Explanation:
\({\rm{t = 600}}\,\,{\rm{s;}}\)
\({\rm{Charge = current}} \times {\rm{time = 1}}{\rm{.5}}\,\,{\rm{A}}\,\, \times \,\,{\rm{600 s = 900 C}}\)
According to the reaction:
\({\rm{C}}{{\rm{u}}^{2 + }}(aq) + 2{e^ - } \to Cu(s)\)
We require \(2F\,\,or\,\,2 \times 96487\,\,C\) to deposit 1 mol or 63 g of Cu.
For 900 C, the mass of Cu deposited
\({\rm{ = }}\frac{{\left( {{\rm{63}}\,\,{\rm{g}}\,\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{ \times 900 C}}} \right)}}{{{\rm{2 \times 96487 Cmo}}{{\rm{l}}^{{\rm{ - 1}}}}}}\)
\({\rm{ = 0}}{\rm{.2938}}\,\,{\rm{g}}{\rm{.}}\)