A) 2 : 3 : 6
B) \[6:2:3\]
C) 6 : 3 : 2
D) \[1:2:3\]
E) 3 : 6 : 2
Correct Answer: A
Solution :
\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] \[3\text{ }F~~~~~\,\,\,\,\,1\text{ }mol\] \[1F~\,\,\,\,~~~~1/3\text{ }mol\] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] \[2\text{ }F\,\,\,\,\,\,~~~~~1\text{ }mol\] \[1F\,\,\,~~~~~~1/2\text{ }mol\] \[N{{a}^{+}}+{{e}^{-}}\xrightarrow{{}}Na\] \[1F~\,\,\,~~~~~~~1\text{ }mol\] \[\therefore \] The mole ratio of\[Al,Cu\]and Na deposited at the respective cathode is \[\frac{1}{3}:\frac{1}{2}:1\]or \[2:3:6\]
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