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Redox Equilibrium

Redox Equilibrium. Objectives. Define oxidation and reduction in terms of electron loss and gain. Deduce the oxidation number of an element in a compound. State the names of compounds using oxidation numbers.

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Redox Equilibrium

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  1. Redox Equilibrium

  2. Objectives • Define oxidation and reduction in terms of electron loss and gain. • Deduce the oxidation number of an element in a compound. • State the names of compounds using oxidation numbers. • Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.

  3. Introduction • Redox reactions are reactions that involve the transfer of electrons between chemical species. • Electron transfer from one place to another is a flow of electrons and a flow of electrons is an electric current. • This study of redox reactions is therefore the combined study of chemistry and electricity. • In fact, an alternative title for this chapter could well be ‘electrochemistry’. The applications of electrochemistry include; • the use of redox reactions inside batteries to generate an electric potential difference (voltage). • predicting whether chosen substances will take part in a redox reaction.

  4. Some terms you need to know • The term redox is a contraction of the words reduction and oxidation. • Oxidation and reduction take place together; simultaneously. • One substance is reduced while the other is oxidised. • A substance that reduces another substance is called a reducing agent, whilst the substance itself will be oxidised. • A substance that oxidises another substance is called an oxidising agent, whilst the substance itself will be reduced.

  5. The Oxidation of Magnesium • An example of a redox reaction is magnesium burning in oxygen to produce magnesium oxide. 2Mg(s) + O2(g) → 2MgO(s) • Magnesium is a metal, oxygen is a diatomic covalent gas, and magnesium oxide is an ionic compound containing Mg2+ and O2- ions. • A simple definition of redox reactions considers oxidation as the gain of oxygen atoms and reduction as the loss of oxygen atoms. Applying this definition to the burning of magnesium: • Magnesium is oxidised by oxygen • Oxygen is reduced by magnesium

  6. Half equations • The balanced chemical reaction does not show electron transfer. Splitting the equation into two half-equations reveals more clearly what is happening in the course of the reaction. • One half equation shows the loss of electrons and the other shows the gain of electrons. • Magnesium loses electrons: 2Mg(s) → 2Mg2+(s) + 4e- • Oxygen gains electrons O2(g) +4e- → 2O2-(s)

  7. Half Equations • The half reaction described by the half equation does not necessarily actually occur; the electrons may not become free in the way suggested. • Half equations are simply a useful way of thinking about the two component parts of a redox reaction. • A half reaction must balance with respect to charge as well as with respect to the number of atoms. • Notice how the sum of the charges on the left equals the sum of the charges on the right. 2Mg(s) → 2Mg2+(s) + 4e- O2(g) + 4e- → 2O2-(s)

  8. Electron Transfer • We can consider the reaction in terms of the underlying electron transfer: • Oxidation is the loss of electrons • Reduction is the gain of electrons • Oxidation is caused by the oxidant (or oxidising agent): the oxidant (oxygen O2) is itself reduced in the process (to O2-) • Reduction is caused by the reductant (Magnesium Mg) is itself oxidised in the process (to Mg2+)

  9. Redox in the absence of oxygen • The example given so far shows that the oxidation of magnesium may be interpreted in two ways: either as a gain of oxygen or as a loss of electrons. • The idea of redox as electron transfer allows many reactions not involving oxygen to be classified as redox reactions.

  10. The reaction between aqueous copper (II) sulphate and zinc metal is a redox reaction: Zn(s) + CuSO4(aq) → ZnSO4(aq) +Cu(s) • Remember that the formulae CuSO4(aq) and ZnSO4(aq) refer to ionic substances dissolved in water: • CuSO4(aq) is therefore present as Cu2+(aq) and SO42-(aq) ions. • ZnSO4(aq) is present as Zn2+(aq) and SO42-(aq) ions. • The sulphate ion SO42-(aq) is a spectator ion; it does not take part in the chemical reaction and so we may disregard it.

  11. The full balanced chemical equation is not very informative. However, the two half-equations reveal the redox nature of the reaction. • The half-equation for zinc shows that the metal atoms are oxidised as the result of electron loss: Zn(s) → Zn2+(aq) + 2e- • The half-equation for copper shows that the aqueous copper (II) ions are reduced as the result of electron gain: Cu2+(aq) + 2e- → Cu(s) • The balanced equation for the reduction of copper (II) ions by zinc results from adding together the two half-equations. • In this example the combination is straightforward because both half-equations involve two electrons (note that we have not included the sulphate spectator ions Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

  12. Using Half Equations – an example • Chlorine gas oxidises iodide ions in solution to iodine, and in the process is reduced itself to chloride ions. Write the ionic equation for the reaction. • All you need to do is to write the two half equations, and then combine them. Chlorine gas is Cl2; chloride ions are Cl-. Write those down on each side of the equation. Cl2 → Cl-

  13. Using Half Equations – an example • Chlorine gas oxidises iodide ions in solution to iodine, and in the process is reduced itself to chloride ions. Write the ionic equation for the reaction. • All you need to do is to write the two half equations, and then combine them. Now balance the atoms Cl2 → 2Cl- Cl2 → Cl-

  14. Using Half Equations – an example • Chlorine gas oxidises iodide ions in solution to iodine, and in the process is reduced itself to chloride ions. Write the ionic equation for the reaction. • All you need to do is to write the two half equations, and then combine them. Finally, the charges need to be balanced by adding electrons where necessary. In this case, you need two on the left hand side: Cl2 + 2e-→ 2Cl- Cl2 → 2Cl-

  15. Using Half Equations – an example • Chlorine gas oxidises iodide ions in solution to iodine, and in the process is reduced itself to chloride ions. Write the ionic equation for the reaction. • All you need to do is to write the two half equations, and then combine them. Repeating this process (first balancing the atoms and then the charges) for the other half equation involving the iodine gives you: 2I- → I2 + 2e-

  16. Using Half Equations – an example • Chlorine gas oxidises iodide ions in solution to iodine, and in the process is reduced itself to chloride ions. Write the ionic equation for the reaction. • All you need to do is to write the two half equations, and then combine them. These two equations can now be literally added together: Cl2 + 2e-→ 2Cl- Because there are two electrons on both sides, they cancel out, and we are left with the ionic equation 2I- → I2 + 2e- --------------------------------- Cl2 + 2I-→ 2Cl- + I2

  17. Rules for constructing half equations • You are only allowed to write certain things into half equations: • The substance you start from and what it is oxidised or reduced to. • Hydrogen ions, H+ (unless the solution is alkaline, in which case these are replaced by hydroxide ions, OH-) • Water • Electrons

  18. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. Start with hydrogen peroxide reducing to water: H2O2 → H2O

  19. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. Balance the oxygens: H2O2 → 2H2O H2O2 → H2O

  20. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. There aren’t enough hydrogen on the left hand side, add two hydrogen: 2H+ + H2O2 → 2H2O H2O2 → 2H2O

  21. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. The atoms balance, but not the charges. There are two positive charges on the left, but zero charge on the right. Add two electrons to the left: 2H+ + H2O2 + 2e- → 2H2O 2H+ + H2O2 → 2H2O

  22. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. Now for the iron, this is much easier; Fe2+ → Fe3+

  23. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. Everything is OK except the charges. Add an electron to supply a negative charge to the right, and cut 3+ down to 2+ Fe2+ → Fe3+ Fe2+ → Fe3+ + e-

  24. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. This leaves 2H+ + H2O2 + 2e- → 2H2O Fe2+ → Fe3+ + e-

  25. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. Warning; you can’t just add these equations together – the electrons will not cancel out, we must multiply the second equation by two. 2H+ + H2O2 + 2e- → 2H2O x2 2Fe2+ →2 Fe3+ + 2e- Fe2+ → Fe3+ + e-

  26. Example • An acidic solution of hydrogen peroxide, H2O2, oxidises iron (II) ions to iron (III) ions. The hydrogen peroxide is reduced to water. Write the ionic equation for the reaction. Now the electrons balance and can cancel out: 2H+ + H2O2 + 2e- → 2H2O 2Fe2+ →2 Fe3+ + 2e- 2H+ + H2O2 + 2Fe2+ → 2H2O + 2Fe3+

  27. Quick questions • For each of the following redox reactions identify the (i) oxidizing agent, (ii) the reducing agent, (iii) the species which is oxidised and (iv) the species which is reduced. • 2Ca(s) + O2(g) → 2CaO(s) • CuO(s) + H2(g) → Cu(s) + H2O(l) • 2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq) • For each of the reactions above write half equations to show (i) electron loss during oxidation and (ii) electron gain during reduction.

  28. Oxidation numbers • Each element in any chemical species may be assigned an oxidation number. The oxidation number of an element is the number of electrons that need to be added to the element to make a neutral atom. • E.g. The iron ion Fe2+ requires the addition of two electrons to make a neutral atom. The oxidation number for iron in the Fe2+ ion form is therefore +2. • The oxidation state of this ion is written as iron(II) or Fe(II); it is expressed in Roman numerals and describes the extent of oxidation of the species.

  29. Assigning oxidation numbers • Uncombined elements always have oxidation number = 0 • Some elements always have the same oxidation numbers in their compounds: • Group I metals always +I • Group II metals always +II • Al always +III • H, +I, except in compounds with Group 1 or 2 metals (metal hydrides) where it is –I • F, always –I • O, -II, except in peroxides and compounds with F, where it is –I • Cl, -I except in compounds with F and O, where it has positive values. • The sum of all the oxidation numbers in a compound = 0 • The sum of the oxidation numbers of a complex ion = the charge of the ion

  30. Quick questions • What is the oxidation number of the metal M in each of the following: • MCl • M2O3 • MSO4 • M

  31. Oxidation of Iron (II) • The Fe3+ ion has an oxidation number of +3. The oxidation state of this ion is written as iron (III) or as Fe(III) , indicating that it is a more highly oxidised species than Fe2+ • When an iron (II) ion,(Fe2+)reacts to form an iron (III) ion, (Fe3+), an electron is lost thus the Fe(II) is oxidised: Fe2+(aq) → Fe3+ (aq) + e- • When it is oxidised, the oxidation number change for iron is from +2 to +3; the oxidation state change for iron is from Fe(II) to Fe(III) • Oxidation involves an increase in oxidation number.

  32. Reduction of Iron (III) • In the reverse process, reduction , an iron (III) ion gains one electron to become an iron (II) ion: Fe3+ (aq) + e- → Fe2+ (aq) • When it is reduced, the oxidation number change for iron is from 3+ to 2+; the oxidation state change for iron is from Fe(III) to Fe(II). • Reduction involves a decrease in oxidation number.

  33. Reactivity Series • As some substances are clearly better at oxidizing than others we can place them in order of their oxidizing ability. • Experimentally this series was arrived at initially by seeing which metal could displace the ions of another metal in aqueous solution. • The most reactive metals are those that give up electrons most readily, so the reactivity series is normally written as a reduction series with the best reducing agents at the top of the table

  34. The Reactivity series Activity: • Carry out Experiment 9.1 or watch the video by clicking on the test-tube. • Record your results in an appropriate table. • Write down the half equations for each of the reactions. • Arrange the half equations in order of reactivity.

  35. The Reactivity Series Increasing reducing ability

  36. Electrical energy from a redox reaction • Activity: Carry out Experiment 9.2 or watch the video.

  37. Voltaic cells • A simple voltaic cell is a device for obtaining electrical energy from a spontaneous chemical reaction. • The oxidation reaction and the reduction reaction, as represented by their two half equations, are separated, and each is carried out experimentally in a separate electrochemical half cell. The simplest half cell consists of a metal in contact with a solution of its own ions e.g. A zinc half cell consists of a strip of zinc metal placed in an aqueous solution of zinc sulphate. An equilibrium will be established between the metal and the ions in solution Zn

  38. Some electrochemical cells • Activity: Carry out Experiment 9.3 or watch the video. • Make a note of the voltage changes and compare the values. • Make a note of the effect of heating and dilution on e.m.f. Values.

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