Question

1. State Henry law
2. van’t Hoff factor for a solution is less than one, what is the conclusion drawn from it.
3. how many faraday of electricity is required to reduce $1$ mole of ${\text{MnO}}_4^ -$ ions to ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$ ions?

Answer 1

Henry’s law relates the partial pressure with mole fraction. Van’t Hoff factor is determined as the degree of dissociation or association of a substance in solution. To determine the Faraday, first we will determine the number of electrons by which the oxidant state of Mn is changing. Then by using the relation of mole of electrons with Faraday, we will determine Faraday.

Complete solution:
$1$)
According to Henry’s law, at constant temperature, the amount of the gas in a liquid is directly proportional to its partial pressure. Or the partial pressure is directly proportional to the mole fraction.
${\text{P}} \propto {\text{x}}$
Where,
P is the partial pressure of the g
X is the mole fraction
${\text{P}}\,{\text{ = }}\,{{\text{K}}_{\text{H}}}{\text{x}}$
Where,
${{\text{K}}_{\text{H}}}$ is the Henry constant.
$2$)
When a substance dissolves into the solution. It undergoes dissociation or association. The van’t Hoff factor tells the actual concentration of substance in solution.
The van’t Hoff for the dissociation is determined as the total number of ions produced by the substance in solution. If the minimum number of the substance is one on dissociation it will give a minimum two ions so the value of van’t Hoff will be two.
When a substance undergoes association, the minimum number of the substance undergoing association will be two.
Consider the example of association of benzoic acid on dissolving into carbon disulphide as follows:
On dissolving benzoic acid in carbon disulphide, the benzoic acid undergoes association which is shown as follows:
${\text{2}}\,{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\,\,\mathop \rightleftharpoons \limits^{{\text{C}}{{\text{S}}_2}} \,\,{({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH)}}_2}$
So, during association the value of the van't Hoff factor decreases so, the value of van’t Hoff can be less than one for association.
$3$)
The oxidation state of Mn in ${\text{MnO}}_4^ -$ is as follows:
${\text{x}}\,{\text{ + }}\,\left( { - {\text{2}} \times {\text{8}}} \right)\,{\text{ + 1}}\,{\text{ = }}\,{\text{0}}$
${\text{x}}\,\,{\text{ = }}\, + 7$
So, the oxidation state of Mn in ${\text{MnO}}_4^ -$ is $+ 7$.
So, we have to convert the Man from $+ 7$ to $+ 2$.
The reaction for the conversion of ${\text{MnO}}_4^ -$ ions to ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$ ions is shown as follows:
${\text{M}}{{\text{n}}^{ + 7}}\, + \,\,5{{\text{e}}^ - } \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}$
For the deposition of one mole of electrons $96500$Faraday is required so, the for the deposition of five mole of electrons the Faraday is required are,
$1\,{\text{mol}}\,\,\,{{\text{e}}^ - }\,\, = \,96500$ Faraday
$5\,{\text{mol}}\,\,\,{{\text{e}}^ - }\,\, = \,482500$ Faraday

So, $482500$ Faraday required to reduce $1$ mole of ${\text{MnO}}_4^ -$ ions to ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$ ions.

Note: At extreme high pressure, the gases do not obey Henry's law. Henry law is only applied at equilibrium state. The chemical reactions change the partial pressure of the particular gas so, when a gas undergoes a chemical reaction it does not obey Henry's law.