1. 8–1 Ibrahim BarryChapter 20-1 Chapter 20 Electrochemistry

2. 8–2 Chapter 20-2 Overview Half-Reactions –Balancing oxidation – reduction in acidic and basic solutions Voltaic cells –Construction of voltaic cells –Notation for voltaic cells –Electromotive force (EMF) –Standard cell potentials –Equilibrium constants from EMFs –Concentration dependence of EMF Electrolytic cells –Aqueous electrolysis –Stoichiometry of electrolysis

3. 8–3 Chapter 20-3 Electrochemistry Electrochemistry - Field of Chemistry dealing with transfer of electrons from one species to another. E.g. Zn in CuSO 4 (aq). Electrochemical cell - combination of two half reactions to produce electricity from reaction. E.g. Danielle cell: Zn and Cu electrodes in salts of these ions.

4. 8–4 Chapter 20-4 Balancing Redox Reactions: Oxidation Number Method Determine oxidation # for each atom- both sides of equation. Determine change in oxidation state for each atom. Left side: make loss of electrons = gain. Balance on other side. Insert coefficients for atoms that don't change oxidation state. E.g. Balance FeS(s)+CaC 2 (s)+CaO(s)  Fe(s)+CO(g)+CaS(s) In acidic or basic solution balance as above, then balance charge with H + or OH  on one side and water on other side. Cancel waters that appear on both sides at end. E.g. Balance which occurs in acidic solution:

5. 8–5 Chapter 20-5 Balancing: Half-Reaction Method Write unbalanced half reactions for the oxidation and the reduction Balance the number of elements except O and H for each. Balance O's with H 2 O to the deficient side. Balance H's with H + to the hydrogen deficient side –Acidic: add H + –Basic: add H 2 O to the deficient side and OH  to the other side. Balance charge by adding e  to the side that needs it. Multiply each half-reaction by integers to make electrons cancel. Add the two half-reactions and simplify. E.g. Balance: –Acidic: Zn(s) + VO 2+ (aq)  Zn 2+ (aq) + V 3+ (aq). –Basic: Ag(s) + HS  (aq) + (aq)  Ag 2 S(s) + Cr(OH) 3 (s).

6. 8–6 Chapter 20-6 Galvanic (Voltaic) and Electrolytic Cells Cell reaction – Redox reaction involved in electrochemical cell. Voltaic (galvanic) cell – reaction is spontaneous and generates electrical current. Electrolytic – non-spontaneous reaction occurs due to passage of current from external power source. E.g. charging of batteries.

7. 8–7 Chapter 20-7 Galvanic Cell 2 anode – electrode where oxidation occurs. cathode – electrode where reduction occurs. salt bridge – ionic solution connecting two half- cells (half-reactions) to prevent solutions from mixing. E.g. Which is the anode and cathode in the cell to the right? Write the half–reactions. Cd(s) + 2Ag + (aq)  2Ag(s) + Cd 2+ (aq) Sign of electrodes (current flows from anode to cathode): E.g. determine direction of electron flow for the reaction for a galvanic cell made from Ni(s) and Fe(s). The reaction is: 2Fe 3+ (aq) + 3Ni  2Fe(s) + 3Ni 2+ (aq)

8. 8–8 Chapter 20-8 Shorthand Notation for Galvanic Cells Shorthand way of portraying electrodes in a voltaic electrochemical cell. Redox couple-oxidized and reduced forms of same element when it is involved in electrochemical reaction. Shorthand: Ox/Red E.g. Cu 2+ /Cu, Zn 2+ /Zn. –2 couples required for electrochemical reaction. Shorthand rules: –Anode reaction-left; reduced form first. –Cathode-right; oxidized form first. –Vertical line drawn between different phases including reaction of gases at metal electrode. –Double vertical drawn where salt bridge separates two half- reactions. E.g. draw cell diagram for Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s). Fe 3+ (aq) + H 2 (g)  Fe 2+ (aq) + 2H + (aq).

9. 8–9 Chapter 20-9 Electrical Work Earlier w =  P  V In electrochemistry electrical pressure = potential difference; w = E  q or charge times the electrical pressure. Units: Coulomb  Volts = Joules (SI Units 1J = 1C  V) ; Also want to relate to # moles. –1 mole e  = 1 Faraday = 1 F F = qe  N = 1.602x10  19 C  6.022x10 23 /mol = 9.65x10 4 C/mol e .

10. 8–10 Chapter 20-10 Cell Potentials for Cell Reactions: Spontaneity of Redox Reactions  G vs E:   G   E   G =  nFE. Use this to calculate  G for electrochemical reaction when cell voltage known. E.g. Determine  G for Zn/Cu cell if E =  1.100V

11. 8–11 Chapter 20-11 Standard Reduction Potentials As with thermodynamic quantities, we list cell potentials at standard state = 1M at 1 atm and usually 25  C. Cell potential is the sum of half-cell potentials using Hess’ law. Half-cell reactions for Daniell cell were Potential at each electrode initially determined relative to SHE = Standard hydrogen electrode. 2H + (aq)+2e   H 2 (g); [H + ] = 1M and P H2 =1atm. Other reaction run to determine if the SHE reaction proceeds spontaneously in direction written when connected to other half-cells. E.g. the cell potential of copper at standard state conditions relative to SHE(acting as the anode) was 0.340 V; determine the half–cell potential for Zn  Zn 2+ if the potential for the Daniell cell (standard state conditions) was 1.100 V. Half-cell reactions reported as reductions.

12. 8–12 Chapter 20-12 Using Standard Reduction Potentials Large negative value means oxidation strongly favored; strong reducing agent. Large positive value means reduction strongly favored; strong oxidizing agent. Relative values in table give an indication that one half-reaction favored over other. Summing half-cell reactions allow determination of standard cell potential. Half-cell potential intensive property  independent of amount of material  we don’t use stoichiometric coefficients for determining standard cell potentials. E.g. determine the cell potential of Br 2 (l) + 2I  (aq)  I 2 (l) + 2Br  (aq) E.g.2 determine the cell potential of 2Ag + (aq) + Cu(s)  2Ag(s) + Cu 2+ (aq) E.g.3 Determine cell potential: (aq) + Fe(s)  Fe 2+ (aq) + Mn 2+ (aq) (balanced?). when it is operated galvanically. Which is the oxidizing agent? reducing agent? E.g. 4 Determine if the reaction below is spontaneous in the direction written. Fe 3+ (aq) + Ag(s)  ?

13. 8–13 Chapter 20-13 Spontaneity of Redox Reactions  G vs E:   G   E   G =  nFE. Use this to calculate  G for electrochemical reaction when cell voltage known. E.g. Determine  G for Zn/Cu cell if E = 1.100V

14. 8–14 Chapter 20-14 Effect of Concentration on Cell EMF: The Nernst Equation Recall that  G =  G o + RTlnQ where  or at 25°C which is called Nernst Equation. E.g. Determine potential of Daniell cell at 25  C if [Zn 2+ ] = 0.100 M and [Cu 2+ ] = 0.00100 M.

15. 8–15 Chapter 20-15 Electrochemical Determination of pH Electrodes can be used to determine acidity of solution by using hydrogen electrode with another one e.g. Hg 2 Cl 2 half-cell. E.g. determine the pH of a solution that develops a cell potential of 0.280 V (at 25°C) given the cell below Pt(s) | H 2 (g) (1atm) | H + (? M) || Pb 2+ (1 M) | Pb(s) E.g. 2 determine the pH of a solution that develops a cell potential of  0.200 V (at 25°C) given the cell below (called a concentration cell). Pt(s)|H 2 (g)(1atm)|H + (1.00M)||H + (?M)|H 2 (g)(1atm)|Pt(s)

16. 8–16 Chapter 20-16 Standard Cell Potentials and Equilibrium Constants Free energy and equilibrium constant for reaction studied can be determined from cell voltage. Cell potential can be determined from  G or from equilibrium constant. –Recall:  G° =  nFE° =  RTlnK. E.g. Determine free energy and equilibrium constant for reaction below (unbalanced). (aq) + Fe(s)  Fe 2+ (aq) + Mn 2+ (aq) E.g.2 Determine cell potential and equilibrium constant of Cl 2 /Br 2 cell. E.g.3 The following cell has a potential of 0.578 V at 25°C; determine K sp. Ag(s)|AgCl(s)|Cl  (1.0 M)||Ag + (1.0 M)|Ag(s).

17. 8–17 Chapter 20-17 Quantitative Aspects of Electrolysis Current, i, measured units: 1 ampere = 1 coulomb per s (1 A = 1 Cs  1). Time of electrolysis, t (s), also measured. Total charge, Q, calculated from the product: Q = i  t Charge on 1 mol of e  : mol of e  in an electrolysis obtained from balanced cell reaction: E.g. determine # mol of electron involved in the electrolysis of the following: –Ag + (aq) + e   Ag(s) –2Cl  (aq) + 2e  Cl 2 (g) Amount deposited given by: E.g. Determine amount of Cu 2+ electrolyzed from solution at constant current of 6.00 A for period of 1.00 hour.

18. 8–18 Chapter 20-18 Electrical Work Maximum electrical work:  G =  w max =  nFE –n = mol of electrons –F = Faraday’s constant –E = cell potential –Units – Joules Electrical Power: 1 watt = 1 J/s. Energy often expressed as kilowatt –hr 1 kW*hr = 1000 W*3600 s = 3.6x10 6 J E.g. determine the maximum work in kW*hr required to produce 1.00 kg of Zn from Zn 2+ in a Daniell cell where the cell potential is  1.100 V for the production of Zn metal. Strategy: –Determine n: mol of Zn 2+ times 2. –Calculate work.