To complete the precipitation of $Ag$ from $400 \ mL$ of $0.04 \ M$ $AgNO_3$ solution,how long should a $4.8 \ A$ current be passed?

(N/A) $1$. Calculate the moles of $Ag^{+}$:
$n = M \times V(L) = 0.04 \ mol \ L^{-1} \times 0.4 \ L = 0.016 \ mol$.
$2$. Determine the charge required using Faraday's law:
The reaction is $Ag^{+} + e^{-} \rightarrow Ag$.
To reduce $1 \ mol$ of $Ag^{+}$,$1 \ mol$ of electrons $(1 \ F)$ is required.
Therefore,charge $Q = 0.016 \ mol \times 96500 \ C \ mol^{-1} = 1544 \ C$.
$3$. Calculate the time $t$ using $Q = I \times t$:
$t = Q / I = 1544 \ C / 4.8 \ A = 321.67 \ s$.