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

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(C) The reduction reaction is: $PtCl_6^{2-} + 4e^{-} ightarrow Pt + 6Cl^{-}$.Number of moles of $Pt = \frac{	ext{given mass}}{	ext{atomic mass}} =

Vedclass · Accepted Answer

$21.44$

Vedclass · Answer

(C) The reduction reaction is: $PtCl_6^{2-} + 4e^{-} ightarrow Pt + 6Cl^{-}$.Number of moles of $Pt = \frac{	ext{given mass}}{	ext{atomic mass}} = \frac{0.195 \ g}{195 \ g/mol} = 0.001 \ mol$.From the stoichiometry,$1 \ mol$ of $Pt$ requires $4 \ mol$ of electrons.Therefore,$0.001 \ mol$ of $Pt$ requires $0.004 \ mol$ of electrons.Total charge $Q = n 	imes F = 0.004 \ mol 	imes 96500 \ C/mol = 386 \ C$.Using the formula $Q = I 	imes t$,where $t = 5.0 \ hours = 5 	imes 3600 \ s = 18000 \ s$.$I = \frac{Q}{t} = \frac{386}{18000} \ A = 0.02144 \ A$.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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