Question

When the same amount of electricity is passed through a solution of silver nitrate and copper sulfate, 0.4 g copper is deposited. The amount of silver deposited is:
(A)- 1.36g
(B)- 2.7g
(C)- 5.1g
(D)- 5.4g

Answer 1

Hint: This question is based on Faraday’s First Law of Electrolysis, which states that the chemical deposition due to the flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

Step by step solution:
-This solution starts by calculating the moles of copper deposited,
Moles of Cu deposited $=\dfrac{\text{given mass}}{\text{molar mass}}=\dfrac{0.4\text{ g}}{63.5\text{ g/mol}}=0.006299\text{ moles}$


-Now let us write the half-reaction for the formation of Ag and Cu -
$A{{g}^{+}}(aq)+{{e}^{-}}\to Ag(s)$
-Now, we will calculate the mole ratio of the copper element,
Mole ratio$=\dfrac{1\text{ mole Cu}}{2\text{ mol }{{\text{e}}^{-}}}$
-We know that,
$\dfrac{\text{moles of Cu deposited}}{\text{moles of Ag deposited}}=\dfrac{\text{mole ratio of Cu half reaction}}{\text{mole ratio of Ag half reaction}}$
$=\dfrac{0.006299\text{ moles}}{\text{moles of Ag deposited}}=\dfrac{\dfrac{1\text{ mol Cu}}{2\text{ mol }{{\text{e}}^{-}}}}{\dfrac{1\text{ mol Ag}}{1\text{ mol }{{\text{e}}^{-}}}}$
$=\dfrac{0.006299\text{ moles}}{\text{moles of Ag deposited}}=\dfrac{1}{2}$

-Hence the number of Ag deposited$=2\times 0.006299\text{ moles=0}\text{.0126 moles}$
-Since,
mole\[=\dfrac{\text{given mass}}{\text{molar mass}}\] , therefore
given mass$=\text{moles}\times \text{molar mass=0}\text{.0126 moles}\times \text{108 g/mol}$

Therefore the amount of silver deposited is 1.36g, i.e option A.

Additional information:
-Faraday’s First Law of Electrolysis is dependent on the mass of the chemical deposited due to electrolysis. The mass deposited is proportional to the quantity of electricity that passes through the electrolyte.
Here we need to remember one thing that the mass of the chemical deposited is not only proportional to the quantity of electricity passing through the electrolyte, it also depends on some other factors like-
(i) Every substance has a unique atomic weight. So, for the same number of atoms, different substances will have different masses.
(ii) If the valency of the substance is more, then for the same amount of electricity, the number of atoms deposited will be less, and if the valency of the substance is less, then for the same quantity of electricity, more number of atoms will be deposited.
Hence, as a conclusion, we can say that for the same quantity of electricity passing through different electrolytes, the mass of deposited chemical is directly proportional to its atomic weight and inversely proportional to its valency.

Note: Faraday’s Second Law of Electrolysis states that, when the same quantity is passed through several electrolytes, the mass of the substances deposited are proportional to their respective chemical equivalent or equivalent weight.