• # question_answer Two Faraday of electricity is passed through a solution of $CuS{{O}_{4}}.$ The mass of copper deposited at the cathode is (at. mass of $Cu=63.5\text{ }u$) A) 0 g B) 63.5 g C) 2 g D) 127 g

Solution :

 According to Faraday's second law Given, $Q=2F$Atomic mass of $Cu=63.5u$ Valency of the metal $Z=2$ We have, $CuS{{O}_{4}}\to C{{u}^{2+}}+S{{O}^{2-}}_{4}$ $\underset{1mol}{\mathop{C{{u}^{2+}}}}\,+\underset{\begin{smallmatrix} 2mol \\ 2F \end{smallmatrix}}{\mathop{2{{e}^{-}}}}\,\to \underset{1mol=63.5g}{\mathop{Cu}}\,$ Alternatively. $W=ZQ=\frac{E}{F}.2F=2E=\frac{2\times 63.5}{2}=63.5g$

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