Redox Reactions And Electrochemistry Question 148
Question: Three faradays of electricity are passed through molten $ Al_2O_3 $ , aqueous solution of $ CuSO_4 $ and molten $ NaCl $ taken in different electrolytic cells. The amount of $ Al,,Cu $ and $ Na $ deposited at the cathodes will be in the ratio of [BHU 1990]
Options:
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
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Answer:
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{Atomic,mass}{2} $
At cathode : $ N{a^{+}}+{e^{-}}\to Na $ $ E_{Na}=\frac{Atomic,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 $ .
BETA
