Question

A current of \[2\;amp\] when passed for $5$ hours through a molten salt deposit \[22.2g\;\] of metal of atomic mass \[177\] . The oxidation state of the metal in the metal salt is:
A. \[ + 1\]
B. \[ + 2\]
C. \[ + 3\]
D. \[ + 4\]

Answer 1

Hint: Faraday described the relationship between the amount of electrical charge flowing through an electrolyte and the quantity of the material stored on the electrodes. This was applicable in expressing the magnitude of electrolytic effects. A faraday was taken as the amount of energy that will bring a chemical shift of one equivalent weight unit. Use the relation between the faraday and the coulomb.

Complete step by step solution:
From Faraday’s laws;
The law defines that (a) the amount of chemical change produced by the current at the electrode-electrolyte boundary is directly proportional to the amount of electricity that is used, and (b) the amount of chemical change produced by the same amount of electricity is directly proportional to the equivalent mass of the material. These are called Faraday’s as first and second laws respectively.
According to the first law ;
$m \propto Q$
$Z = \dfrac{m}{Q}$
Here $Z$ is the electrochemical equivalent and $m$ is mass and $Q$ is charge deposited
According to the second law;
$m \propto E$
$E = $the equivalent weight, and $m = $ mass,
So,
$E = \dfrac{{Molar\;mass}}{{Valence}}$
There is a relation that links 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, proceeding into our calculations;
$m = \dfrac{{QE}}{F} = \dfrac{{ItE}}{F}$
Then, $E = \dfrac{{F \times m}}{{It}}$
We have $F = 96500C$, \[m{\text{ }} = 22.2g\;\], \[I{\text{ }} = {\text{ }}2amp\], $t = 5 \times 3600s = 18000s$
By replacing the above values we get;
$E = \dfrac{{22.2 \times 96500}}{{2 \times 5 \times 3600}} = 59.5g$
And also Equivalent mass $ = \dfrac{m}{v}$
where $m$ is the atomic mass and $v$ is the valency or the oxidation state
We take, $E = 59.5g,m = 177g$
By substituting in the above equation we get;
$59.5 = \dfrac{{177}}{v}$
Hence, $v = \dfrac{{177}}{{59.5}} = 3$.
Hence, Option C is correct.

Note:
The electrolysis laws of Faraday are only used in the following circumstances;
If the whole conduction is electrolytic in nature, where only the ions bear the current.
After the electrode reaction has happened, no other side-reaction will take place.