A current of 10.0 , A flows for 2.00 , h thro | vedclass

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(C) The total charge $Q$ passed through the cell is given by $Q = i 	imes t$.Given $i = 10.0 \, A$ and $t = 2.00 \, h = 2.00 	imes 3600 \, s = 7200

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$3$

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(C) The total charge $Q$ passed through the cell is given by $Q = i 	imes t$.Given $i = 10.0 \, A$ and $t = 2.00 \, h = 2.00 	imes 3600 \, s = 7200 \, s$.$Q = 10.0 \, A 	imes 7200 \, s = 72,000 \, C$.According to Faraday's law,the number of moles of electrons required to deposit $n$ moles of metal $X$ with oxidation state $x$ is $n 	imes x = \frac{Q}{F}$.Here,$n = 0.250 \, mol$ and $F = 96,500 \, C \, mol^{-1}$.$0.250 	imes x = \frac{72,000}{96,500}$.$0.250 	imes x \approx 0.746$.$x = \frac{0.746}{0.250} \approx 2.98 \approx 3$.Therefore,the oxidation state of metal $X$ is $3+$.

$A$ current of $10.0 \, A$ flows for $2.00 \, h$ through an electrolytic cell containing a molten salt of metal $X$. This results in the decomposition of $0.250 \, mol$ of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is $(F = 96,500 \, C \, mol^{-1})$ (in $+$)

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