For how much time,10 A electric current shou | vedclass

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(A) The reduction reaction for $Ni^{2+}$ is: $Ni^{2+} + 2e^- ightarrow Ni(s)$. The number of moles of $Ni$ produced is $n = \frac{	ext{mass}}{	ext

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$1930$

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(A) The reduction reaction for $Ni^{2+}$ is: $Ni^{2+} + 2e^- ightarrow Ni(s)$. The number of moles of $Ni$ produced is $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$. According to the reaction,$1 \ mol$ of $Ni$ requires $2 \ mol$ of electrons. Therefore,$0.1 \ mol$ of $Ni$ requires $0.2 \ mol$ of electrons. Total charge $Q = n 	imes F = 0.2 	imes 96500 \ C = 19300 \ C$. Since $Q = I 	imes t$,we have $t = \frac{Q}{I} = \frac{19300 \ C}{10 \ A} = 1930 \ s$. Thus,the correct option is $A$.

For how much time,$10 \ A$ electric current should be passed through a dilute aqueous $NiSO_4$ solution during electrolysis using inert electrode,in order to get $5.85 \ g$ Nickel (in $s$)?
[At. mass of $Ni = 58.5 \ g/mol$]

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For how much time,$10 \ A$ electric current should be passed through a dilute aqueous $NiSO_4$ solution during electrolysis using inert electrode,in order to get $5.85 \ g$ Nickel (in $s$)? [At. mass of $Ni = 58.5 \ g/mol$]