A) 2 : 3 : 1
B) 2 : 1 : 1
C) 2 : 1 : 3
D) 2 : 2 : 1
Correct Answer: A
Solution :
From the Faradays Und law, we know that \[W\propto Zit,\,\,\] \[\,W\propto \frac{E}{96500}\]it or it \[\propto \frac{1}{E}\]or electricity \[\propto \frac{1}{E}\] Thus, the amount of electricity required for \[\text{ZnS}{{\text{O}}_{\text{4}}}\text{,}\,\text{AlC}{{\text{l}}_{\text{3}}}\]and\[\text{AgN}{{\text{O}}_{\text{3}}}\] is in the ratio of 2:3:1. (Here, 2, 3 and 1 are the valencies of metal atom in the given compounds). Alternatively In \[\text{ZnS}{{\text{O}}_{4}},\] \[\underset{\text{1mol}}{\mathop{Z{{n}^{2+}}}}\,+\underset{\begin{smallmatrix} \text{2}\,\text{mol} \\ \text{or}\,\text{Faraday} \end{smallmatrix}}{\mathop{2{{e}^{-}}}}\,\xrightarrow{{}}Zn\] In \[AlC{{l}_{3}},\] \[\underset{\text{1}\,\text{mol}}{\mathop{A{{l}^{3+}}}}\,+\underset{3F}{\mathop{3{{e}^{-}}}}\,\xrightarrow{{}}Al\] In \[AgN{{O}_{3}},\] \[\underset{1\,mol}{\mathop{A{{g}^{+}}}}\,+\underset{1\,F}{\mathop{{{e}^{-}}}}\,\xrightarrow{{}}Ag\] Thus, the ratio of amounts of electricity required is 2 : 3 : 1
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