High school basics / Redox reactions 5 / 6

Electrolysis and Industrial Applications

Electrolysis is the operation of supplying electrical energy from outside to force a redox reaction that would not occur spontaneously. It plays an important role in industrial production.

Basic Knowledge

Electrodes in electrolysis (note that the terms are used differently from batteries):
・Cathode: Connected to the negative terminal of the power source, where reduction occurs (positive ions receive electrons).
・Anode: Connected to the positive terminal of the power source, where oxidation occurs (negative ions give electrons, or the electrode itself dissolves).
Representative examples:
・Electrolysis of water (dilute H2SO4, etc.): H2 is produced at the cathode and O2 at the anode in a volume ratio of 2:1.
・Electrolysis of saltwater: H2 is produced at the cathode, Cl2 at the anode, and NaOH remains in the solution (ion-exchange membrane process).
・Electrorefining of copper: Electrolysis is carried out in CuSO4 using an impure copper plate as the anode and a pure copper plate as the cathode → 99.99% pure copper is obtained.

Important Terms
Electrolysis(The operation of forcing oxidation and reduction using an external power source)
Cathode(The negative side of the power source. Reduction occurs. Positive ions gather)
Anode(The positive side of the power source. Oxidation occurs. Negative ions gather)
Faraday’s law(The amount produced is proportional to the quantity of electricity (C) that flows)
Quantity of electricity(Q (C) = current I (A) × time t (s). 1 mol of electrons = 96500 C)
Molten-salt electrolysis(A refining method for metals with a large ionization tendency, such as Al and Na)

Further Study (Background・Significance)

The key to calculations involving electrolysis is the Faraday constant F = 96500 C/mol. Remember that the quantity of electricity carried by 1 mol of electrons is 96500 C.
Example: When a current of 1.0 A flows for 32 minutes 10 seconds (=1930 seconds), the quantity of electricity is Q = 1.0×1930 = 1930 C. The amount of substance of electrons that flowed = 1930/96500 = 0.020 mol.
Industrial applications:
・Production of aluminum: Aluminum oxide Al2O3 is reduced by molten-salt electrolysis together with cryolite. It consumes a large amount of electricity (an “electricity can”).
・Production of NaOH (electrolysis of salt): The ion-exchange membrane process is central to the chlor-alkali industry.
・Electrorefining of copper: Impurities Au and Ag settle as anode slime and are recovered.
These are core technologies supporting familiar metals and chemical products.

Key Points
  • Cathode = reduction; anode = oxidation
  • The terminology differs from batteries (battery: negative/positive electrodes; electrolysis: cathode/anode)
  • Faraday constant F = 96500 C/mol
  • Quantity of electricity Q = I × t (C = A × s)
  • Amount of substance of electrons = Q ÷ 96500
  • Electrolysis of water: H2:O2 = 2:1 (volume ratio)
  • Aluminum production = molten-salt electrolysis (an “electricity can”)

Points to Note (Easily Confused)

① The positive electrode of a battery and the anode in electrolysis have opposite roles (battery positive electrode = reduction; electrolysis anode = oxidation). Be careful not to confuse them. ② The anode reaction changes depending on the electrode material. With a copper electrode, the copper itself dissolves; with a platinum electrode, the negative ions in the electrolyte are oxidized. ③ The anode slime from copper electrorefining contains Au and Ag. ④ Remember carefully that 1 mol of electrons = 96500 C.

Practice

  1. At which electrode does reduction occur during electrolysis?
  2. Find the quantity of electricity and the amount of substance of electrons that flow when a current of 1.0 A flows for 16 minutes 5 seconds (965 seconds). (F=96500 C/mol)
  3. What method is used industrially to produce aluminum?

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