Question

State Faraday’s second law of electrolysis. How much electricity is required in terms of Faraday for the reduction of 1 mole of $Cr_2\text{O}_7^{2-} \text{ to Cr}^{3+}$?

Hint:

For electrolysis problems, always calculate the number of electrons involved (n) and use Faraday’s constant (F = 96500 C/mol) to find the total charge (Q).

Answer 1

Faraday’s second law of electrolysis states that the masses of substances deposited or liberated at the electrodes are directly proportional to the quantity of electricity passed through the electrolyte.
The quantity of electricity required for the reduction of 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\) can be calculated using Faraday's law, where n = 6 (since 6 electrons are involved): \[ Q = nF = 6 \times 96500 = 579000 \, \text{C} \]