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Chapter 20

Chapter 20. Electrochemistry. 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

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Chapter 20

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  1. Chapter 20 Electrochemistry

  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. Electrochemistry • Electrochemistry - Field of Chemistry dealing with transfer of electrons from one species to another. E.g. Zn in CuSO4(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. 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)+CaC2(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. 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 H2O to the deficient side. • Balance H's with H+ to the hydrogen deficient side • Acidic: add H+ • Basic: add H2O 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) + VO2+(aq)  Zn2+(aq) + V3+(aq). • Basic: Ag(s) + HS(aq) + (aq)  Ag2S(s) + Cr(OH)3(s).

  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. Galvanic Cell 2 • anode – electrode whereoxidation occurs. • cathode – electrode wherereduction 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) + Cd2+(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: 2Fe3+(aq) + 3Ni  2Fe(s) + 3Ni2+(aq)

  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. Cu2+/Cu, Zn2+/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) + Cu2+(aq)  Zn2+(aq) + Cu(s). Fe3+(aq) + H2(g)  Fe2+(aq) + 2H+(aq).

  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19C6.022x1023/mol = 9.65x104C/mol e-.

  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. 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 H2(g); [H+] = 1M and PH2=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  Zn2+ if the potential for the Daniell cell (standard state conditions) was 1.100 V. • Half-cell reactions reported as reductions.

  12. Using Standard Reduction Potentials • Largenegative value meansoxidation strongly favored; strong reducing agent. • Largepositive value meansreduction 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 Br2(l) + 2I(aq)  I2(l) + 2Br(aq) E.g.2 determine the cell potential of 2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq) E.g.3 Determine cell potential: (aq) + Fe(s)  Fe2+(aq) + Mn2+(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. Fe3+(aq) + Ag(s)  ?

  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. Effect of Concentration on Cell EMF: The Nernst Equation • Recall that G = Go + RTlnQ where  or at 25°C which is called Nernst Equation. • E.g. Determine potential of Daniell cell at 25C if [Zn2+] = 0.100 M and [Cu2+] = 0.00100 M.

  15. Electrochemical Determination of pH • Electrodes can be used to determine acidity of solution by using hydrogen electrode with another one e.g. Hg2Cl2 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) | H2(g) (1atm) | H+(? M) || Pb2+(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)|H2(g)(1atm)|H+(1.00M)||H+(?M)|H2(g)(1atm)|Pt(s)

  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)  Fe2+(aq) + Mn2+(aq) E.g.2 Determine cell potential and equilibrium constant of Cl2/Br2 cell. E.g.3 The following cell has a potential of 0.578 V at 25°C; determine Ksp. Ag(s)|AgCl(s)|Cl(1.0 M)||Ag+(1.0 M)|Ag(s).

  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 Cl2(g) • Amount deposited given by: E.g. Determine amount of Cu2+ electrolyzed from solution at constant current of 6.00 A for period of 1.00 hour.

  18. Electrical Work • Maximum electrical work: G = wmax = 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 = 3.6x106 J • E.g. determine the maximum work in kW*hr required to produce 1.00 kg of Zn from Zn2+ in a Daniell cell where the cell potential is 1.100 V for the production of Zn metal. Strategy: • Determine n: mol of Zn2+ times 2. • Calculate work.

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