A solution of `CuSO_(4)` is elecrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ?

Explanation

Quantity of electricity Q=It
`therefore Q=10xx60sxx1.5A|"where, "I=1.5A`
`t=10xx60s`
Deposition reaction of copper and its mole
Cathode reaction: `Cu_((aq))^(2+)+2e^(-)toCu_((s))`
According to stoichiometry, for production of 1 mol of Cu, 2 mol of electron=2F electricity is used.
So, 2F=`2xx96500F` will gives 1 mol
=63.5 g Cu
So, 900 C will reduced Cu`=(900Cxx63g)/((2xx96500)C)`
`=0.2938g" "Cu`