The time required to remove electrolytically | vedclass

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(C) $1$. Calculate the total moles of $Ag^+$ in the solution: $n = M 	imes V = 0.1 \ mol/L 	imes 0.2 \ L = 0.02 \ mol$. $2$. Calculate the moles of

Vedclass · Accepted Answer

$80.4$

Vedclass · Answer

(C) $1$. Calculate the total moles of $Ag^+$ in the solution: $n = M 	imes V = 0.1 \ mol/L 	imes 0.2 \ L = 0.02 \ mol$. $2$. Calculate the moles of $Ag^+$ to be removed: $n_{removed} = \frac{1}{4} 	imes 0.02 \ mol = 0.005 \ mol$. $3$. The reduction reaction is $Ag^+ + e^- ightarrow Ag(s)$. Thus,$1 \ mol$ of $Ag^+$ requires $1 \ mol$ of electrons. $4$. Total charge required $Q = n 	imes F = 0.005 \ mol 	imes 96500 \ C/mol = 482.5 \ C$. $5$. Using $Q = I 	imes t$,where $I = 0.1 \ A$: $t = \frac{Q}{I} = \frac{482.5 \ C}{0.1 \ A} = 4825 \ s$. $6$. Convert time to minutes: $t = \frac{4825}{60} \approx 80.4 \ min$.

The time required to remove electrolytically one fourth of $Ag^+$ from $0.2 \ L$ of $0.1 \ M \ AgCl$ solution by a current of $0.1 \ A$ is approximately $.......... \ min$.

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