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(B) Using Faraday's law of electrolysis: $W = \frac{M 	imes I 	imes t}{n 	imes F}$Where:$W = 22.2 \ g$ (mass deposited)$M = 119 \ g/mol$ (atomic ma

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$2$

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(B) Using Faraday's law of electrolysis: $W = \frac{M 	imes I 	imes t}{n 	imes F}$Where:$W = 22.2 \ g$ (mass deposited)$M = 119 \ g/mol$ (atomic mass of $Sn$)$I = 2 \ A$ (current)$t = 5 \ hours = 5 	imes 3600 \ s = 18000 \ s$ (time)$F = 96500 \ C/mol$ (Faraday's constant)$n$ = oxidation state (number of electrons transferred)Rearranging for $n$:$n = \frac{M 	imes I 	imes t}{W 	imes F}$$n = \frac{119 	imes 2 	imes 18000}{22.2 	imes 96500}$$n = \frac{4284000}{2142300} \approx 2.00$Therefore,the oxidation state of tin in the salt is $2$.

$A$ current of $2 \ A$ passing for $5 \ hours$ through molten tin salt deposited $22.2 \ g$ of tin. The oxidation state of tin in the salt is . . . . . . . (Atomic mass of $Sn = 119$)

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