne} \rightarrow \mathrm{Cu}$
$965 \mathrm{C}$ charge deposit $0.32 \mathrm{~g}$
$\begin{aligned} & \therefore \quad 96500 \mathrm{n} \text { C charge deposit }=\frac{0.32 \times 96500 \mathrm{n}}{965}=63.5 \\ & \therefore \mathrm{n} \simeq \frac{64}{32}=2\end{aligned}$
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