Some problems of electrolysis

Some problems of electrolysis

Q 1. Find the charge of an electron.

Ans: The charge of an electron (e) is calculated using the formula:

e = F / N_A

Where:

• F = Faraday’s constant = 96,485 C/mol
• N_A = Avogadro’s number = 6.022 × 10²³ mol⁻¹

Substituting the values:

e = 96485 ÷ 6.022 × 10²³

e ≈ 1.602 × 10⁻¹⁹ C

Therefore, the charge of an electron is approximately −1.602 × 10⁻¹⁹ coulombs.

Q 2. Find the charge of the following ions:
i) Cu²⁺    ii) PO₄³⁻

Ans:
The charge (q) of an ion is calculated using the formula:
q = n × e = n × (F / N_A)
Where:
    n = number of electrons gained or lost
    F = Faraday constant = 96500 C/mol
    N_A = Avogadro’s number = 6.022 × 10²³ mol⁻¹

i) For Cu²⁺ (n = 2):
q = 2 × (96500 / 6.022 × 10²³) C ≈ 3.204 × 10⁻¹⁹ C (positive charge)

ii) For PO₄³⁻ (n = 3):
q = 3 × (96500 / 6.022 × 10²³) C ≈ −4.806 × 10⁻¹⁹ C (negative charge)