Question

The cost of electricity required to deposit 1g of Mg is $Rs.5.00$. How much would it cost to deposit $10g$ of aluminium?
A. $Rs.10.00$
B. $Rs.27.00$
C. $Rs.44.44$
D. $Rs.66.67$

Answer 1

Hint: By writing the equation for the reduction of magnesium, we can find the quantity of electricity passed. As the cost of this value is given, we can then find the cost per unit quantity of charge passed. Then we can write the equation for the deposition for aluminium, find the charge passed in this case and multiply with the cost we got from the previous step to get the final answer.

Formulas used: $n = \dfrac{W}{M}$
Where $n$ is the number of moles, $W$ is the given mass and $M$ is the molar mass

Complete step by step answer:
Magnesium gets deposited by reduction according to the following equation:
$M{g^{2 + }} + 2{e^ - } \to Mg$
Thus, each mole of magnesium will require two moles of electrons to get reduced. As we know,
$n = \dfrac{W}{M}$
Where $n$ is the number of moles, $W$ is the given mass and $M$ is the molar mass
Here the given mass is $W = 1g$ and the molar mass of magnesium is $M = 24g$. Substituting these values, we get:
$n = \dfrac{1}{{24}}mol$. Hence, the number of electrons needed will be the double of this value, that is,
Number of moles of electrons required = $\dfrac{{1 \times 2}}{{24}} = \dfrac{1}{{12}}$
We are given that the cost of passing this many electrons (electricity) is $Rs.5.00$.
Now let us look at what happens with aluminium:
$A{l^{3 + }} + 3{e^ - } \to Al$
Hence, to deposit one mole of aluminium, we need three moles of electrons, or the number of electrons required will be thrice the number of moles of aluminium.
Given mass of aluminium is $W = 10g$ and its molar mass is $M = 27g$. Therefore, we have:
$n = \dfrac{{10}}{{27}}mol$. The number of electrons required is thus, thrice this value, that is,
Number of moles of electrons required $ = 3 \times \dfrac{{10}}{{27}} = \dfrac{{30}}{{27}}mol$
Now let us find how much this would cost using the help of our previous observations:

Moles of electrons	Cost in Rupees
$\dfrac{1}{{12}}$	$5.00$
$\dfrac{{30}}{{27}}$	$x$

Here $x$ is the cost required in the case of aluminium. Using cross-multiplication rule, we have:
$x \times \dfrac{1}{{12}} = 5 \times \dfrac{{30}}{{27}}$
Rearranging, we get:
$x = 5 \times \dfrac{{30}}{{27}} \times 12$
$ \Rightarrow x = \dfrac{{1800}}{{27}} = 66.67$
Thus, the cost required for depositing $10g$ of aluminium is $Rs.66.67$.

So, the correct answer is Option D .

Note: The charge possessed by one mole of electrons is approximately equal to $96,500C$, and is termed as one Faraday. Note that the deposition of metals through electrolysis is governed by the Faraday’s Laws of electrolysis, which states that the amount of metal deposited is directly proportional to the quantity of electricity passed.