A) 1 mole : 2 mole : 3 mole
B) 3 mole : 2 mole : 1mole
C) 1 mole : 1.5 mole : 3 mole
D) 1.5 mole : 2 mole : 3 mole
Correct Answer: C
Solution :
At cathode : \[A{{l}^{3+}}+3{{e}^{-}}\to Al\] \[{{E}_{Al}}=\frac{\text{Atomic mass}}{3}\] At cathode : \[C{{u}^{2+}}+2{{e}^{-}}\to Cu\] \[{{E}_{Cu}}=\frac{\text{Atomic}\,\text{mass}}{2}\] At cathode : \[N{{a}^{+}}+{{e}^{-}}\to Na\] \[{{E}_{Na}}=\frac{\text{Atomic}\,\text{mass}}{1}\] For the passage of 3 faraday; mole atoms of \[Al\] deposited = 1 mole atoms of Cu deposited \[=\frac{1\times 3}{2}=1.5\] mole atoms of Na deposited \[=1\times 3=3\].You need to login to perform this action.
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