Question

Calculate the mass of Hg which can be prepared by the reduction of mercury (II) ions in the presence of chloride ions by the passage of 5.0 ampere current for 3.0 hours.

Answer 1

Hint: In reduction reaction, the gain of the electron and hydrogen takes place with the loss of an oxygen atom. The number of moles is defined as the $6.022\text{ }\times \text{ 1}{{\text{0}}^{23}}$ particles of an atom, element, ions, etc.

Complete Step by Step Answer:
-Firstly, let's understand Faraday's law of electrolysis.
-The law states that the quantity of the electricity passed is directly proportional to the mass of any substance that is deposited at any electrode during electrolysis.
-Now, let's write the reaction of the reduction of mercury:
$\text{H}{{\text{g}}^{2+}}\text{ + 2e }\to \text{ Hg}$
-Now, to calculate the mass of the mercury that can be produced by the reduction of mercury (II) we will use the formula: $\text{Moles of Hg produced = }\dfrac{\text{I }\times \text{ t}}{2\text{ }\times \text{ 96500cmo}{{\text{l}}^{1-}}}\text{ }....\text{(1)}$
-Here, 'I' is the current in amperes and t is the time taken whereas 96500 is the value of one faraday constant.
-As we know that no. of moles will be equal to the mass of mercury divided by the atomic mass of mercury.
-So, equation 1 will become: $\dfrac{\text{Mass of mercury}}{\text{Atomic mass}}\text{ = }\dfrac{\text{5 }\cdot \text{ 10800}}{2\text{ }\cdot \text{ 96500}}$
$\text{Mass = }\dfrac{\text{5 }\cdot \text{ 10800}}{2\text{ }\cdot \text{ 96500}}\text{ }\cdot \text{ 200}\text{.59}$
$\text{Mass = 56}\text{.1g}$
-So, the mass of mercury that can be prepared by the reduction of mercury (II) is 56.1g.

Note: 96500 c/mol is the value of one Faraday. It tells us about the amount of electricity which one mole of the electrons carries and it is also known as Faraday's constant.