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 o\[Al\] \[\because \] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] \[\therefore \] 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.
You will be redirected in
3 sec