Question

State Faraday's first law of electrolysis. How much charge in Faraday is required for the reduction of 1 mol of MnO\(_4^-\) to Mn\(^{2+}\)?

Hint:

- Faraday's law helps determine the amount of substance deposited during electrolysis based on the charge passed through the electrolyte. - The number of electrons involved in the reduction or oxidation process is key to calculating the total charge.

Answer 1

Faraday's first law of electrolysis states that:
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte. The equation is: \[ m = ZIt \] Where: - \(m\) is the mass of the substance deposited, - \(Z\) is the electrochemical equivalent, - \(I\) is the current, - \(t\) is the time. For the reduction of 1 mole of MnO\(_4^-\) to Mn\(^{2+}\), the number of electrons required is 5 (because MnO\(_4^-\) is reduced to Mn\(^{2+}\) by gaining 5 electrons). Thus, the charge required to reduce 1 mole of MnO\(_4^-\) to Mn\(^{2+}\) is given by: \[ Q = n \times F = 5 \times 96500 = 482500 \, \text{Coulombs}. \] Therefore, the charge required for the reduction of 1 mole of MnO\(_4^-\) to Mn\(^{2+}\) is 482500 Coulombs.