# Oxidation number. The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred.

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1 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 = 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 = The oxidation number of oxygen is usually 2. In H 2 O 2 and O 2 2- it is

2 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 Group IA metals are +1, IIA metals are +2 and fluorine is always 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. Oxidation numbers of all the atoms in HCO 3-? HCO 3 - O = -2 H = +1 3x(-2) + 1 +? = -1 C =

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

4 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 H + + Cr 2 O 7 2-2Cr H 2 O 2Cr H 2 O 5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction. 6e H + + Cr 2 O 7 2- Fe 2+ Fe e - 2Cr H 2 O 6. If necessary, equalize the number of electrons in the two halfreactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe e - 6e H + + Cr 2 O 7 2-2Cr H 2 O 19.1

5 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. Oxidation: Reduction: 6e H + + Cr 2 O 7 2-6Fe 2+ 6Fe e - 2Cr H 2 O 14H + + Cr 2 O Fe 2+ 6Fe Cr H 2 O 8. Verify that the number of atoms and the charges are balanced. 14x x2 = 24 = 6x3 + 2x3 9. Note: For reactions in basic solutions, add H 2 O to the side having every an excess of O atom and add 2OH - to other side of the equation to balance both H and O atoms, and so forth. 19.1

6 Electrochemistry Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7 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 Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 19.1

8 Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 19.2

9 The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Galvanic Cells Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M & [Zn 2+ ] = 1 M Zn (s) Zn 2+ (1 M) Cu 2+ (1 M) Cu (s) anode cathode 19.2

10 Standard Reduction Potentials Zn (s) Zn 2+ (1 M) H + (1 M) H 2 (1 atm) Pt (s) Anode (oxidation): Cathode (reduction): 2e - + 2H + (1 M) Zn (s) Zn 2+ (1 M) + 2e - H 2 (1 atm) Zn (s) + 2H + (1 M) Zn 2+ + H 2 (1 atm) 19.3

11 Standard Reduction Potentials Standard reduction potential (E ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction 2e - + 2H + (1 M) H 2 (1 atm) E = V Standard hydrogen electrode (SHE) 19.3

12 Standard Reduction Potentials Ecell =.76 V Standard emf (E ) cell Ecell = E cathode - E anode Zn (s) Zn 2+ (1 M) H + (1 M) H 2 (1 atm) Pt (s) Ecell = E H /H - E + Zn 2+ /Zn 2.76 V = - E Zn 2+ /Zn E Zn 2+ /Zn = -.76 V Zn 2+ (1 M) + 2e - Zn E = -.76 V 19.3

13 Standard Reduction Potentials Ecell =.34 V Ecell = E cathode - E anode E cell = E Cu 2+ /Cu E H + /H 2.34 = E Cu 2+ /Cu - E Cu 2+ /Cu =.34 V Pt (s) H 2 (1 atm) H + (1 M) Cu 2+ (1 M) Cu (s) Anode (oxidation): H 2 (1 atm) 2H + (1 M) + 2e - Cathode (reduction): 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) 19.3

14 X X

15 E is for the reaction as written The more positive E the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 19.3

16 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1. M Cd(NO 3 ) 2 solution and a Cr electrode in a 1. M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E = -.4 V Cr 3+ (aq) + 3e - Cr (s) E = -.74 V Anode (oxidation): Cathode (reduction): 2e - + Cd 2+ (1 M) Cd is the stronger oxidizer Cd will oxidize Cr Cr (s) Cr 3+ (1 M) + 3e - Cd (s) x 3 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 Ecell = E cathode - E anode E = -.4 (-.74) cell Ecell =.34 V 19.3

17 Spontaneity of Redox Reactions DG = -nfe cell DG = -nfe cell n = number of moles of electrons in reaction F = 96,5 J V mol = 96,5 C/mol DG = -RT ln K = -nfe cell E cell = RT nf (8.314 J/K mol)(298 K) ln K = n (96,5 J/V mol) ln K E cell E cell =.257 V n =.592 V n ln K log K 19.4

18 Spontaneity of Redox Reactions DG = -RT ln K = -nfe cell 19.4

19 What is the equilibrium constant for the following reaction at 25 C? Fe 2+ (aq) + 2Ag (s) Fe (s) + 2Ag + (aq) E cell =.257 V n ln K Oxidation: Reduction: 2e - + Fe 2+ 2Ag 2Ag + + 2e - Fe n = 2 E = E Fe /Fe E 2+ Ag + /Ag E = -.44 (.8) E = V K = exp Ecell x n.257 V = exp V x V K = 1.23 x

20 The Effect of Concentration on Cell Emf DG = DG + RT ln Q DG = -nfe DG = -nfe -nfe = -nfe + RT ln Q Nernst equation E = E - RT nf ln Q At 298 E = E V n ln Q E = E V n log Q 19.5

21

22 Will the following reaction occur spontaneously at 25 C if [Fe 2+ ] =.6 M and [Cd 2+ ] =.1 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) Oxidation: Reduction: 2e - + Fe 2+ E = E Fe /Fe E 2+ Cd 2+ /Cd Cd Cd e - 2Fe n = 2 E = -.44 (-.4) E = -.4 V E = E E = -.4 V V n ln Q V 2 ln.1.6 E =.13 E > Spontaneous 19.5

23 LATIMER DIAGRAM Potential Reduction for each reduction: Latimer puts them in a diagram to be more informative as follows:

24 FROST DIAGRAM Frost depicted a plot of oxidation number vs DG (-ne )

25 Fig. 5.2 Frost diagram for Mn in acid solution -ne o / V mol e 6 MnO 4-4 MnO Mn Mn 2+ Mn 3+ MnO Oxydation State

26 What is the E cell of this redox: Ag + (aq) + Cu(s) Ag(s) + Cu 2+ (aq) Ag + (aq) + e Ag(s) E = +.8 V Cu(s) Cu 2+ (aq) + 2e E = -.34 V 2Ag + (aq) + 2e 2Ag(s) E = +.8 V + 2Ag + (aq) + Cu(s) 2Ag(s) + Cu 2+ (aq) E = +.46 V By calculating of ΔG of each half-reaction Cu(s) Cu 2+ (aq) + 2e ΔG = -nfe = -2F(-.34V) = -.68 FV 2Ag + (aq) + 2e 2Ag(s) ΔG = -nfe = -2F(+.8V)= -1.6 FV + 2Ag + (aq) + Cu(s) 2Ag(s) + Cu 2+ (aq) ΔG = -.92 FV E = - ΔG /nf = - (-.92 FV)/2F = +.46 V

27

28 Batteries Dry cell Leclanché cell Anode: Zn (s) Zn 2+ (aq) + 2e - + 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) 19.6

29 Batteries Mercury Battery Anode: Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) 19.6

30 Batteries Lead storage battery Anode: Pb (s) + SO 2- (aq) PbSO 4 (s) + 2e - 4 Cathode: PbO 2 (s) + 4H + (aq) + SO 2- (aq) + 2e - 4 PbSO 4 (s) + 2H 2 O (l) Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 2- (aq) 4 2PbSO 4 (s) + 2H 2 O (l) 19.6

31 Batteries Solid State Lithium Battery 19.6

32 Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + O 2 (g) 2H 2 O (l) 19.6

33 Chemistry In Action: Bacteria Power CH 3 COO - + 2O 2 + H + 2CO 2 + 2H 2 O

34 Corrosion 19.7

35 Cathodic Protection of an Iron Storage Tank 19.7

36 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. 19.8

37 Electrolysis of Water 19.8

38 Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e - = 96,5 C 19.8

39 How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of.452 A is passed through the cell for 1.5 hours? Anode: 2Cl - (l) Cl 2 (g) + 2e - Cathode: Ca 2+ (l) + 2e - Ca (s) Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) 2 mole e - = 1 mole Ca mol Ca =.452 C s x 1.5 hr x 36 s hr x 1 mol e - 96,5 C x 1 mol Ca 2 mol e - =.126 mol Ca =.5 g Ca 19.8

40 Chemistry In Action: Dental Filling Discomfort 2+ Hg 2 /Ag 2 Hg 3.85 V 2+ Sn /Ag 3 Sn -.5 V 2+ Sn /Ag 3 Sn -.5 V

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

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