A) \[29.3\]
B) \[58.7\]
C) \[117.4\]
D) \[176.1\]
Correct Answer: C
Solution :
\[NiC{{l}_{2}}\xrightarrow{Electrolysis}N{{i}^{2+}}+2C{{l}^{-}}\] At anode, \[2C{{l}^{-}}\xrightarrow{{}}C{{l}_{2}}+2{{e}^{-}}\] At cathode, \[N{{i}^{2+}}+\underset{2F}{\mathop{\underset{2F}{\mathop{2{{e}^{-}}}}\,}}\,\xrightarrow{{}}\underset{58.6g}{\mathop{\underset{1\,mol}{\mathop{Ni}}\,}}\,\] \[\because \]2F electricity deposits \[Ni=58.6\text{ }g\] \[\therefore \] 4F electricity will deposit \[Ni=\frac{58.6}{2F}\times 4F=117.2g\]
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