Chapter 18. Electrochemistry. Hill, Petrucci, McCreary & Perry 4 th Ed.

Transcription

1 Chapter 18 Electrochemistry Hill, Petrucci, McCreary & Perry 4 th Ed.

2 Oxidation Reduction Reactions Oxidation = Loss of electrons from a chemical species = Increase in Oxidation State. Na Na + 1 electron a "+1" oxidation state Reduction = Gain of electrons by a chemical species = Decrease in Oxidation State. Cl electrons 2 Cl a "-1" oxidation state

3 Balancing RedOx Equations The Oxidation Number Method a 5 electron loss per iodine 0 +5 I 2 + HNO 3 2 HIO 3 + NO a 1 electron gain per nitrogen Can you balance this equation? You must balance the electron transfer and the atoms. This reaction occurs in aqueous acidic solution.

4 Balancing RedOx Equations The Oxidation Number Method a 5 electron loss per iodine 0 +5 I 2 + HNO 3 2 HIO 3 + NO a 1 electron gain per nitrogen The atoms which change oxidation state are balanced first! The Nitrogen change must occur 10 times to change one Iodine Molecule: I HNO 3 2 HIO NO 2 Now we must balance the atoms which do not change oxidation state! I HNO 3 2 HIO NO 2 Being in acidic aqueous solution means we have lots of H+ and H 2 O to balance hydrogens and oxygens! + 4 H 2 O

5 Rules for Oxidation Number Method 1. Assign Oxidation Numbers. 2. Initially Balance atoms that change oxidation state. 3. Determine the electron transfer for like atoms. 4. Alter coefficients (only!) so electron transfer balances for the entire equation. 5. Balance Hydrogen and Oxygen and other atoms or ions which do not change oxidation state.

6 Balancing RedOx Equations Half-Reaction Method a 2 electron loss per Zn 0 Zn + NO Zn 2+ + NH an 8 electron gain per nitrogen Balance this reaction by first dividing the reaction into an oxidation half-reaction and a reduction half-reaction. Using coefficients only, balance all atoms, using H 2 O and H+/OH- to balance H & O, balance charge using electrons in the oxidation half-reaction. Balance all atoms and charge in the reduction half-reaction the same way. Multiply each balanced half-reaction by coefficients which cause the electrons to cancel, add the half-reactions together finally reducing any species occurring on both sides of arrow.

7 Half-Reaction Method a 2 electron loss per Zn 0 Zn + NO Zn 2+ + NH an 8 electron gain per nitrogen Balance this reaction by first dividing the reaction into an oxidation half-reaction and a reduction half-reaction. Oxidation: Zn Zn 2+ Reduction: + 2 e NO e NH H+ x H 2 O Using coefficients only, balance all atoms, using H 2 O and H+/OH- to balance H & O, balance charge using electrons in the oxidation half-reaction /the reduction half-reaction. Multiply each half-reaction by coefficients to cancel electrons, add the half-reactions together, reduce species occurring on both sides 4 Zn + NO H+ 4 Zn 2+ + NH H 2 O

8 Rules for Half-Reaction Method 1. Assign Oxidation Numbers. 2. Separate into Oxidation and Reduction halfreactions. 3. Initially balance atoms that change oxidation state by changing only coefficients. 4. Balance any spectator ions. 5. Balance Hydrogen & Oxygen: H + /H 2 O acidic; H 2 O/OH - basic solution.

9 Rules for Half-Reaction Method continued 6. Balance the charges on each half-reaction by using electrons. 7. Multiply half-reactions by whole numbers to balance the electron transfer. 8. Add the half-reactions together and cancel species occurring on both sides of the chemical equation.

10 Electrochemical Cells An application of Oxidation/Reduction Reactions Voltaic Cell Produces electric current by means of a spontaneous RedOx reaction. Electrolytic Cell Uses an electric current to cause a non-spontaneous reaction to occur.

11 A Voltaic Cell Zn + Zn 2+ + Cu e- Cu 2+ e- salt bridge Zn 2+ Cu 2+ Anode Compartment Cathode Compartment Zn Cu 2+ + Zn e- 2 e- Cu Oxidation Half-reaction Reduction Half-reaction

12 Voltaic Cell Notation ANODE CATHODE salt bridge Anode Terminal Anode Electrolyte Cathode Electrolyte Cathode Terminal phase separator Zn Zn 2+ Cu 2+ Cu

13 Standard Hydrogen Electrode Oxidation Reduction ANODE CATHODE salt bridge salt bridge Pt Anode Terminal Gas Anode Electrolyte Cathode Electrolyte Gas Pt Cathode Terminal phase separator phase separator Pt H 2(g) H + (aq) H + (aq) H 2(g) Pt Half Reaction o E red 0 H 2 H + 2(g) (aq) + 2e- Half Reaction 2 H + (aq) + 2e- o E red 0 H 2(g)

14 Standard Reduction Potentials Cathode Potentials Table 18.1 p 762: emf in Volts vs SHE Table Layout: Red. Half-Reactions vs SHE Strongest Oxidizing Agents at Top Strongest Reducing Agents at Bottom For any pair of couples (half-reactions) the Anodes will always be lower in the table in the text.

15 Reduction Half-Reaction - Acidic Solution - E o, volt H 2 O 2 (aq) + 2 H + (aq) + 2 e- => 2 H 2 O(l) Cl 2 (g) + 2 e- => 2 Cl - (aq) O 2 (g) + 4 H + (aq) + 4 e- => 2 H 2 O(I) Ag + (aq) + e- => Ag(s) I 2 (s) + 2 e- => 2 I - (aq) Cu 2+ (aq) + 2 e- => Cu(s) H + (aq) + 2 e- => H 2 (g) 0 Pb 2+ (aq) + 2 e- => Pb(s) Sn 2+ (aq) + 2 e- => Sn(s) Zn 2+ (aq) + 2 e- => Zn(s) Al 3+ (aq) + 3 e- => Al(s) Mg 2+ (aq) + 2 e- => Mg(s) Na + (aq) + e- => Na(s) Li+(aq) + e- => Li(s)

16 Cell Potentials E o cell = E o red cathode - E o red anode Higher in Table Lower in Table The Anode Reaction is the Reverse of the reaction in the Table. The Anode potential is opposite in sign of that tabulated, hence the subtraction. Zn 2+ (aq) + 2e- Zn (s) Zn (s) Zn 2+ (aq) + 2e- E o red = E o cathode = V E o ox = E o anode = V

17 Half-Reactions are commonly represented by Reaction Couples Half Reaction Abbreviated "Couple" Zn 2+ (aq) + 2e- Zn (s) Zn 2+ (aq) Zn (s) Zn (s) Zn 2+ (aq) + 2e- Zn (s) Zn 2+ (aq) The species Oxidized or Reduced are shown in the couples in the same order they occur in the reaction.

18 Reduction Half-Reaction - Acidic Solution Couple H 2 O 2 (aq) + 2 H + (aq) + 2 e- => 2 H 2 O(l) H 2 O 2 (aq)/ H 2 O(l) Cl 2 (g) + 2 e- => 2 Cl - (aq) Cl 2 (g)/ Cl - (aq) O 2 (g) + 4 H + (aq) + 4 e- => 2 H 2 O(I) O 2 (g)/ H 2 O(I) Ag + (aq) + e- => Ag(s) Ag + (aq)/ Ag(s) I 2 (s) + 2 e- => 2 I - (aq) I 2 (s)/ I - (aq) Cu 2+ (aq) + 2 e- => Cu(s) Cu 2+ (aq)/cu(s) 2 H + (aq) + 2 e- => H 2 (g) H + (aq)/ H 2 (g) Pb 2+ (aq) + 2 e- => Pb(s) Pb 2+ (aq)/ Pb(s) Sn 2+ (aq) + 2 e- => Sn(s) Sn 2+ (aq)/ Sn(s) Zn 2+ (aq) + 2 e- => Zn(s) Zn 2+ (aq)/ Zn(s) Al 3+ (aq) + 3 e- => Al(s) Al 3+ (aq)/ Al(s) Mg 2+ (aq) + 2 e- => Mg(s) Mg 2+ (aq)/ Mg(s) Na + (aq) + e- => Na(s) Na + (aq)/ Na(s) Li+(aq) + e- => Li(s) Li+(aq)/ Li(s)

19 Gibb s Free Energy & Cell Potential G o o = - n F E cell A positive E cell spontaneous A negative G o A negative E cell n = # electrons transferred F = Faraday Constant = 96,500 Coulombs/mole "charge of a mole of electrons" "one Faraday of charge" non-spontaneous A positive G o

20 Cell Potentials & the Equilibrium o o G = - n F E cell Constant Rearranging: o Ecell = R T n F Nernst Equation: For Zn + E cell = Cu 2+ = - R T ln K ln K o E cell - R T n F Zn 2+ + Cu An E cell < n V will give an equilibrium o mixture: G < kj ln Q Q = [product] [reactant] Q = [Zn2+ ] [Cu 2+ ]

21 Calculating Cell Potentials Calculate the cell potential and write the cell notation and net chemical equation for the half reactions below. I 2(s) + 2 e- => 2 I - (aq) Sn 2+ (aq) + 2 e- => Sn (s)

22 Calculating a Cell Potential - Process Specified Given the following reaction, determine the cell potential and whether the reaction is spontaneous. Write the cell notation for the reaction as written. Sn(s) + Cu 2+ (aq) => Sn 2+ (aq) + Cu(s)

23 Electrolysis of Molten Salts A molten salt is the simplest case. NaCl (s) melting Na+ (l) + Cl- (l) electrolysis 2 Na+ (l) + 2 Cl- (l) 2 Na (l) + Cl 2(g) Anode: Cathode: 2 Cl- (l) Cl 2(g) + 2 e- 2 mol 1 mol 2 Faradays Na (l) 1 mol 1 Faraday 1 mol Na+ (l) + e- 1 mole electrons = 1 Faraday Charge = 96,500 Coulombs. 1 ampere = 1 Coulomb/second The amount of material produced depends directly on the amount of current passed through the cell.

24 Some Salts can be Electrolysed in Water CuBr 2(aq) : Cu 2+ (aq) + 2 Br - (aq) Electrolysis Products Cu (s) + Br 2(l) Anode Rxn: 2 Br- (aq) Br 2(l) + 2 e- - ( E red = 1.07 V) Cathode Rxn: Cu 2+ (aq) + 2 e- Cu (s) E red = 0.34 V E cell = V An applied voltage > 0.73 V will cause: the reduction of Copper ion to Copper metal. and the oxidation of Bromide ion to Bromine..

25 How much copper metal can be plated by a current of 5.0 amperes in 1 hour? You will need to calculate the number of Faradays delivered by this current and relate that to the numbers of moles of copper.