Given, Current (i)=50×10−3 amp. Time (t)=60min=3600sec (F)=96500C ∵ charge (Q)=i×t Q=i×t=50×10−3×3600=180C Now, molar mass of copper (Cu)=63.5g and to convert to Cu(s) ∵96500C charge is required to deposit =63.5g of copper (s) and to convert Cu+→Cu(s) : ∵1C charge will deposit =
63.5
96500
gof Cu(s) ∴180C charge will deposit =
63.5×180
96500
=0.118g of Cu Hence, option (a) is the correct answer.