Calculate the current (in mA) required to dep | vedclass

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(C) The reaction for the deposition of platinum from $[PtCl_6]^{2-}$ is: $Pt^{4+} + 4e^- \rightarrow Pt(s)$.Here,the $n$-factor (valency) is $4$.The e

Vedclass · Accepted Answer

$21.44$

Vedclass · Answer

(C) The reaction for the deposition of platinum from $[PtCl_6]^{2-}$ is: $Pt^{4+} + 4e^- ightarrow Pt(s)$.Here,the $n$-factor (valency) is $4$.The equivalent weight of $Pt$ is $E = \frac{195}{4} = 48.75 \ g/eq$.Using Faraday's law: $W = \frac{I 	imes t 	imes E}{96500}$.Given: $W = 0.195 \ g$,$t = 5 	imes 3600 \ s = 18000 \ s$.$0.195 = \frac{I 	imes 18000 	imes 48.75}{96500}$.$I = \frac{0.195 	imes 96500}{18000 	imes 48.75} = \frac{18817.5}{877500} \approx 0.02144 \ A$.Converting to $mA$: $0.02144 	imes 1000 = 21.44 \ mA$.

Calculate the current (in $mA$) required to deposit $0.195 \ g$ of platinum metal in $5.0 \ hours$ from a solution of $[PtCl_6]^{2-}$ : (atomic weight: $Pt = 195$)

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