Explanation:
\({\rm{Cathode}}:2{\rm{A}}{{\rm{g}}^ + } + {\rm{2}}{{\rm{e}}^ - } \to 2{\rm{Ag}}\)
\({\rm{Anode}}:2{\rm{O}}{{\rm{H}}^ - } - 2{{\rm{e}}^ - } \to 2{\rm{OH}}\)
\(2{\rm{OH}} \to {{\rm{H}}_{\rm{2}}}{\rm{O}} + \frac{1}{2}{{\rm{O}}_2}\)
\({\rm{2}}\,{\mkern 1mu} {\rm{moles}}{\mkern 1mu} \,{\rm{of}}{\mkern 1mu} {\rm{electrons}}{\mkern 1mu} {\mkern 1mu} {\rm{or}}{\mkern 1mu} {\mkern 1mu} {\rm{2}}\,{\rm{ \times }}\,{\rm{96500}}{\mkern 1mu} \,{\rm{C}}\)
\({\rm{liberates}}{\mkern 1mu} \,{\rm{16}}{\mkern 1mu} \,{\rm{g}}\,{\mkern 1mu} {\rm{of}}{\mkern 1mu} {{\rm{O}}_{\rm{2}}}\)
\(\therefore 1.6\;{\rm{g}}\,{\rm{of}}\,{{\rm{O}}_{\rm{2}}}{\rm{will}}\,{\rm{be}}\,{\rm{liberated}}\,{\rm{by}}\)
\(\frac{{2 \times 96500}}{{16}} \times 1.6\,{\rm{C}} = 19300\,{\rm{C}}\)
\({\rm{Now}},2 \times 96500\,{\rm{C}}\,{\rm{deposits}}\,2 \times 107.8\;{\rm{g}}\,{\rm{of}}\,{\rm{Ag}}\)
\(\therefore 19300\,{\rm{C}}\,{\rm{will}}\,{\rm{deposit}}\)
\(\frac{{2 \times 107.8}}{{2 \times 96500}} \times 19300 = 21.56\;{\rm{g}}\,{\rm{Ag}}\)