A) 0 g
B) 63.5 g
C) 2 g
D) 127 g
Correct Answer: B
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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