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  • The use of electric current to stimulate a non-spontaneous reaction. Electrolysis can be used to separate a substance into its original components/elements and it was through this process that a number of elements have been discovered and are still produced in today's industry. In Electrolysis, an electric current it sent through an electrolyte and into solution in order to stimulate the flow of ions necessary to run an otherwise non-spontaneous reaction. Processes involving electrolysis include: electro-refining, electro-synthesis, and the chloro-alkali process.

    Electrolytic Cell vs Voltaic Cell

    Example: When we electrolyze water by passing an electric current through it, we can separate it into hydrogen and oxygen.

    \[ 2 H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]

    More information : The Electrolysis of Water

    An electrolytic cell is essentially the non-spontaneous reaction's voltaic cell, (in fact if we reversed the flow of electricity within a voltaic cell by exceeding a required voltage, we would create an electrolytic cell). Electrolytic cells consist of two electrodes (one that acts as a cathode and one that acts as an anode), and an electrolyte. Unlike a voltaic cell, reactions using electrolytic cells must be electrically induced and it's anode and cathode are reversed (anode on the left, cathode one the right).

    Voltaic Electrolytic
    Oxidation: X → X+ + e- (Negative Anode) Y → Y+ + e- (Positive Anode)
    Reduction: Y+ + e- → Y (Positive Cathode) X+ + e- → X (Negative Cathode)
    Overall: X + Y+→ X+ + Y (G<0) X+ + Y → X + Y+ (G>0)
    This reaction is spontaneous and will release energy This reaction is non-spontaneous and will absorb energy


    Factors Affecting Electrolysis Reactions

    1. Overpotential- The generated voltage is significantly higher than expected. An overpotential may be necessary to overcome interactions taking place on the electrode itself (especially for gasses).
    2. Electrode type- An inert electrode acts as a surface for a reaction to occur on and is not involved in the chemical reaction whereas an active electrode becomes a part of the half reaction.
    3. Simultaneous electrode reactions- If two different pairs of half-reactions take place at once. Some half reactions should be eliminated in order to determine a single pair of half reactions best suited for the electrolysis to occur.
    4. The state of reactants- If reactants are in nonstandard states, the voltage of half cells may differ from that of the standard amount. In this case, the solution for the anode half cell may have a pH that is either higher or lower than the standard pH of 4 which may lead to a nonstandard voltage as well.


    Faraday's Constant-The amount of electric charge associated with one mole of electrons.

    Faraday's Constant: 1 mole e- = 96,485 C (where C measures Coulombs)

    To find the amount of moles of electrons that have been involved in an electrolysis reaction use the following equation:

    Charge (C)= current (C/s) x time (s) x (1 mole e-/96,485 C)


    Several processes of electrolysis are used in today's industry:

    1. Electrorefining

    the anode is the impure metal and any impurities are removed during the process of electrolysis when the metal travels from anode to cathode. During the electrorefining of metals, the cathode has a decomposition of pure metals from a solution containing the metal ion. For example copper is purified through electrolysis in order to be used for applications that require high electrical conductivity. During this process, the cathode is a pure piece of copper, while the anode is an impure piece of copper. The Cu2+ from the anode moves through a sulfuric acid-copper(II) sulfate solution into the cathode where it becomes solid copper. While this is occurring, the impurities are left at the bottom of the tank. This leftover residue is called anode mud. The electrolysis is carried out at 0.15 - 0.30V (low voltage) in order to make sure that Ag, Au, and Pt impurities are not oxidized while in the anode and become anode mud. Whereas most of the other components become oxides or hydroxides and form water-soluble species.

    1. Electrosynthesis

    a method which uses electrolysis reactions to produce certain products. For example MnO2 needs to undergo electrolysis in order to be used for alkaline batteries. The solution for the electrosynthesis of MnO2is MnSO4in H2SO4. The anode is graphite, where Mn2+ is oxidized. While at the cathode, hydrogen is reduced from H+ to H2.

    Overall Reaction: Mn2+(aq) + 2H2O(l)--> MnO2(s) + 2H+(aq) + H2(g)

    1. Chloro-Alkali Process

    Electrolysis of seawater which leads to the production of chlorine and the alkali, sodium hydroxide. There are 3 different methods in which these two components are produced: membrane cell, diaphragm cell, and mercury cell process.

    1. Mercury Cell Process

    Electrolysis of seawater in a mercury cell leads to the production of chlorine and sodium hydroxide at the same time. This method involves using mercury as the cathode and graphite as the anode.The mercury attracts either sodium or potassium cations and the mercury forms an amalgam with it. However when the amalgam is introduced to water it forms sodium hydroxide and hydrogen leaving the mercury to be reused later. The chlorine gas is left to form at the anode.

    1. Diaphragm Cell Process

    The diaphragm cell has Cl2 being released from the anode section, while there is H2 being released from the cathode section. If Cl2 manages to mix with NaOH, the Cl turns into other products. Therefore the diaphragm cell has a bigger amount of NaCl,and a smaller amount of solution in the cathode in order for the NaCl to come in contact with the other solution gradually, while simultaneously preventing backflow of NaOH.

    1. Membrane Cell Process

    This process is more efficient than others as it does not use mercury and does not require a significant amount of energy. Contains a cation-exchange membrane which is usually made from flourocarbon polymer. This membrane allows hydrated cations to pass in between the anode and cathode compartments, but does not allow the backflow of the ions, Cl- and OH-. This allows the sodium hydroxide produced to have less contamination by chloride ions.


    1. Miller, Fredrick P.,McBrewster, John, Vandome, Agnes F. Cathode: Electrode, Electric current, Anode, Battery (electricity), Cathode bias, Electrolysis, Electrolytic cell, Electron. United States: Alphascript Publishing, 2009.
    2. Hale, Arthur J. The Applications of Electrolysis in Chemical Industry. New York:General Books LLC, 2010.
    3. Petrucci, Ralph H., Harwood, William S., Herring, F. G., and Madura Jeffrey D. General Chemistry: Principles and Modern Applications. 9th ed. Upper Saddle River: Pearson Education, Inc., 2007.
    4. Hoffman, Brittany; Mitchell, Elizabeth; Roulhac, Petra; Thomes, Marc; Stumpo, Vincent M. "Determination of the Fundamental Electronic Charge via the Electrolysis of Water." J. Chem. Educ. 2000 77 95.


    1. What is electrolysis?
    2. How does an electrolytic cell differ from a voltaic cell? Explain.
    3. Which requires a higher voltage for electrolysis to occur? O2 (g) or Sn (s).
    4. Where does reduction occur in a voltaic cell and an electrolytic cell?
    5. At the cathode, the half reaction(reduction) for zinc is Zn2+(aq) + 2e- --> Zn(s).
    6. What is the mass of zinc which will be deposited after an hour with a current of 1.5 A?


    1. Electrolysis is the use of an electric current through an electrolyte to create a non-spontaneous chemical reaction.
    2. Unlike the voltaic cell, the electrolytic cell uses electric energy for a non-spontaneous reaction instead of a spontaneous reaction. The cathodes and anodes are reversed from each other.
    3. Oxygen requires a higher voltage because it is a gas, whereas Sn is in a solid state and requires less voltage for electrolysis to occur.
    4. Reduction occurs in the cathode for the voltaic cell, while reduction occurs in the cathode for the electrolytic cell.
    5. 1 hour X (60 min/1 hour) X (60 sec/1 min) X (1.62 C/ 1sec) X (1 mol e-/ 96,485 C)= .0604 mol e-
    6. .0604 mol e- X (1 mol Zn/2 mol e-) X ( 65.38 g Zn/ 1 mol Zn)= 1.975 g Zn


    • Kimberly Song (UC Davis)