Question

A certain quantity of electricity is passed through aqueous solution $Al{(S{O_4})_3}$ and $CuS{O_4}$ solutions connected in series. $0.09g$ of $Al$ is deposited on cathode during electrolysis. The amount of copper deposited on cathode in gram is: (At. Wt. of $Al = 27;Cu = 63.6$ )
A.$31.8$
B.$0.636$
C.$3.18$
D.$0.318$

Answer 1

We must remember that the decomposing of a compound into the elements by passing an electric current is defined as electrolysis. Electrolysis is done in a vessel, named electrolytic cell. In electrolysis the compound is either loss or gain weight, which is one compound is reduced by another compound while the other compound is oxidized by the other compound.

Complete answer:
Given,
Weight of $Al$ is $0.09g$.
The aqueous solutions of $Al{(S{O_4})_3}$ and $CuS{O_4}$ on electrolysis,
$A{l^{3 + }} + 3{e^ - } \to Al$
$C{u^{2 + }} + 2{e^ - } \to Cu$
The electrolysis redox reaction depends upon the quantity of electricity is flowing through the cell. This is given by Micheal Faraday's law of electrolysis.
According to the second law of electrolysis,
$\dfrac{{{W_1}}}{{{E_1}}} = \dfrac{{{W_2}}}{{{E_2}}}$
Where, $W\& E$ are the changed mass and equivalent mass of the compound respectively.
From the above equation,
$\dfrac{{{W_{Cu}}}}{{{E_{Cu}}}} = \dfrac{{{W_{Al}}}}{{{E_{Al}}}}$
At. Wt. of $Al = 27;Cu = 63.6$
${E_{Al}} = \dfrac{{27}}{3}$ & ${E_{Cu}} = \dfrac{{63.6}}{2}$
The above equation is applied to the second law of electrolysis,
${W_{Cu}} = \dfrac{{{W_{Al}}}}{{{E_{Al}}}} \times {Z_{Cu}}$
On substituting the known values we get,
$\Rightarrow {W_{Cu}} = \dfrac{{0.09 \times 3}}{{27}} \times \dfrac{{63.6}}{2}$
On simplification we get,
${W_{Cu}} = 0.318$
So,Option D. $0.318$ is the correct option from the above data.

Note:
We must know that the electrolytic cell contains two electrodes (used to make electric contact with the same part of a circuit), cathode and anode, connected to a DC source. An ionic compound which is going to decompose in either dissolved state in a suitable solvent or in molten form. In electrolysis, the anions are oxidized at the anode and the cations are reduced at cathode. Because of the addition or removal of electrons from the circuit, there is an interchange of ions and atoms. Electrolysis is a redox reaction, it involves a simultaneous reduction reaction at the cathode and oxidation reaction at the anode.