Question

The element indium is to be obtained by electrolysis of a molten halide of the element. Passage of a current of $3.20A$ for a period of $40.0$ min results in formation of $3.05$ g of In. What is the oxidation state of indium in the halide melt? (Atomic weight of In = $114.8$ )
A. 3
B. 2
C. 5
D. 1

Answer 1

Hint:The valence or oxidation state of Indium in its halide state can be obtained by dividing the product of charge and atomic mass to the product of mass of indium formed at the electrode and electro-chemical equivalent.

Complete answer:
Indium is an element of atomic number 49 which can be obtained by the process of electrolysis of a molten halide (compound formed from the combination of halogen) of the same element. The current $\left( I \right)$ of magnitude $3.20$ A is passed through the electrolyte for a time period $\left( t \right)$ of 40 min which results in the formation of Indium of mass $m = 3.05$ g.
According to Faraday’s first law of electrolysis, the mass of an element deposited at an electrode is directly proportional to the charge in Coulombs and Faraday’s second law of electrolysis states that the amount of substance deposited at the electrodes by the passage of same magnitude of electric current is proportional to the equivalent weights and equivalent weight is equal to the molar mass divided by the valence of the element.
From the above two laws, we get $\dfrac{m}{E} = \dfrac{Q}{Z}$ where E is the equivalent weight, Q is the charge and Z is the constant of proportionality known as Electro-Chemical Equivalent of the substance (Indium).
$\dfrac{m}{{\dfrac{M}{n}}} = \dfrac{{I \times t}}{Z}$
where M is the molar or atomic mass of Indium, n is the valence or oxidation state of Indium.
$\Rightarrow \dfrac{{mn}}{M} = \dfrac{{It}}{Z}$
$\Rightarrow n = \dfrac{{ItM}}{{mZ}}$
$\Rightarrow n = \dfrac{{3.20 \times 40 \times 60 \times 114.8}}{{3.05 \times 96500}}$ [$1\min = 60\sec $ and $40\min = 40 \times 60\sec $ , $Z = 96500C$ (rounded off)]
$\Rightarrow n = \dfrac{{881664}}{{294325}}$
$\Rightarrow n = 2.99 \,i.e.,3$ (rounded off)

Therefore, option A is correct i.e., the oxidation state of Indium in its halide state is 3.

Note:
We should remember the two important Faraday’s laws of electrolysis because the formula i.e., $\dfrac{m}{E} = \dfrac{Q}{Z}$ has been derived from the combination of these two laws and we have rounded off the value of Z to make our calculations easier.