Discussion Forum

 Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode  is [at .mass of Cu=63.5 u]

Answer:

Explanation:

Given, Q=2F

  Atomic  mass of Cu= 63.5 u

 Valancy of the metal Z= 2

 we have,   $CuSo_{4}\rightarrow Cu^{2+}+SO_{4}^{2-}$

  $Cu^{2+}+2e^{-}\rightarrow Cu$

   1 mol     2 mol            1 mol=63.5 g

                 2F

Alternatively,  W= ZQ

                     =  $\frac{E}{F}.2F=2E$

                     =   $\frac{2\times 63.5}{2}=63.5$