When a current of 10800 C is passed through | vedclass

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Question

(A) According to Faraday's law of electrolysis,the mass deposited $w$ is given by $w = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the molar mass,$Q$

Vedclass · Accepted Answer

$+4$

Vedclass · Answer

(A) According to Faraday's law of electrolysis,the mass deposited $w$ is given by $w = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the molar mass,$Q$ is the charge in Coulombs,$n$ is the valency (charge on the cation),and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.Given: $w = 2.977 \ g$,$Q = 10800 \ C$,$M = 106.4 \ g \ mol^{-1}$,$F = 96500 \ C \ mol^{-1}$.Rearranging the formula for $n$: $n = \frac{M 	imes Q}{w 	imes F}$.Substituting the values: $n = \frac{106.4 	imes 10800}{2.977 	imes 96500}$.$n = \frac{1149120}{287280.5} \approx 4$.Therefore,the charge on the metal cation is $+4$.

When a current of $10800 \ C$ is passed through an electrolyte,$2.977 \ g$ of a metal is deposited. If the molar mass of the metal is $106.4 \ g \ mol^{-1}$,what is the charge on the metal cation?

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