Question

In an electrolysis experiment, a current $\text{I}$passes through two different cells in series, one containing a solution of $\text{CuS}{{\text{O}}_{\text{4}}}$and the other a solution of $AgN{{O}_{3}}$. The rate of increase of the weight of the cathodes in the two cells will be
A. In the ratio of the densities of $Cu$ and $Ag$
B. In the ratio of the weights of $Cu$ and $Ag$
C. In the ratio of half the atomic weight of$Cu$to the atomic weight of $Ag$
D. In the ratio of half the atomic weight of $Cu$to half the atomic weight of $Ag$

Answer 1

Hint: According to Faraday’s second law of electrolysis, When the same amount of current is passed through different electrolytes, the mass of the deposited substances is directly proportional to their respective chemical equivalent. To approach this question we can use the second law of electrolysis to calculate the weight of the chemical substance deposited at the cathode.

Formula used: The mathematical expression of the second law of electrolysis,
$\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}}$
Here ${{m}_{1}}$= Mass of substance $1$
${{m}_{2}}=$ Mass of substance $2$
${{E}_{1}}=$ The electrochemical equivalent of substance $1$
${{E}_{2}}=$ The electrochemical equivalent of substance $2$
Electrochemical equivalent,$E=\dfrac{M}{{{n}_{f}}}$
Here $M=$ atomic weight
${{n}_{f}}=$ n factor

Complete Step by Step Answer:
Here in this question, there are two different electrolytes of copper sulphate $(CuS{{O}_{4}})$and silver nitrate $(AgN{{O}_{3}})$over which an electrolysis experiment is performed.
$A{{g}^{+}}(aq.)+{{e}^{-}}\to Ag(s)$
${{n}_{f}}$of $Ag$$=1$
$C{{u}^{2+}}(aq.)+2{{e}^{-}}\to Cu(s)$
${{n}_{f}}$of $Cu$$=2$

According to Faraday’s second law of electrolysis, $\dfrac{{{m}_{Cu}}}{{{m}_{Ag}}}=\dfrac{{{E}_{Cu}}}{{{E}_{Ag}}}$
Or, $\dfrac{{{m}_{Cu}}}{{{m}_{Ag}}}=\dfrac{\dfrac{{{M}_{Cu}}}{{{n}_{f}}}}{\dfrac{{{M}_{Ag}}}{{{n}_{f}}}}=\dfrac{\dfrac{{{M}_{Cu}}}{2}}{\dfrac{{{M}_{Ag}}}{1}}=\dfrac{\dfrac{1}{2}Cu}{{{M}_{Ag}}}$
Therefore the rate of increase of weight of the cathodes in the two cells will be in the ratio of half atomic weight $Cu$ to half the atomic weight $Ag$.
Thus, option (C) is correct.

Additional information: Faraday’s law of electrolysis explains the relationship between the amount of electric charge passed through an electrolyte and the amount of deposited substance at any electrode. Faraday’s first law tells us chemical deposition caused by the flow of an electricity is directly proportional to the amount of passing current through an electrolyte.

Note: To approach the problem regarding electrolysis we should remember both the first and second laws of Faraday. We should know how to calculate chemical equivalents. Thus to determine the Chemical equivalent, the atomic weight of the substances is divided by their n-factor.