Electrochemistry. Oxidation number. Chapter 18. 2Mg 2Mg e - Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - )

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1 Electrochemistry Chapter 18 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 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 2+ 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg e - O 2 + 4e - 2O Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 2 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, O = The oxidation number of oxygen is usually 2. In H 2 O 2 and O 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. 4 Balancing Redox Equations The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 in acid solution, also yielding Cr 3+ ion? 1. Write the unbalanced equation for the reaction in ionic form. Fe 2+ + Cr 2 O 7 Fe 3+ + Cr Separate the equation into two half-reactions. Oxidation: Reduction: Fe 2+ Fe Cr 2 O 7 Cr Balance the atoms other than O and H in each half-reaction. Cr 2 O 7 2Cr 3+ 5 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 7 14H + + Cr 2 O 7 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 6e H + + Cr 2 O 7 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 - 2Cr H 2 O 6 2

3 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 14H + + Cr 2 O 7 + 6Fe 2+ 6Fe 2+ 6Fe e - 2Cr H 2 O 6Fe Cr H 2 O 8. Verify that the number of atoms and the charges are balanced. 14x 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 Write a balanced ionic equation to represent the oxidation of iodide ion (I - ) by permanganate ion ( MnO 4- ) in basic solution to yield molecular iodine (I 2 ) and manganese(iv) oxide (MnO 2 ). Strategy We follow the preceding procedure for balancing redox equations. Note that the reaction takes place in a basic medium. Solution Step 1: The unbalanced equation is MnO 4- + I - MnO 2 + I Step 2: The two half-reactions are -1 Oxidation: I - I Reduction: MnO - 4 MnO 2 Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction: We first balance the I atoms: 2I - I 2 9 3

4 18.1 To balance charges, we add two electrons to the right-hand side of the equation: 2I - I 2 + 2e - Reduction half-reaction: To balance the O atoms, we add two H 2 O molecules on the right: MnO - 4 MnO 2 + 2H 2 O To balance the H atoms, we add four H + ions on the left: MnO H + MnO 2 + 2H 2 O There are three net positive charges on the left, so we add three electrons to the same side to balance the charges: MnO H + + 3e - MnO 2 + 2H 2 O Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows: 3(2I - I 2 + 2e - ) 2(MnO H + + 3e - MnO 2 + 2H 2 O) 6I - + 2MnO H + + 6e - 3I 2 + 2MnO 2 + 4H 2 O + 6e - The electrons on both sides cancel, and we are left with the balanced net ionic equation: 6I - + 2MnO H + 3I 2 + 2MnO 2 + 4H 2 O This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic medium, for every H + ion we need to add equal number of OH - ions to both sides of the equation: 6I - + 2MnO 4 + 8H + + 8OH - 3I 2 + 2MnO 2 + 4H 2 O + 8OH - Finally, combining the H + and OH - ions to form water, we obtain 6I - + 2MnO 4 + 4H 2 O 3I 2 + 2MnO 2 + 8OH

5 Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 13 The difference in electrical potential between the anode and cathode is called: voltage electromotive force (emf) potential Galvanic Cells Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M and [Zn 2+ ] = 1 M Cell Diagram phase boundary Zn (s) Zn 2+ (1 M) Cu 2+ (1 M) Cu (s) anode salt bridge cathode 14 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) 15 5

6 Standard Reduction Potentials Zn (s) Zn 2+ (1 M) H + (1 M) H 2 (1 atm) Pt (s) Anode (oxidation): Zn (s) Zn 2+ (1 M) + 2e - Cathode (reduction): 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) + 2H + (1 M) Zn 2+ (1 M) + H 2 (1 atm) 16 E Standard Reduction Potentials =.76 V Standard emf (E ) E = E cathode - E anode Zn (s) Zn 2+ (1 M) H + (1 M) H 2 (1 atm) Pt (s) E = E H /H - E Zn /Zn.76 V = - E Zn 2+ /Zn E Zn 2+ /Zn = -.76 V Zn 2+ (1 M) + 2e - Zn E = -.76 V 17 Standard Reduction Potentials E =.34 V E = E cathode - E anode E = E Cu 2+ /Cu E H + /H.34 = E Cu 2+ /Cu - E Cu 2+ /Cu =.34 V 2 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) 18 6

7 E is for the reaction as written The more positive E the greater the tendency for the substance to be reduced The half- reactions are reversible The sign of E changes when the reaction is reversed Changing the stoichiometric coefficients of a half- reaction does not change the value of E 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. Strategy To predict what redox reaction(s) will take place, we need to compare the standard reduction potentials of Cl 2, Br 2, and I 2 and apply the diagonal rule. Solution From Table 18.1, we write the standard reduction potentials as follows: Cl 2 (1 atm) + 2e - 2Cl - (1 M) E = 1.36 V Br 2 (l) + 2e - 2Br - (1 M) E = 1.7 V I 2 (s) + 2e - 2I - (1 M) E =.53 V Applying the diagonal rule we see that Br 2 will oxidize I - but will not oxidize Cl -. Therefore, the only redox reaction that will occur appreciably under standard-state conditions is Oxidation: 2I - (1 M) I 2 (s) + 2e - Reduction: Br 2 (l) + 2e - 2Br - (1 M) Overall: 2I - (1 M) + Br 2 (l) I 2 (s) + 2Br - (1 M) Check We can confirm our conclusion by calculating E. Try it. Note that the Na + ions are inert and do not enter into the redox reaction. 21 7

8 18.3 A galvanic consists of a Mg electrode in a 1. M Mg(NO 3 ) 2 solution and a Ag electrode in a 1. M AgNO 3 solution. Calculate the standard emf of this at 25 C. Strategy At first it may not be clear how to assign the electrodes in the galvanic. From Table 18.1 we write the standard reduction potentials of Ag and Mg and apply the diagonal rule to determine which is the anode and which is the cathode. Solution The standard reduction potentials are Ag + (1. M) + e - Ag(s) E =.8 V Mg 2+ (1. M) + 2e - Mg(s) E = V Applying the diagonal rule, we see that Ag + will oxidize Mg: Anode (oxidation): Mg(s) Mg 2+ (1. M) + 2e - Cathode (reduction): 2Ag + (1. M) + 2e - 2Ag(s) Overall: Mg(s) + 2Ag + (1. M) Mg 2+ (1. M) + 2Ag(s) Note that in order to balance the overall equation we multiplied the reduction of Ag + by 2. We can do so because, as an intensive property, E is not affected by this procedure. We find the emf of the by using Equation (18.1) and Table 18.1: E = E cathode - E anode = E Ag +/Ag - E Mg 2+/Mg =.8 V - (-2.37 V) = 3.17 V 23 DG = -nfe DG = -nfe Spontaneity of Redox Reactions n = number of moles of electrons in reaction F = 96,5 DG = -RT ln K = -nfe E = RT nf E E =.257 V n =.592 V n J V mol = 96,5 C/mol (8.314 J/K mol)(298 K) ln K = ln K n (96,5 J/V mol) ln K log K 24 8

9 Spontaneity of Redox Reactions DG = -RT ln K = -nfe Calculate the equilibrium constant for the following reaction at 25 C: Sn(s) + 2Cu 2+ (aq) Sn 2+ (aq) + 2Cu + (aq) Strategy The relationship between the equilibrium constant K and the standard emf is given by Equation (18.5): E = (.257 V/n)ln K Thus, if we can determine the standard emf, we can calculate the equilibrium constant. We can determine the E of a hypothetical galvanic made up of two couples (Sn 2+ /Sn and Cu 2+ /Cu + ) from the standard reduction potentials in Table Solution The half- reactions are Anode (oxidation): Sn(s) Sn 2+ (aq) + 2e - Cathode (reduction): 2Cu 2+ (aq) + 2e - 2Cu + (aq) E = E cathode - E anode = E Cu 2+/Cu+ - E Sn 2+/Sn =.15 V - (-.14 V) =.29 V Equation (18.5) can be written o ne ln K =.257 V 27 9

10 18.4 In the overall reaction we find n = 2. Therefore, (2)(.29V) ln K = = V 22.6 K = e = Calculate the standard free-energy change for the following reaction at 25 C: Strategy 2Au(s) + 3Ca 2+ (1. M) 2Au 3+ (1. M) + 3Ca(s) The relationship between the standard free energy change and the standard emf of the is given by Equation (18.3): ΔG = -nfe. Thus, if we can determine E, we can calculate ΔG. We can determine the E of a hypothetical galvanic made up of two couples (Au 3+ /Au and Ca 2+ /Ca) from the standard reduction potentials in Table Solution The half- reactions are Anode (oxidation): 2Au(s) 2Au 3+ (1. M) + 6e - Cathode (reduction): 3Ca 2+ (1. M) + 6e - 3Ca(s) E = E cathode - E anode = E Ca 2+/Ca - E Au 3+/Au = V V = V 3 1

11 18.5 Now we use Equation (18.3): ΔG = -nfe The overall reaction shows that n = 6, so G = -(6) (96,5 J/V mol) (-4.37 V) = 2.53 x 1 6 J/mol = 2.53 x 1 3 kj/mol Check The large positive value of ΔG tells us that the reaction favors the reactants at equilibrium. The result is consistent with the fact that E for the galvanic is negative. 31 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 At 298 K ln Q E = E V n ln Q E = E V n log Q 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+ ] =.15 M and [Fe 2+ ] =.68 M

12 18.6 Strategy Because the reaction is not run under standard-state conditions (concentrations are not 1 M), we need Nernst s equation [Equation (18.8)] to calculate the emf (E) of a hypothetical galvanic and determine the spontaneity of the reaction. The standard emf (E ) can be calculated using the standard reduction potentials in Table Remember that solids do not appear in the reaction quotient (Q) term in the Nernst equation. Note that 2 moles of electrons are transferred per mole of reaction, that is, n = Solution The half- reactions are Anode (oxidation): Co(s) Co 2+ (aq) + 2e - Cathode (reduction): Fe 2+ (aq) + 2e - Fe(s) E = E cathode - E anode = E Fe 2+/Fe - E Co 2+/Co = -.44 V (-.28 V) = -.16 V From Equation (18.8) we write o.257 V E = E - lnq n 2+ = o.257 V [Co ] E - ln [Fe 2+ n ].257 V.15 = -.16 V - ln 2.68 = -.16 V +.19 V = -.14 V Because E is negative, the reaction is not spontaneous in the direction written

13 18.7 Consider the galvanic shown in Figure 18.4(a). In a certain experiment, the emf (E) of the is found to be.54 V at 25 C. Suppose that [Zn 2+ ] = 1. M and P H2 = 1. atm. Calculate the molar concentration of H Strategy The equation that relates standard emf and nonstandard emf is the Nernst equation. The overall reaction is Zn(s) + 2H + (? M) Zn 2+ (1. M) + H 2 (1. atm) Given the emf of the (E), we apply the Nernst equation to solve for [H + ]. Note that 2 moles of electrons are transferred per mole of reaction; that is, n = 2. Solution As we saw earlier, the standard emf (E ) for the is.76 V. From Equation (18.8) we write V n o E = E - lnq 2+ o.257 V [Zn ] PH 2 = E - ln + 2 n [H ].257 V (1.)(1.).54 V=.76 V - ln [H ].257 V V = - ln [H ] = ln [H + ] 2 e = [H + ] [H ]= = M 39 13

14 18.7 Check The fact that the nonstandard-state emf (E) is given in the problem means that not all the reacting species are in their standard-state concentrations. Thus, because both Zn 2+ ions and H 2 gas are in their standard states, [H + ] is not 1 M. 4 Concentration Cells Galvanic from two half-s composed of the same material but differing in ion concentrations