Question

A solution of $CuS{{O}_{4}}$is electrolyzed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Answer 1

Hint: The charge required for the electrolysis can be calculated by multiplying the current with the time taken in seconds. Now to calculate the mass deposited by this charge we have to multiply it with the charge required for 1 mole of copper to deposit at the cathode.

Complete step by step solution:
Faraday’s first law of electrolysis:
It states that the amount of mass liberated at the electrode is always directly proportional to the quantity of the current which is passed into the solution.
\[W\text{ }=\text{ }Z\text{ x }Q\]
W = mass,
 Z = proportionality constant called electrochemical equivalent.
Q = amount of electricity.
 Faraday’s second law of Electrolysis.
It states that when electricity is passed to a solution which contains electrolyte that is connected in series, the masses produced at the electrode is directly proportional to their equivalent weight.
When electrolysis occurs, the charge carried by one mole of an electron is obtained by the product of charge present on one electron with Avogadro's number which equals to 96500coloumbs. This quantity is called one faraday.
So, by combining both the laws we can formulate:
\[W=\dfrac{QM}{Fz}\]
The charge required for the electrolysis can be calculated by multiplying the current with the time taken in seconds
$Q=I\text{ x t}$
Given time is 10 minutes, and it has to be converted into seconds,
$t=10\text{ x 60 = 600 sec}$
$I=1.5\text{ amp}$
So, the charge (Q) will be:
$Q=I\text{ x t}=\text{ 1}\text{.5 x 600 = 900 C}$
F = 1 faraday
Z = valency of the metal.
The reaction at the cathode will be:
$C{{u}^{2+}}+2{{e}^{-}}\to Cu$
M = Mass of the element = $63.5\text{ g / mol}$
Q = amount of electricity = $900\text{ C}$
F = \[96500\text{ }C/mol\]
z = valency of copper = 2
By combining all these, we get
$W=\dfrac{63.5\text{ x 900}}{96500\text{ x 2}}=\text{ 0}\text{.296 g}$

So, the mass of copper deposited will be 0.296 g.

Note: The actual value of the faraday is 96487 but for numerical and easier calculation 96500 is used. If the time is not given we can directly use $W=\dfrac{QM}{Fz}$the formula. And if the time is mentioned then always take the value in seconds.