Doc Brown's A Level Chemistry Revision Notes

Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7.1

7.1 Half cell equilibria and half–cell reactions and electrode potential

The equilibrium between a metal and its cation solution is first considered to introduce the idea of a half–cell potential and examples of half–cell reactions given.

Chemical Equilibrium Notes Part 7 Index


7.1 Examination of some redox equilibrium situations

  • (i) A metal atom and metal ion

    • When a piece of metal (M) is dipped into an aqueous solution of its ions (Mn+(aq)) an equilibrium exists between the two chemical states of the metal, and importantly, in different oxidation states (0 for the element and +n for the cation).

    • Mn+(aq) + ne (c) doc b M(s)

      • Note the use of IUPAC convention in writing half-equations for electrode reactions.

        • The half-reaction is written as an electron gain i.e. the reduction direction of chemical change.

        • You will see this notation used in tables of electrode potential (V) data later.

    • n is the numerical value of the positive charge on the cation and the number of electrons involved in the oxidation state change.

    • This is an example of a half–equation which represents the particular oxidation (electron loss) or reduction (electron gain) of an atom/ion related by electron transfer connecting two different oxidation states of the element.

    • The same idea of an equilibrium between two species of the same element in two different oxidation states expressed as a half–cell equation also applies to many other situations e.g.

  • (ii) A non–metal atom/molecule and associated anion e.g.

    • Cl2(g/aq) + 2e (c) doc b 2Cl(aq)

    • Oxidation states: chlorine molecule, 0, and chlorine in chloride ion, –1.

      • Note again the use of IUPAC convention in writing half-equations for electrode reactions, the half-reaction is written as an electron gain i.e. the reduction direction of chemical change.

  • (iii) Two different cations of the same metal e.g.

    • Fe3+(aq) + e (c) doc b Fe2+(aq)

    • Interchange of iron(II) and iron(III) ions, i.e. interchange of the +2 and +3 ox. states of iron.

  • (iv) An anionic and cationic species of the same metal e.g.

    • MnO4(aq) + 8H+(aq) + 5e (c) doc b Mn2+(aq) + 4H2O(l)

    • Interchange of manganate(VII) and manganese(II) ions in acid solution, +7 and +2 ox. states.

    • Note that half–cell equations are often, but not always, presented as a reduction (electron gain). For examples (ii) to (iv), a chemically inert platinum electrode is dipped into the solution when used in electrochemical cell construction (see later).

  • For an example of situation (i) and its consequences consider the diagram Fig.1 below comparing the relative positions of the metal–metal ion equilibrium for copper and zinc.

  • cell4 Fig.1

  • In the case of the copper atom/copper(II) ion equilibrium, you get a positive charge on a piece of copper when it is dipped into an aqueous copper(II) ion solution. This is due to an electron deficiency on the copper because the right–hand side of the equilibrium, the solid copper, is favoured, and copper(II) ions remove negative electrons to form copper atoms on the surface (reduction process). The +'s represent the electron deficiency in the copper metal, which in cells would make it the positive pole, and the 's represent surplus anion charge in the solution.

  • However, in the case of the zinc atom/zinc(II) ion equilibrium, you get a negative charge on a piece of zinc when it is dipped into an aqueous zinc(II) ion solution. This is due to an electron surplus on the zinc metal because the left–hand side of the equilibrium, the aqueous zinc(II) ion, is favoured, and zinc atoms lose negative electrons to form zinc(II) ions in the solution (oxidation process). For Zn/Zn2+: The +'s represent the extra zinc ion charge in solution, the 's represent the surplus electrons in the zinc metal which in cells would be the negative pole.

  • Quite simply, thermodynamically, zinc atoms have a much greater tendency or potential to lose electrons than copper atoms and copper(II) ions have a much greater tendency or potential to accept electrons than zinc ions. Therefore the two half–cell reactions would be

  • Zn(s) ==> Zn2+(aq) + 2e  and  Cu2+(aq) + 2e ==> Cu(s)

  • and the overall reaction is  Zn(s) +  Cu2+(aq) ==> Zn2+(aq) + Cu(s)

  • This overall reaction can be accomplished in two very different ways.

  • (a) by adding zinc metal to blue copper(II) sulphate solution when the blue colour fades and a red–brown deposit of copper forms. The reaction is exothermic which can be observed as a significant temperature rise if zinc powder is added to 1M copper(II) sulphate solution.

  • (b) By setting up a simple chemical cell ('battery') in which the two half–reactions are kept physically separated. To complete the electrical circuit, the zinc and copper(II) ion solutions are connected by a salt bridge (a supported ion solution). External wires connect strips of zinc and copper metal to a voltmeter or other device. The energy release from the exothermic reaction is now in the form of electrical energy. This, so–called Daniel cell is described in detail in the next section 7.2 Simple cells notation and construction.


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