Question

Two faradays of electricity are passed through a solution of $CuS{{O}_{4}}$ . The mass of copper deposited at the cathode is:
(at. Mass of $Cu=63.5amu$ )
a.) 0g
b.) 63.5g
c.) 2g
d.) 127g

Answer 1

Hint: Faraday's laws of electrolysis of electrolysis states that amount of element deposited on
electrode is directly proportional to the amount of current passed and secondly if same amount of
charge is passed through different electrolytes then quantity of charge deposited at different
electrodes are directly proportional to the ratio of their equivalent masses.
Complete answer: Faradays gave two laws:
Faraday's first law of electricity states that the amount of electrolyte deposited is equal to the amount
of current passed through it.
Faraday's second law of electrolysis states that when same quantity of charge is passed through
different electrolytes the masses of different substances deposited at respective electrodes will be in ratio of their equivalent masses.
Half-cell reaction;
$C{{u}^{2+}}+2{{e}^{-}}\to Cu.$
Two moles of electrons react with 1 mole of copper (II) by which it deposits one mole of copper.
So, the amount of electrolyte deposited will be equal to the amount of current passed through it.
Amount of electricity passed is 2F. Therefore, 2 faradays of electricity are required to form one mole of copper.
So, the number of moles of copper came out to be 1.
Molecular mass of copper is 63.5g.
Number of moles of copper is equal to the given mass divided by the molecular mass of copper.

& No\text{ }of\text{ }moles=\text{ }\dfrac{given\text{ }mass}{molar\text{ }mass}\text{ } \\\ & \text{1=}\dfrac{given\text{ }mass}{63.5} \\\ & given\text{ }mass=63.5\times 1=63.5g \\\ & \\\ \end{aligned}$$ So, the mass of the copper deposited at the cathode is 63.5 g. So, option B is correct. Note: Faraday gave two laws of electrolysis: First law of electrolysis states that amount of element deposited on electrode is directly proportional to the amount of current passed $$W=Z\times I\times T$$ Second law of electrolysis states that if same amount of charge is passed through different electrolytes then quantity of charge deposited at different electrodes is directly proportional to the ratio of their equivalent masses. $$\dfrac{Mass~of~firstelectrolyte~deposited}{Mass~of~\sec ondelectrolyte~deposited}=\dfrac{Eq.~mass~of~firstelectrolyte}{\begin{aligned} & Eq.~mass~of~\sec ondelectrolyte \\\ & \\\ \end{aligned}}$$