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

According to Faraday's second law of electrolysis, the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity that flows through the electrolyte. To determine the quantity of electricity needed to reduce 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\), we apply Faraday's law. The calculation uses n = 6, representing the 6 electrons involved in the reaction: \[ Q = nF = 6 \times 96500 = 579000 \, \text{C} \]