1 Electrochemistry Chapter 18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur

3 Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1.

4 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

5 Balancing Redox Equations 1.Write the unbalanced equation for the reaction in ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr Reduction: Fe 2+ Fe Balance the atoms other than O and H in each half-reaction. Cr 2 O Cr 3+

6 Balancing Redox Equations 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O Cr H 2 O 14H + + Cr 2 O Cr H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe e - 6e H + + Cr 2 O Cr H 2 O 6.If necessary, equalize the number of electrons in the two half- reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe e - 6e H + + Cr 2 O Cr H 2 O

7 Balancing Redox Equations 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 6e H + + Cr 2 O Cr H 2 O 6Fe 2+ 6Fe e - Oxidation: Reduction: 14H + + Cr 2 O Fe 2+ 6Fe Cr H 2 O 8.Verify that the number of atoms and the charges are balanced. 14x1 – x 2 = 24 = 6 x x 3 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation.

Example Write a balanced ionic equation to represent the oxidation of iodide ion (I - ) by permanganate ion ( ) in basic solution to yield molecular iodine (I 2 ) and manganese(IV) oxide (MnO 2 ).

9 Galvanic Cells spontaneous redox reaction anode oxidation cathode reduction

10 Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M and [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anode cathode salt bridge phase boundary

11 Standard Reduction Potentials Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. E 0 = 0 V Standard hydrogen electrode (SHE) 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

12 Standard Reduction Potentials Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (s) + 2H + (1 M) Zn 2+ (1 M) + H 2 (1 atm)

13 E 0 = 0.76 V cell Standard emf (E 0 ) cell 0.76 V = 0 - E Zn /Zn 0 2+ E Zn /Zn = V 0 2+ Zn 2+ (1 M) + 2e - Zn E 0 = V E 0 = E H /H - E Zn /Zn cell Standard Reduction Potentials E 0 = E cathode - E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s)

14 Standard Reduction Potentials Pt (s) | H 2 (1 atm) | H + (1 M) || Cu 2+ (1 M) | Cu (s) 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) 2H + (1 M) + 2e - Anode (oxidation): Cathode (reduction): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) E 0 = E cathode - E anode cell 00 E 0 = 0.34 V cell E cell = E Cu /Cu – E H /H = E Cu /Cu E Cu /Cu = 0.34 V 2+ 0

15 E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

Example Predict what will happen if molecular bromine (Br 2 ) is added to a solution containing NaCl and NaI at 25°C. Assume all species are in their standard states.

Example A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO 3 ) 2 solution and a Ag electrode in a 1.0 M AgNO 3 solution. Calculate the standard emf of this cell at 25°C.

18 Spontaneity of Redox Reactions  G = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = V n ln K E cell 0 = V n log K E cell 0

19 Spontaneity of Redox Reactions  G 0 = -RT ln K = -nFE cell 0

Example Calculate the equilibrium constant for the following reaction at 25°C: Sn(s) + 2Cu 2+ (aq) Sn 2+ (aq) + 2Cu + (aq)

Example Calculate the standard free-energy change for the following reaction at 25°C: 2Au(s) + 3Ca 2+ (1.0 M) 2Au 3+ (1.0 M) + 3Ca(s)

22 The Effect of Concentration on Cell Emf  G =  G 0 + RT ln Q  G = -nFE  G 0 = -nFE 0 -nFE = -nFE 0 + RT ln Q E = E 0 - ln Q RT nF Nernst equation At 298 K V n ln Q E 0 E = V n log Q E 0 E =

Example Predict whether the following reaction would proceed spontaneously as written at 298 K: Co(s) + Fe 2+ (aq) Co 2+ (aq) + Fe(s) given that [Co 2+ ] = 0.15 M and [Fe 2+ ] = 0.68 M.

Example Consider the galvanic cell shown in Figure 18.4(a). In a certain experiment, the emf (E) of the cell is found to be 0.54 V at 25°C. Suppose that [Zn 2+ ] = 1.0 M and P H 2 = 1.0 atm. Calculate the molar concentration of H +.

25 Concentration Cells Galvanic cell from two half-cells composed of the same material but differing in ion concentrations.

26 Batteries Leclanché cell Dry cell Zn (s) Zn 2+ (aq) + 2e - Anode: Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) + Zn (s) + 2NH 4 + (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s)

27 Batteries Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Anode: Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) Mercury Battery

28 Batteries Anode: Cathode: Lead storage battery PbO 2 (s) + 4H + (aq) + SO 2- (aq) + 2e - PbSO 4 (s) + 2H 2 O (l) 4 Pb (s) + SO 2- (aq) PbSO 4 (s) + 2e - 4 Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 2- (aq) 2PbSO 4 (s) + 2H 2 O (l) 4

29 Batteries Solid State Lithium Battery

30 Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - 2H 2 (g) + O 2 (g) 2H 2 O (l)

31 Corrosion Corrosion is the term usually applied to the deterioration of metals by an electrochemical process.

32 Cathodic Protection of an Iron Storage Tank

33 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. Electrolysis of molten NaCl

34 Electrolysis of Water

Example An aqueous Na 2 SO 4 solution is electrolyzed, using the apparatus shown in Figure If the products formed at the anode and cathode are oxygen gas and hydrogen gas, respectively, describe the electrolysis in terms of the reactions at the electrodes.

36 Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mol e - = 96,500 C

Example 37 A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions and calculate the volume of gases generated at STP. 18.9