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Electrochemistry Review

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    Chemical reactions involving transfer of electrons are called oxidation and reduction reactions or redox reactions. When properly set up, these reactions generate power in a Battery, or galvanic cell. A galvanic cell consists of at least two half cells, each of which consists of an electrode and an electrolyte solution.

    A redox reaction can be divided into two Half Reactions, an oxidation reaction and a reduction half reaction.

    Each half reaction can be set up as a Half Cell and putting two half cells together makes a simple battery. A galvanic cell can be considered as a battery, but batteries include packages of galvanic cells in series to supply higher voltages than a single galvanic cell.

    Oxidation of a species (atom, ion, or molecule) is a loss of electron(s), and reduction of a species a gain of electron(s). Some conventions are used to define the Oxidation States, and oxidation of an atom causes an increase in its oxidation state. Conversely, reduction causes a decrease in its oxidation state.

    Balancing Redox Equations is a complicated task, but the use of half reactions is a very good strategy.

    Energy is the driving force for chemical reactions. Energies of oxidation and reduction reactions are related to the electrochemical potentials (E), Cell EMF, of the galvanic cells. The standard reduction potentials for half cells are values that make sensible comparisons, because a set of conventions have been followed. The energy (called Gibb's free energy G) is the electric work of the reaction. Therefore,

    \(\ce \Delta G^\circ = n F \, \ce D E^\circ\).

    The difference of Gibb's free energy (\(\ce \Delta G^\circ\)) between products and reactants is equal to the charge n F times the potential difference DEo. Since n electrons are transferred in the equation, n F (F is the Faraday constant of 96485 C) is the charge involved for the equation in terms of moles.

    The equilibrium constant K is related to \(\ce \Delta G^\circ\),

    \(\ce \Delta G^\circ = - R T \ln K\).

    For the reaction,

    \(\mathrm{a\: A + b\: B \rightleftharpoons c\: C + d\: D}\),

    the Nernst Equation is a natural result:

    \(\ce \Delta E = \ce \Delta E^\circ - \dfrac{R T}{n F} \ln \mathrm{\dfrac{[C]^c[D]^d}{[A]^a[B]^b}}\)

    Notations Used

    In the summary given above, we have used many energy and potential related notations. For completeness, the notations are given as follows:

    • DG - difference in Gibb's free energy
    • DGo - difference in Gibb's free energy at standard condition
    • Eo - standard reduction cell potential
    • E* - standard oxidation cell potential
    • emf or EMF - Electromotive force
    • DEo - standard potential of a battery or galvanic cell
    • DE - Potential of a galvanic cell not necessarily at standard conditions
    • F = 96485 C, Faraday constant
    • Q - charge, unit coulomb (C)
    • I = Q / t, current Ampere
    • R = 8.3145 J/mole.K, gas constant
    • T - temperature in K
    • ln(10) R T / F = 0.0592 at 298 K

    Electrochemistry Skills Expected in Quiz

    1. Explain the meaning of terms used in electrochemistry and properly use the cell notations. For example, galvanic cells or voltaic cells are devices that use spontaneous chemical reactions to produce electric currents. Other terms are reduction potentials, oxidation potentials, oxidation state, redox reactions, cathode, anode, cell potentials, Nernst equations, etc.

      The definitions for anode and cathode apply to both galvanic cells and electrolytic cells.
      Oxidation takes place on anodes, for example,

      &\textrm{reaction:}\: \ce{Zn \rightarrow Zn^2+ + 2 e^-} &&E^* = 0.762\\
      &\textrm{notation:}\: \ce{Zn \,|\, Zn^2+} &&E^* = - E^\circ

      Note that we have defined and used E* to represent the oxidation potential which has the same absolute value as the reduction potential Eo, but a different sign. This notation is not used in most text books.

      Reduction takes place on cathodes, for example,

      &\textrm{reaction:}\: \ce{Hg2^2+ + 2 e^- \rightarrow 2 Hg} &&E^\circ = 0.796\\
      &\textrm{notation:}\: \ce{Hg^2+ \,|\, Hg} &&

      The details regarding oxidation and reduction half reactions have been given on the page of Half-Cell Reactions.

    2. Calculate DEo and DE for galvanic cells according to the conventions used for cell notations.

      The standard cell potential DEo is the sum of the reduction potential of the cathode and the oxidation potential of the anode,

      \(\ce D E^\circ = E^\circ + E^*\).

      The data for calculating DEo may be given using cell notations or in the form of equations. If a cell notation is given, you must be able to search for proper reduction and oxidtation couples for the cell and then evaluate Eo's from a table of standard reduction potentials. Proper use of such tables is a basic skill, and some usages have been given in the modules Battery or Galvanic Cells, and Electromotive Force.
    3. If the cells are not at standard conditions, you will be required to calculate the Gibb's free energy, DG, or the cell potential DE.

      The standard Gibb's free energy, DGo, is the negative of the maximum available electrical work. The electrical work We is the product of the charge and the potential of the cell.

      W_{\ce e} &= q E\\
      \ce D G &= - q E\\
      &= - n F E

      where q is the charge in C and E is the potential in V. Note also that q = n F, and F = 96485 C is the Faraday constant, whereas n is the number of moles of electrons in the reaction equation.

      The change of Gibbs free energy at standard condition is derived in the following way:

      \ce D G^\circ &= - W_{\ce e}\\
      &= - q E^\circ\\
      &= - n F E^\circ

      This equation is for a galvanic cell under the standard conditions. The notations Eo and DGo are used for cells under standard conditions. When the cells are not at standard conditions, DE and DGo are used for these quantities.

      Note also that DGo is the theoretical amount of energy per reaction equation as written. The amount of energy for a given system depends on the quantities of the reactants and their conditions (concentration or pressure etc.).

    4. The Nernst equation allows us to evaluate the cell potentials when the condition of the cell is not at standard conditions. The details have been given on the Nernst Equation page.
    5. For the calculation of Gibb's energies and quantities generated in Electrolysis and electroplating, skills in solving stoichiometric problems are also required.


    1. How many minutes will be required for a 1.5 A current to electroplate 1.97 g of gold (at. mass, 197.0)?
    2. For the Daniell cell,

      \(\ce{Zn \,|\, Zn^2+ \,||\, Cu^2+ \,|\, Cu}\),

      calculate the dE0. Look up data from a table of standard reduction potential. You need to have the skill. That is why no data is given.
    3. For the Daniell cell,

      \(\ce{Zn \,|\, Zn^2+ \,||\, Cu^2+ \,|\, Cu}\),

      dE0 = 1.10 V. Calculate the Kc for the reaction,

      \(\ce{Cu^2+ + Zn \rightarrow Zn^2+ + Cu}\)

    4. For the Daniell cell,

      \(\mathrm{Zn \,|\, Zn^{2+}\: (0.010\: M) \,||\, Cu^{2+}\: (1.0\: M) \,|\, Cu}\),

      calculate dE.
    5. For the Daniell cell,

      \(\ce{Zn \,|\, Zn^2+ \,||\, Cu^2+ \,|\, Cu}\),

      dE0 = 1.10 V. Calculate dG0 in J.
    6. For the Daniell cell,

      \(\mathrm{Zn \,|\, Zn^{2+}\: (0.010\: M) \,||\, Cu^{2+}\: (1.0\: M) \,|\, Cu}\),

      calculate dG in J.
    7. What is the dE0 for the reaction

      \(\ce{Pb_{\large{(s)}} + Co^2+ \rightarrow Pb^2+ + Co_{\large{(s)}}}\)?

    8. What is the equilibrium constant for the reaction

      \(\ce{Pb_{\large{(s)}} + Co^2+ \rightarrow Pb^2+ + Co_{\large{(s)}}}\)?


    1. Hint...

      \(\mathrm{1.97\: g\: Au \:\dfrac{1\: mol\: Au}{197\: g\: Au} \:\dfrac{1\: mol\: e^-}{1\: mol\: Au} \:\dfrac{96485\: C}{1\: mol\: e^-} \:\dfrac{1\: s}{1.5\: C} \:\dfrac{1\: min}{60\: s} =\: ?\: min}\)

      Answer 10.72

      The reaction is

      \(\ce{Au(CN)- + e^- \rightarrow Au + 2 CN-}\)

      In exams, the reaction

      \(\ce{Al^3+ + 3 e^- \rightarrow Al}\)

      is usually used.
      Pay attention to the number of electrons involved in the reaction.

    2. Hint...
      Did you get these data?

      &\ce{Zn \rightarrow Zn^2+ + 2 e^-} &&E^* = 0.762\\
      &\ce{Zn \,|\, Zn^2+} &&\textrm{Note: } E^* = -E^\circ\\
      &\ce{Cu \rightarrow Cu^2+ + 2 e^-} &&E^* = -0.339\\
      &\ce{Cu \,|\, Cu^2+} &&\textrm{Note: } E^* = -E^\circ

      dE = E0 + E* Note sign for convention? Calculate dE0

      Answer 1.10 V
      The reaction is \(\ce{Zn + Cu^2+ \rightarrow Zn^2+ + Cu}\)

    3. Hint...
      At equilibrium, dE = 0;

      \(\ce d E^\circ = \dfrac{0.0592}{2}\log K_{\ce c}\)

      Kc = antilog 37.2 = ?

      Answer 2e37
      You may be asked to calculate

      \(\mathrm{[Zn^{2+}] / [Cu^{2+}] =\: ?}\)


      \(\mathrm{[Cu^{2+}] / [Zn^{2+}] =\: ?}\)

      The large Kc value means that the reaction is almost quantitative. The \(\ce{Zn}\) metal almost causes all \(\ce{Cu^2+}\) ions to deposit as copper metal.

      \(\ce{Cu^2+ + Zn \rightarrow Zn^2+ + Cu}\)

    4. Hint...
      Assume you've done the previous problems.

      \ce d E &= \ce d E^\circ - \dfrac{0.0592}{2}\log \dfrac{0.01}{1.0}\\
      &=\: ?

      Answer 1.16
      The voltage is higher than the standard potential of 1.10 because \(\ce{Zn^2+}\) concentration is less than 1.0 M in this case. The Nernst equation enables us to give a quantitative value when the conditions change.

    5. Hint...

      \ce d G^\circ &= - n F \:\ce d E^\circ\\
      &= - 2 \times 96485 \times 1.10 =\: ?

      Answer -212267 J
      More often, you'll see -212 kJ rather than -212000 J

    6. Hint...
      From \(\ce d E = \ce dE^\circ - \dfrac{0.0592}{2} \log \ce K = 1.26\: \ce V\: \ce d G = - n F \:\ce d E = - 2 \times 96485 \times 1.16 =\: ?\)

      Answer -223845 J
      dG0 = - 212 kJ; dG for the said condition is -224 kJ. Concentration makes a difference.

    7. Hint...
      The purpose of this problem is to ask you to search the table for the proper reduction potentials. If you get dE0 = -0.151 V, you are probably right. The negative potential indicates that the reaction should be reversed.

    8. Hint...
      Use the Nernst equation to derive the equilibrium constant. If you get a value of K = 7.6x10-6, you have acquired the skill.

    We have used mostly the \(\ce{Zn/Cu}\) cell in these questions. A similar set of questions can be set up using any two couples of reduction potentials.

    Next page: Sulfur

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