Answer
3.22 g
Work Step by Step
Time, t= 20.0 min= $(20.0\times60)s$= 1200 s
Current, I= 2.40 A
Charge=$ I\times t$=$ 2.40 A\times 1200 s$= 2880 C
According to the reaction: $Ag^{+}+ e^{-}→Ag $
we require 1F or 96487 C to deposit 1 mol or 107.87 g of Ag at the cathode.
For 2880 C, the mass of Ag obtained=
$\frac{107.87gmol^{-1}\times2880 C}{1\times96487Cmol^{-1}}$ = 3.22 g
