Question

A current strength of 3.863 amp was passed through molten calcium oxide for 41 minutes and 40 seconds. The mass of calcium in grams deposited at the cathode is –
(Atomic mass of Ca is $40\text{ g/mol}$, $1\text{ F}=96500\text{ C}$)
(A) 4
(B) 2
(C) 6
(D) 8
(E) 1

Answer 1

Hint: The mass of substance deposited (m) at the electrode when Q amount of charge is passed can be calculated using Faraday’s law:
\[\text{m}=\dfrac{\text{E}\times \text{Q}}{96500}=\dfrac{\text{E}\times \left( \text{I}\times \text{t} \right)}{96500}\]

Complete answer:
The electrolysis is a process of breaking down a compound through a chemical reaction by passing an electric current through it. Michael Faraday gave two laws describing the quantitative aspect of electrolysis and are known as Faraday’s laws of electrolysis.
(1) – First law: The amount of substance undergoing chemical change by the flow of current is directly proportional to the quantity of electricity used. Mathematically,
\[\begin{align}
  & \text{m}\propto \text{Q} \\
 & \text{m}=\text{ZQ}=\text{Z}\times \left( \text{I}\times \text{t} \right) \\
\end{align}\]
Where m is the mass of substance deposited or produced at the electrode, Q is the amount of charge passed and Z is proportionality constant known as electrochemical equivalent.
\[\begin{align}
  & \text{Z}=\dfrac{\text{E}}{96500} \\
 & \Rightarrow \text{m}=\dfrac{\text{E}\times \text{Q}}{96500}=\dfrac{\text{E}\times \left( \text{I}\times \text{t} \right)}{96500}\text{ }..............\text{ (1)} \\
\end{align}\]
(2) – The amount of substances deposited on each electrode during a chemical reaction when the same amount of electricity is passed through different substances is proportional to their equivalent weights. Mathematically,
\[\begin{align}
  & \text{w}\propto \text{E} \\
 & \dfrac{{{\text{w}}_{1}}}{{{\text{w}}_{2}}}=\dfrac{{{\text{E}}_{1}}}{{{\text{E}}_{2}}}\text{ }..........\text{ (2)} \\
\end{align}\]
Where w is the mass of the substance and E is the equivalent weight of the substance.
In the given question, we have to find the mass of calcium deposited at the electrode when 3.863-ampere current was passed through molten calcium oxide for 41 minutes and 40 seconds.
The equivalent weight of calcium, in this case, will be:
     \[\begin{align}
  & \text{E}=\dfrac{\text{Molar mass }}{\text{No}\text{. of electrons transferred}} \\
 & \text{E}=\dfrac{40}{2}=20 \\
\end{align}\]
Time in seconds will be: $\text{t}=\left( 41\times 60\text{s} \right)+40\text{s}=2500\text{ s}$
Putting all known values in equation (1), we get:
\[\begin{align}
  & \text{m}=\dfrac{\text{20}\times \left( 3.863\times 2500 \right)}{96500} \\
 & \Rightarrow \text{m}=2.00\text{ g} \\
\end{align}\]

Hence, the correct answer is (B) 2.

Note:
During the electrolysis of calcium oxide, positively charged calcium ions will move toward the negatively charged electrode, that is, cathode, and gain two electrons to deposit calcium at the cathode.