Question

A solution of $CuS{O_4}$ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Answer 1

Hint:We can calculate the mass of copper using the molar mass of copper, charge, number of electrons transferred, Faraday’s constant. The charge is calculated by multiplying the time and current.

Complete step by step answer:

Given data contains,
Time taken is ten minutes.
Current is 1.5 amperes.
Faraday’s constant is $96485\,C\,mo{l^{ - 1}}$.
First, let us convert the time in minutes to seconds. This can be done by multiplying the given time with 60.
We can convert time in minutes to seconds as,
Time in minutes=$10\,\min X \dfrac{{60\,\sec }}{{1\,\min }}$
Time in minutes=$600\,\sec $
The time taken is seconds in $600\,\sec $.
Now let us calculate the charge. We can obtain the charge by multiplying the time in seconds with current in amperes.
Charge=${\text{Current X Time}}$
Let us now substitute the values of current and time in the equation of charge.
We obtain charge as,
Charge=${\text{Current X Time}}$
Charge=${\text{1}}{\text{.5}}\,{\text{amp X 600}}\,{\text{sec}}$
Charge=$900\,{\text{C}}$
The charge is $900\,{\text{C}}$.
Based on the reaction, \[C{u^{2 + }}\left( {aq} \right) + 2{e^ - }\xrightarrow{{}}Cu\left( s \right)\]
The number of electrons taken part in the chemical reaction is two.
Molar mass of copper is $63\,g\,mo{l^{ - 1}}$.
Let us now calculate the mass of copper using the values of charge, molar mass, number of electrons and faradays’ constant.
We can obtain the mass of copper deposited as,
Mass of copper deposited=$\dfrac{{{\text{Molar}}\,{\text{mass X Charge}}}}{{{{\text{e}}^{\text{ - }}}\,{\text{transferred X Faraday's}}\,{\text{constant}}}}$
Mass of copper deposited=$\dfrac{{63\,g\,mo{l^{ - 1}} X 900\,C}}{{2 X 96487\,C\,mo{l^{ - 1}}}}$
Mass of copper deposited=$0.2938\,g$
The mass of copper deposited is $0.2938\,g$.
If we assume the take of Faradays’ constant as $96500\,C\,mo{l^{ - 1}}$ and the molar mass of copper as $63.5\,g\,mo{l^{ - 1}}$
We can obtain the mass of copper deposited as,
Mass of copper deposited=$\dfrac{{{\text{Molar}}\,{\text{mass X Charge}}}}{{{{\text{e}}^{\text{ - }}}\,{\text{transferred X Faraday's}}\,{\text{constant}}}}$
Mass of copper deposited=$\dfrac{{63.5\,g\,mo{l^{ - 1}} X 900\,C}}{{2 X 96500\,C\,mo{l^{ - 1}}}}$
Mass of copper deposited=$0.296g$
The mass of copper deposited is $0.296g$.


Note:
We must know that electrolysis is one of the many methods which are widely used in industry, it is used not only for extraction of metals but also to reduce the impurities in metal. This process is used in the medical field as well like for permanent removal of hair, etc. Even though such an application has some side effects but for long run it is good.