## Chemistry (7th Edition)

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