Question

A current of 10 A flows for 2 hours through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is: (F = 96500 C)
(A)- $+1$
(B)- $+2$
(C)- $+3$
(D)- $+4$

Answer 1

Hint: In order to find the oxidation state of the reduced metal in the metal salt, the amount of metal deposited on the electrode is related to both the amount of charge passing through the solution of the metal salt and the equivalent weight, which we can determine, using the Faraday’s law.

Complete step by step solution:
It is given that in the electrolytic cell, on passing current of I = 10 A for time, t = 2 hours $\text{=2 }\!\!\times\!\!\text{ 60 }\!\!\times\!\!\text{ 60}\,\text{seconds}$, it causes n= 0.250 moles of molten salt of the metal to decompose at the cathode. Then, in order to find the oxidation state of the metal that got reduced at the cathode, we will use the Faraday’s law of electrolysis.
According to which the amount of metal (W) deposited on the electrode is directly proportional to the amount of charge (q) passed through the solution. So, we have,
$W=Z\times q$ ------------ (a)
where, Z is the proportionality constant, that is the equivalent mass deposited by one coulomb charge $=\dfrac{Equivalent\,mass}{96500}$
Then, $W=\dfrac{E}{96500}\times q$ ------------ (b)
Also, we know, charge, $q=I\times t$
On substituting the value of charge, q in equation (b), we get,
$W=\dfrac{E\times I\times t}{96500}$
$W=\dfrac{E\times 10\times 2\times 60\times 60}{96500}$
$W=\dfrac{M\times 10\times 2\times 60\times 60}{x\times 96500}$ , as the equivalent $mass$ $=\dfrac{molar\,mass\,(M)}{valency\,(x)}$
As the moles of the molten metal salt that is reduced, is given to be n= 0.250 moles and $n=\dfrac{weight}{molar\,mass}=0.25\,moles$. Substituting this value in above equation, we get,
\[x=\dfrac{M\times 10\times 2}{W\times 96500}=\dfrac{10\times 2\times 60\times 60}{0.25\times 96500}=2.98\approx 3\]

Therefore, the valence or oxidation state of the metal X in molten salt is option (C)- $+3$.

Note: The units of all the quantities are in SI units, so the conversion must be done carefully, like for the time, it is converted from hours to seconds. We are using both Faraday’s first and second law above.