A) 3:4:6
B) 2:1:6
C) 3: 2: 1
D) 2: 3 : 6
Correct Answer: D
Solution :
\[\because \]\[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] \[\therefore \]3F of electricity will deposite 1 mole of Al. \[\because \] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] \[\therefore \]3F of electricity will deposite 1.5 mole of Cu. \[\because \] \[N{{a}^{+}}+{{e}^{-}}\xrightarrow{{}}Na\] \[\therefore \]3F of electricity will deposite 3 mole of Na. Hence, the mole ratio of Al, Cu and Na deposited on the cathode will be 1 : 1.5 : 3 or 2:3:6.You need to login to perform this action.
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