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Question: A current of \[2\;amp\] when passed for \(5\) hours through a molten salt deposit \[22.2g\;\] of met...

A current of 2  amp2\;amp when passed for 55 hours through a molten salt deposit 22.2g  22.2g\; of metal of atomic mass 177177 . The oxidation state of the metal in the metal salt is:
A. +1 + 1
B. +2 + 2
C. +3 + 3
D. +4 + 4

Explanation

Solution

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 ;
mQm \propto Q
Z=mQZ = \dfrac{m}{Q}
Here ZZ is the electrochemical equivalent and mm is mass and QQ is charge deposited
According to the second law;
mEm \propto E
E=E = the equivalent weight, and m=m = mass,
So,
E=Molar  massValenceE = \dfrac{{Molar\;mass}}{{Valence}}
There is a relation that links all this as follows;
m=QEF=ItEFm = \dfrac{{QE}}{F} = \dfrac{{ItE}}{F} since, Q=I×tQ = I \times t
Where
m=m = mass
Q=Q = charge deposited
E=E = equivalent mass
F=96500CF = 96500C which is the Faraday constant
Now, proceeding into our calculations;
m=QEF=ItEFm = \dfrac{{QE}}{F} = \dfrac{{ItE}}{F}
Then, E=F×mItE = \dfrac{{F \times m}}{{It}}
We have F=96500CF = 96500C, m =22.2g  m{\text{ }} = 22.2g\;, I = 2ampI{\text{ }} = {\text{ }}2amp, t=5×3600s=18000st = 5 \times 3600s = 18000s
By replacing the above values we get;
E=22.2×965002×5×3600=59.5gE = \dfrac{{22.2 \times 96500}}{{2 \times 5 \times 3600}} = 59.5g
And also Equivalent mass =mv = \dfrac{m}{v}
where mm is the atomic mass and vv is the valency or the oxidation state
We take, E=59.5g,m=177gE = 59.5g,m = 177g
By substituting in the above equation we get;
59.5=177v59.5 = \dfrac{{177}}{v}
Hence, v=17759.5=3v = \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.