Question

Current of 2A when passed for 5 hours through molten salt deposits 22.2g of metal of atomic mass $177$. Then calculate the positive oxidation state of the metal in the metal salt:

Answer 1

Hint: Faraday defined the relationship between the amount of electrical charge passing through an electrolyte and the amount of the material accumulated on the electrodes. This was used to express the magnitude of electrolytic effects. A faraday was designated as the quantity of energy that will induce a chemical shift of one equivalent weight unit.
$F = 96500C$

Formula used: $m = \dfrac{{QE}}{F}$
Where, $m = $ mass
$Q = $ charge deposited
$E = $ equivalent mass
$F = 96500C$ which is the Faraday constant

Complete step by step answer:
Let us first understand Faraday’s laws;
The law states that (a) the quantity of chemical change caused by the current at the electrode-electrolyte boundary is proportional to the quantity of electricity used, and (b) the quantity of chemical change caused by the same quantity of electricity is proportional to the equivalent mass of the substance. These are termed Faraday’s as first and second laws respectively.
As per first law ;
$m \propto Q$
$Z = \dfrac{m}{Q}$
Where $Z$ is the electrochemical equivalent and $m$ is mass and $Q$ is charge deposited
As per second law;
$m \propto E$ where $E$ is the equivalent weight and $m$ is the mass and
$E = \dfrac{{Molar\;mass}}{{Valence}}$
There is a formula that connects all this as follows;
$m = \dfrac{{QE}}{F} = \dfrac{{ItE}}{F}$ since, $Q = I \times t$
Where, $m = $mass
$Q = $ charge deposited
$E = $ equivalent mass
$F = 96500C$ which is the Faraday constant

Now, moving into our calculations;
$m = \dfrac{{QE}}{F} = \dfrac{{ItE}}{F}$
Then, $E = \dfrac{{F \times m}}{{It}}$
We have \[F = 96500C\], $m = 22.2g,I = 2amp,t = 5 \times 3600s = 18000s$
By substituting the above values we get;
$E = \dfrac{{22.2 \times 96500}}{{2 \times 5 \times 3600}} = 59.5g$
Also Equivalent mass $ = \dfrac{m}{v}$ where $m$ is the atomic mass and $v$is the valency or the oxidation state
We have, $E = 59.5g,m = 177g$
By substituting in the above equation we get;
$59.5 = \dfrac{{177}}{v}$
Therefore, $v = \dfrac{{177}}{{59.5}} = 3$

Note: The electrolysis laws of Faraday are only applicable in the following conditions;
- When the entire conduction is electrolytic in nature, i.e., only the ions bear the current.
- After the electrode reaction has occurred, no other side-reaction takes place.