How many seconds of current at 241.25 A must | vedclass

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(B) The number of moles of $Ag^+$ in $125 \ mL$ of $1 \ M \ AgNO_3$ is calculated as: $n = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 1 \ M 	imes

Vedclass · Accepted Answer

$50$

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(B) The number of moles of $Ag^+$ in $125 \ mL$ of $1 \ M \ AgNO_3$ is calculated as: $n = 	ext{Molarity} 	imes 	ext{Volume (in L)} = 1 \ M 	imes 0.125 \ L = 0.125 \ mol$. The reduction reaction is: $Ag^+ + e^- 	o Ag_{(s)}$. According to the reaction,$1 \ mol$ of $Ag^+$ requires $1 \ F$ $(96500 \ C)$ of charge. Therefore,the total charge $Q$ required for $0.125 \ mol$ of $Ag^+$ is: $Q = 0.125 \ mol 	imes 96500 \ C/mol = 12062.5 \ C$. Using the formula $Q = I 	imes t$,where $I = 241.25 \ A$: $t = Q / I = 12062.5 \ C / 241.25 \ A = 50 \ s$.

How many seconds of current at $241.25 \ A$ must be passed to deposit all the silver present in $125 \ mL$ of $1 \ M \ AgNO_3$ solution? $(1 \ F = 96500 \ C)$

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