Question

The cost of electricity required to deposit 1 gm of \[Mg\] is Rs. 5. How much would it cost to deposit 10gm of \[Al\]? (atomic weight of \[Al\]= 27, \[Mg\]= 24)
a.) Rs. 10.00
b.) Rs. 27.00
c.) Rs. 44.44
d.) Rs. 66.67

Answer 1

Hint: The mass of the substance (m) deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge (Q) passed. Mathematically, it can be written as
$m=Z\times I\times T$
Where, m=mass of the deposited element, I = current passes through it, t=time duration for which current was passed, Z = electrochemical equivalent. So to deposit the same number of moles of a metal will be the same and depends only on the electron involved in the reaction. Or we can say that electricity used is directly proportional to the moles of electrons involved during the deposition reaction.

Complete step by step answer:
\[M{{g}^{2+}}+2{{e}^{-}}\to Mg\]
1g of Mg=\[\dfrac{1}{24}\]​ mole of Mg
∴ Amount of electricity=\[\dfrac{1}{24}\]​×2 electricity

∴\[\dfrac{1}{24}\]​ mole of 2\[{{e}^{-}}\] will cost us Rs. 5.
Then, 1 mole of \[{{e}^{-}}\] will cost us =Rs. 5×12=60

Now, for aluminium the following reaction will take place,
\[A{{l}^{3+}}+3{{e}^{-}}\to Al\]
10g Al = \[\dfrac{10}{27}\]mole ofAl

Therefore, the moles of \[{{e}^{-}}\]used = \[\dfrac{10}{27}\times 3\] ​=\[\dfrac{10}{9}\,mole\]
Hence, the cost of electricity will be =\[\dfrac{10\times 60}{9}=\dfrac{600}{9} = Rs.66.67\]​
The cost of electricity required to deposit 10gm of \[Al\] is Rs 66.67 .
So, the correct answer is “Option D”.

Note: Michael Faraday reported that the quantity of elements separated by passing an electric current through a molten or dissolved salt is proportional to the quantity of electric charge passed through the circuit. This is known as the basis of the first law of electrolysis.