Lecture Notes 3 rd Series: Electrochemistry Oxidation number or states When atoms gain or lose electrons they are said to change their oxidation number or oxidation state. If an element has gained electrons in a chemical reaction it is said to be reduced and its oxidation number becomes more negative (or less positive). If an element loses electrons in a chemical reaction it is said to be oxidized and its oxidation number becomes more positive (or less negative). Consider the following chemical reaction : Ca(s) Ca 2+ (aq) Ca(s) has an effective charge of zero. Therefore its oxidation number is set at zero. Ca 2+ (aq) has an effective charge of two plus. Its oxidation number is given as +2. Thus the oxidation number has increased (become more positive) during the reaction. This is called an oxidation. Oxidation refers to any process in which the oxidation number of an atom becomes more positive For the above reaction we can write a reaction halfreaction. Ca(s) Ca 2+ (aq) + 2e For a reaction ½ equation it is important to note that the overall charge on the LHS of the equation (zero) is equal to the overall charge on the RHS of the equation. [(2 + ) + (2 x 1)] = 0 from an electron If we take another example. Cl 2 (g) 2Cl (aq) Here the oxidation number of Cl 2 (g) is zero and the oxidation number of the chloride ion is minus one. Therefore the oxidation number has become more negative during the reaction and therefore chlorine must have gained electrons and become reduced. We can therefore write a balanced ½ reaction as Cl 2 (g) + 2e 2Cl (aq) Reduction refers to any process that leads to a decrease in the oxidation number of an atom There is one most important thing to remember about ½ equations. You cannot have a reduction taking place unless some other species is being oxidised. ½ reactions therefore are only theoretical representations of what is happening to one of the reactants. Electrons are actually transferred from one species to another in a chemical reaction. Free electrons do not exist in chemistry. Therefore oxidation is always accompanied by reduction. These reactions are called redox reactions. A redox reaction is one in which electrons are transferred between species and thus changes in oxidation number occur 38
Balancing Redox Equations A redox reaction, as we have seen, consists of two ½reactions; a REDuction and an OXidation. It is vitally important that you learn how to balance these equations as they are one of the most common, and most important, classes of reactions in chemistry. As well as balancing the number of electrons involved in both ½reactions, one must also balance the elements which are not directly involved in the redox reaction. Remember the Law of Conservation of Mass! A further consideration is whether the redox reaction is taking place in acidic media or basic media. Acidic media contains H + /H 2 O Basic media contains OH /H 2 O Example 3.1 Question: Balance the following redox equation in acidic media MnO 4 + Fe 2+ Fe 3+ + Mn 2+ Answer: 39
You might well be asking yourself How can we have a charged solution when all matter is electrically neutral? The answer is quite straightforward. We don t! What chemists tend to do is miss out socalled SPECTATOR IONS, i.e. those that remain unchanged in the reaction. Therefore for the above equation we could easily add these spectator ions and the charge would be zero on both sides of the equation. For instance we could react KMnO 4 with FeCl 2 in the presence of H 2 SO 4. KMnO 4 + 5 FeCl 2 + 4 H 2 SO 4 K + + Mn 2+ + 10 Cl + 5 Fe 3+ + 4 H 2 O + 4 SO 4 2 The charge is now zero on both sides but the equation looks clumsy as there are now far more species involved which do not add to our understanding of the chemistry. We will continue to omit the spectator ions. Example 3.2 Question: Balance the following reaction in basic media: MnO 4 + Br BrO 3 Answer: + MnO 2 40
Exercise 3.1 Balance the following redox equation in acidic media. Cr 2 O 7 2 + Sn 2+ Cr 3+ + Sn 4+ Exercise 3.2 Balance the following ½reaction equations in acidic media and identify each as either an oxidation or a reduction. (a) I IO 3 (b) NO NO 3 (c) VO 2+ V 3+ (d) PbSO 4 PbO 2 + SO 4 2 Exercise 3.3 Balance the following ½reaction equations in basic media and identify each as either an oxidation or a reduction. (a) ClO Cl (b) IO 3 IO (c) NO 3 NO 2 (d) CrO 4 2 Cr(OH) 3 Exercise 3.4 Balance the following redox reaction equations. (a) Al + Cr 2 O 3 Al 2 O 3 + Cr (acidic) 2 (b) Cr 2 O 7 + Fe 2+ Cr 3+ + Fe 3+ (acidic) (c) MnO 2 + BrO MnO 4 + Br (basic) Oxidising and Reducing Agents In a redox reaction one (or more) species is reduced and one (or more) species is oxidised. If we look at a simple redox equation: Fe 2+ + Ce 4+ Ce 3+ + Fe 3+ where Ce is the element cerium (atomic number 58). Fe 2+ is oxidised to Fe 3+ by Ce 4+ Ce 4+ is reduced to Ce 3+ by Fe 2+ This means that the species that is oxidised is acting as a reducing agent and the species that is reduced is acting as an oxidizing agent. An oxidising agent is itself reduced A reducing agent is itself oxidized The Relative Strength of Oxidising and Reducing Agents A very useful concept in chemistry is the Standard Electrode Potential. This is given the symbol E 0 and is measured in volts. E 0 values give us an idea as to how much one species wants to gain electrons relative to another species. You can think of it as a tugofwar for the electrons. The more positive the E 0 value is, 41
the more a species wants to gain electrons and therefore the stronger the oxidising power of the species. If we look at the following two redox ½ reactions: Fe 3+ + e Fe 2+ MnO 4 + 8H + + 5e Mn 2+ + 4H 2 O we can immediately tell which is the stronger oxidising agent by looking at the table of E 0 values on the next page. The Fe 3+ /Fe 2+ redox couple has an E 0 value of + 0.77 V while the MnO 4 /Mn 2+ redox couple has an E 0 value of + 1.51 V. Thus MnO 4 is a stronger oxidising agent than Fe 3+. We can also use the E 0 values to determine how strong the reducing agents are relative to each other. You will notice that the E 0 values in the table are all quoted for the ½ reactions written as reductions. This means that the oxidising agents are on the LHS of the equation and the reducing agents are on the RHS. e.g. Na + + e Na Na + oxidizing agent and Na reducing agent The general rule is that the more negative the E o value, the stronger is the reducing agent on the RHS of the equation. Standard Electrode Potentials Reduction half reactions E o /V Reduction half reactions E o /V F 2 + 2e 2F +2.87 Cu 2+ + 2e Cu +0.34 Co 3+ + e Co 2+ +1.81 Hg 2 Cl 2 + 2e 2Hg + 2Cl +0.27 H 2 O 2 + 2H + + 2e 2H 2 O +1.78 AgCl + e Ag + Cl +0.22 Au + + e Au +1.69 Cu 2+ + e +0.16 Pb 4+ + 2e Pb 2+ +1.67 Sn 4+ +2e Sn 2+ +0.15 Ce 4+ + e Ce 3+ +1.61 2H + + 2e H 2 0, by definition MnO 4 + 8H + + 5e Mn 2+ + 4H 2 O +1.51 Fe 3+ + 3e Fe 0.04 Mn 3+ + e Mn 2+ +1.51 Pb 2+ + 2e Pb 0.13 Au 3+ + 3e Au +1.40 Sn 2+ + 2e Sn 0.14 Cl 2 + 2e 2Cl +1.36 Ni 2+ +2e Ni 0.23 Cr 2 O 7 2 + 14H + + 6e Cr 3+ + 7H 2 O +1.33 Co 2+ + 2e Co 0.28 O 2 + 4H + + 4e 2H 2 O +1.23 Tl + + e Tl 0.34 ClO 4 + 2H + + 2e ClO 3 + H 2 O +1.23 Cd 2+ + 2e Cd 0.40 MnO 2 + 4H + + 3e Mn 2+ + 2H 2 O +1.23 Cr 3+ + e Cr 2+ 0.41 Br 2 + 2e 2Br +1.09 Fe 2+ + 2e Fe 0.44 NO 3 + 4H + + 3e NO + 2H 2 O +0.96 S + 2e S 2 0.48 2Hg 2+ + 2e 2+ Hg 2 +0.92 Cr 3+ + 3e Cr 0.74 ClO + H 2 O + 2e Cl + 2OH +0.89 Zn 2+ + 2e Zn 0.76 Hg 2+ + 2e Hg +0.86 2H 2 O + 2e H 2 + 2OH 0.83 NO 3 + 2H + + e NO 2 + H 2 O +0.80 Cr 2+ + 2e Cr 0.91 Ag + + e Ag +0.80 Mn 2+ + 2e Mn 1.18 2Hg 2 2+ + 2e 2Hg +0.79 Al 3+ + 3e Al 1.66 Fe 3+ + e Fe 2+ +0.77 Mg 2+ + 2e Mg 2.36 MnO 4 + 2H 2 O + 2e MnO 2 + 4OH +0.60 Ce 3+ + 3e Ce 2.48 42
MnO 4 + e MnO 4 2 +0.56 La 3+ + 3e La 2.52 I 2 + 2e 2I +0.54 Na + + e Na 2.71 Cu + + e Cu +0.52 Ca 2+ + 2e Ca 2.87 I 3 + 2e 3I +0.53 K + + e K 2.93 ClO 4 + 2H 2 O + 2e ClO 3 + 2OH +0.36 Li + + e Li 3.05 Question 3.1 (a) Which is the stronger oxidising agent, Na + or K +? (b) Which is the stronger reducing agent, Na or K? Predicting the Direction of Spontaneous Change Knowing the relative values of E o will also allow us to predict whether certain reactions will occur or not. Take for instance the following two reactions. Cu 2+ + Zn Zn 2+ + Cu Zn 2+ + Cu Cu 2+ + Zn Only one of these reactions can occur spontaneously. A spontaneous reaction is one which goes in the energetically favourable direction. We can work out which one from the E 0 values. Zn 2+ + 2e Zn E o = 0.76 V Cu 2+ + 2e Cu E o = +0.34 V We can see that Cu 2+ is a much stronger oxidising agent than Zn 2+ and therefore the copper ion grabs the electrons before zinc can. Thus Cu 2+ will oxidise zinc metal to Zn 2+ and Cu 2+ + Zn Zn 2+ + Cu will be the spontaneous reaction. Another way of looking at the above reaction is to recognise Zn is a stronger reducing agent than Cu (i.e. it has a more negative E 0 value) and will thus reduce Cu 2+ to Cu. For the reaction above we can calculate the difference in E 0 values, ΔE 0. By definition ΔE o = E o (for the reduction ½reaction) E o (for the oxidation ½reaction). ΔE o = E o red E o ox Thus for the above reaction ΔE o = E o (Cu 2+ /Cu) E o (Zn 2+ /Zn) because Cu 2+ is being reduced and Zn is being oxidised. E o = +0.34 (0.76) V = +1.10 V It is a general rule that E o will always be positive for a spontaneous reaction. Example 3.3 Question Which is the direction of spontaneous change for the following reaction to the left or to the right? 2 Cr 3+ + 6 Fe 3+ + 7 H 2 O Cr 2 O 7 2 + 14 H + + 6 Fe 2+ 43
Answer Exercise 3.5 Calculate the E o value for the following reactions and predict whether they are spontaneous or not. (a) 2 Li + Mg 2+ 2 Li + + Mg (b) Sn 4+ + 2 Fe 2+ Sn 2+ + 2 Fe 3+ The following two ½reactions Zn 2+ + 2e Zn E o = 0.76 V Cu 2+ + 2e Cu E o = + 0.34 V represents halfcells where a strip of metal, Zn or Cu called electrodes are immersed in solutions containing ions of the same metal. Electrodes are classified according to whether oxidation or reduction takes place there. If oxidation takes place, the electrode is called an anode, if reduction takes place, the electrode is called the cathode. Charge is carried through solutions by the migration of ions through a U tube called a salt bridge and the connected combination of two halfcells is called an electrochemical cell. The halfcells are represented as Zn(s)/Zn 2+ (aq) ane Cu 2+ (aq)/cu(s). The spontaneous reaction in the electrochemical cell as shown in the figure below is Oxidation: Zn(s) Zn 2+ (aq) + 2 e Reduction: Overall: Cu 2+ (aq) + 2 e Cu(s) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) 44
The cell diagram corresponding to the above figure and reaction is written as: anode Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) cathode E cell = 1.103 V Standard Electrode Potential (E cell ), Free Energy (ΔG) and Equilibrium Constant (K) ΔG o = nfe o cell where n is the number of electrons transferred in the reaction F is Faraday s constant, the charge of 1 mole of electrons which can be found on the data sheet as 96485 C mol 1 We must of course remember that when we use E o values to predict the direction of spontaneous change, we are not saying that the reaction goes to completion. E o values are a measure of the relative demand between two species for elections. This tugofwar for electrons results in an equilibrium being established. If we have another look at example 3 then we have the following equilibrium. Cr 2 O 7 2 + 14 H + + 6Fe 2+ 2 Cr 3+ + 6 Fe 3+ + 7 H 2 O E o = +0.56 V The positive sign of the E o value tells us the forward reaction is spontaneous. This in turn means that the position of the equilibrium is to the RHS of the equation. If the E o value tells us something about the equilibrium state then it must, somehow, be linked to the equilibrium constant, K, for the reaction. If E o is large and positive then the reaction favours the products and therefore we would expect the equilibrium constant to be a large number. The mathematical relationship between the two is given by the following equation: where: n is the number of electrons transferred in the reaction F is Faraday s constant, the charge of 1 mole of electrons which can be found on the data sheet as 96485 C mol 1 R is the gas constant, 8.315 J mol 1 K 1 T is the temperature, which for standard conditions is 298 K (25 o C) 45
Example 3.4 Question: Calculate the equilibrium constant for the reaction between MnO 4 and Fe 2+ in acidic aqueous solution at 25 o C Answer: Exercise 5.6 Calculate the K C value for the following reactions at 25 o C (a) Cr 2 O 7 2 ions with Mn 2+ ions to give Cr 3+ ions and MnO 4 ions in acidic media. 46
(b) Co 3+ ions with I ions to give Co 2+ ions and I 2 molecules in acidic solution. Exercise 5.7 Which of the equilibria equations in Exercise 6 are spontaneous and which favour the products? Explain your answer. E cell as a function of concentrations Nernst equation: Example 3.5 Question: Consider the electrochemical cell represented by : Zn(s) Zn 2+ Fe 3+ Fe(s) Determine E for the cell when the concentration of Fe 3+ is 10.0 M and that of Zn 2+ is 1.00 x 10 3 M. Answer: 47