Question

Two electrolytic cells containing $CuS{{O}_{4}}$ and $AgN{{O}_{3}}$respectively are connected in series and a current is passed through them until $1$ mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately [Atomic weights of copper and silver are respectively $63.57$and$107.88$]
A. $1.7$mg
B. $3.4$mg
C. $5.1$ mg
D. $6.8$mg

Answer 1

Hint: Faraday’s second law of electrolysis states that when the same amount of current is passed through different electrolytes, the mass of different ions liberated at the electrodes are proportional to chemical equivalents of ions. The same law also can be applied here to determine the amount of silver.

Formula used:The mathematical expression for Faraday’s second law is given by
$\dfrac{{{m}_{2}}}{{{m}_{1}}}=\dfrac{{{E}_{2}}}{{{E}_{1}}}$
${{m}_{1}}$and ${{m}_{2}}$are masses of the elements $1$and $2$deposited on electrodes
${{E}_{1}}$and ${{E}_{2}}$are chemical equivalents of the deposited elements $1$ and $2$

Complete Step by Step Answer:
Michael Faraday who discovered the electrolysis law is also known as the first and second law of electrolysis. These laws govern the deposition of substances in the forms of ions on the electrodes by applying electricity.

Generally according to the second law of Faraday the masses of different ions deposited are directly proportional to their chemical equivalents.
Then we can write $\dfrac{{{m}_{2}}}{{{m}_{1}}}=\dfrac{{{E}_{2}}}{{{E}_{1}}}$
Here m represents the mass of elements $1\And 2$
And $E$ represents the chemical equivalents of elements $1\And 2$
For this particular problem the second law of Faraday can be applied.

According to the second law the masses of ions of different substances deposited by the same amount of current are directly proportional to their chemical equivalents.
Chemical equivalent, $E = \dfrac{M}{n}$
Here $M$ is atomic weight and $n$ is number of valency
$Cu^{2+}+2e^{-} \to Cu$
$\therefore n=2$
Chemical equivalent of copper, $E_1 = \dfrac{63.57}{2}$ = 31.79
$Ag \to Ag^{+}+e^{-}$
$\therefore n = 1$
Chemical equivalent of silver, $E_2 = \dfrac{107.88}{1} = 107.88$
Mass of copper, $m_1 = 1 \ mg$
Therefore $m_2 = \dfrac{E_2}{E_1} \times m_1$
Or, $m_2 = \dfrac{107.88}{31.79} \times 1 = 3.393 \ mg$
Or, $m_2 \approx 3.4 \ mg$
Hence the amount of silver deposited in the second cell is $3.4$ mg.
Thus, option (B) is correct.

Note: To solve this type of numerical regarding Electrolysis, we should remember both the laws of Micheal Faraday. Also we should know how to calculate chemical equivalents by using atomic weight and valency of deposited different ions on electrodes.