A) 2 g
B) 127 g
C) 0 g
D) 63.5 g
Correct Answer: D
Solution :
\[1F\xrightarrow[{}]{{}}1eq.wt.\] \[C{{u}^{2+}}+2_{2F}^{{{e}^{-}}}\xrightarrow[{}]{{}}\underset{1mol}{\mathop{Cu}}\,\] 2 Faraday’s electricity deposit\[\to \] 63.5 g of Cu.You need to login to perform this action.
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