Electrolysis and Faraday's Laws

Write the electrode half-reaction. Count electrons needed per mole of product. Convert time to seconds. Calculate charge using Q = It. Convert charge to moles of electrons using Q/F. Use the half-reaction ratio to find product amount and then mass or volume.

A diagram is not compulsory for every question, but a simple electrolytic cell showing battery, anode, cathode, electrolyte, cation movement to cathode, and anion movement to anode helps distinguish it from a galvanic cell.

A common wrong answer is using n = 1 for copper deposition instead of n = 2 for Cu₂⁺ + 2e- → Cu.

Electrolysis is the use of electrical energy to drive a non-spontaneous chemical reaction, and Faraday's laws relate the amount of substance deposited or liberated to the quantity of electricity passed.

Electrolysis is the use of electrical energy to drive a non-spontaneous chemical reaction, and Faraday's laws relate the amount of substance deposited or liberated to the quantity of electricity passed. Faraday relation: Q = It. Mass deposited: m = ZIt. Also, moles of electrons = Q/F, where F = 96500 C mol-1 approximately. For Mn⁺ + ne- → M, moles of metal deposited = Q/(nF), and mass = molar mass x Q/(nF). During electrolysis of molten NaCl, sodium ions are reduced at the cathode to sodium metal, and chloride ions are oxidised at the anode to chlorine gas.

In electrolysis, oxidation occurs at the anode and reduction occurs at the cathode, but the electrode signs differ from a galvanic cell because an external power source drives the reaction. The mass deposited is directly proportional to charge passed. Charge is current multiplied by time. One faraday, approximately 96500 coulombs, corresponds to one mole of electrons. The product formed during electrolysis depends on electrode material, ion concentration, and electrode potentials. Faraday relation: Q = It. Mass deposited: m = ZIt. Also, moles of electrons = Q/F, where F = 96500 C mol-1 approximately. For Mn⁺ + ne- → M, moles of metal deposited = Q/(nF), and mass = molar mass x Q/(nF). Calculate mass of copper deposited when 1.93 A current passes through CuSO₄ solution for 500 s. Q = It = 1.93 x 500 = 965 C. For Cu₂⁺ + 2e- → Cu, n = 2. Moles of Cu = 965/(2 x 96500) = 0.005 mol. Mass = 0.005 x 63.5 = 0.3175 g. The small mass is reasonable because the charge is only one-hundredth of a faraday. Questions may ask mass deposited, gas volume liberated, time required, current required, products at electrodes, or comparison of products using electrode potentials. A common wrong answer is using n = 1 for copper deposition instead of n = 2 for Cu₂⁺ + 2e- → Cu.