Question

The element indium is to be obtained by electrolysis of a molten halide of the element. Passage of a current of 3.20 A 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 mass of In=114.8)

A.3
B.2
C.5
D.1

Answer 1

The oxidation state of indium in the halide melt is +2. (Option B)

Step-by-step explanation:

Faraday's Law:

Faraday's Law of electrolysis relates the amount of substance produced at an electrode to the quantity of electricity passed through the electrolyte. It can be expressed as:

m = zQ / F

where:

m is the mass of substance deposited

z is the number of electrons transferred per ion

Q is the quantity of electricity passed

F is Faraday's constant (96485.33 C/mol)

Calculate the moles of indium formed:

Mass of indium (m) = 3.05 g

Atomic mass of indium (M) = 114.8 g/mol

Moles of indium (n) = m / M

n = 3.05 g / 114.8 g/mol

n ≈ 0.0266 mol

Calculate the quantity of electricity passed:

Current (I) = 3.20 A

Time (t) = 40.0 min = 2400 s

Quantity of electricity (Q) = I * t

Q = 3.20 A * 2400 s

Q = 7680 C

Calculate the number of electrons transferred:

Let z be the oxidation state of indium in the halide melt.

For every ion of indium deposited, z electrons are transferred.

Therefore, the number of electrons transferred is equal to the moles of indium multiplied by the oxidation state:

z * n = Q / F

Substitute the known values:

z * 0.0266 mol = 7680 C / 96485.33 C/mol

z ≈ 0.22

Determine the oxidation state:

Since the oxidation state of an element is an integer, the closest integer to 0.22 is +2. Therefore, the oxidation state of indium in the halide melt is +2.