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(C) According to Faraday's law of electrolysis,the mass of metal deposited is given by $W = \frac{Z 	imes I 	imes t}{n 	imes F}$,where $Z$ is the a

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$+3$

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(C) According to Faraday's law of electrolysis,the mass of metal deposited is given by $W = \frac{Z 	imes I 	imes t}{n 	imes F}$,where $Z$ is the atomic weight,$I$ is current,$t$ is time,$n$ is the oxidation state (valency),and $F$ is Faraday's constant $(96500 \ C/mol)$.Given: $W = 22.2 \ g$,$I = 2.0 \ A$,$t = 5 \ hours = 5 	imes 3600 \ s = 18000 \ s$,$At. \ wt. = 177$.Using the formula $W = \frac{At. \ wt. 	imes I 	imes t}{n 	imes 96500}$,we rearrange to find $n$:$n = \frac{At. \ wt. 	imes I 	imes t}{W 	imes 96500}$$n = \frac{177 	imes 2.0 	imes 18000}{22.2 	imes 96500}$$n = \frac{6372000}{2142300} \approx 2.97 \approx 3$.Therefore,the oxidation state of the metal is $+3$.

$A$ current of $2.0 \ A$ is passed for $5 \ hours$ through a molten metal salt,which deposits $22.2 \ g$ of metal (At. wt. $= 177$). The oxidation state of the metal in the metal salt is:

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