John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition Chapter 17 Electrochemistry © 2012 Pearson Education, Inc.

Galvanic Cells Electrochemistry: The area of chemistry concerned with the interconversion of chemical and electrical energy Galvanic (Voltaic) Cell: A spontaneous chemical reaction which generates an electric current Electrolytic Cell: An electric current which drives a nonspontaneous reaction

Galvanic Cells Cu(s)Cu 2+ (aq) + 2e  Reduction half-reaction: Oxidation half-reaction: Zn 2+ (aq) + Cu(s)Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + 2e  Zn(s)

Galvanic Cells Zn 2+ (aq) + Cu(s)Zn(s) + Cu 2+ (aq)

Galvanic Cells Anode: The electrode where oxidation occurs. The electrode where electrons are produced. Is what anions migrate toward. Has a negative sign.

Galvanic Cells Cathode: The electrode where reduction occurs. The electrode where electrons are consumed. Is what cations migrate toward. Has a positive sign.

Galvanic Cells Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 17/8 Salt Bridge: a U-shaped tube that contains a gel permeated with a solution of an inert electrolytes Maintains electrical neutrality by a flow of ions Anions flow through the salt bridge from the cathode to anode compartment Cations migrate through salt bridge from the anode to cathode compartment

Galvanic Cells Why do negative ions (anions) move toward the negative electrode (anode)?

Galvanic Cells Zn 2+ (aq) + Cu(s)Zn(s) + Cu 2+ (aq) Cu(s)Cu 2+ (aq) + 2e  Zn 2+ (aq) + 2e  Zn(s) Overall cell reaction: Anode half-reaction: Cathode half-reaction: No electrons should be appeared in the overall cell reaction

Shorthand Notation for Galvanic Cells Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) Phase boundary Electron flow Salt bridge Cathode half-cellAnode half-cell

17.2 Shorthand Notation for Galvanic Cells Cell involving gas Additional vertical line due to presence of addition phase List the gas immediately adjacent to the appropriate electrode Detailed notation includes ion concentrations and gas pressure E.gCu(s) + Cl 2 (g)  Cu 2+ (aq) + 2 Cl - (aq) Cu(s)|Cu 2+( aq)||Cl 2 (g)|Cl - (aq)|C(s)

Example Consider the reactions below Write the two half reaction Identify the oxidation and reduction half Identify the anode and cathode Give short hand notation for a galvanic cell that employs the overall reaction Pb 2+( aq) + Ni(s)  Pb(s) + Ni 2+ (aq)

Example Given the following shorthand notation, sketch out the galvanic cell Pt(s)|Sn 2+,Sn 4+ (aq)||Ag + (aq)|Ag(s)

Cell Potentials and Free-Energy Changes for Cell Reactions Electromotive Force (emf): The force or electrical potential that pushes the negatively charged electrons away from the anode (  electrode) and pulls them toward the cathode (+ electrode). It is also called the cell potential (E) or the cell voltage.

Cell Potentials and Free-Energy Changes for Cell Reactions 1 J = 1 C x 1 V volt (V) SI unit of electric potential joule (J) SI unit of energy coulomb (C) Electric charge 1 coulomb is the amount of charge transferred when a current of 1 ampere (A) flows for 1 second.

Cell Potentials and Free-Energy Changes for Cell Reactions  G ° =  nFE ° Cell potentialFree-energy change Number of moles of electrons transferred in the reaction faraday or Faraday constant The electric charge on 1 mol of electrons and is equal to 96,500 C/mol e   G =  nFE or

Cell Potentials and Free-Energy Changes for Cell Reactions Calculate the standard free-energy change for this reaction at 25 ° C. Is the reaction spontanous at this condition? Zn 2+ (aq) + Cu(s)Zn(s) + Cu 2+ (aq) The standard cell potential at 25 ° C is 1.10 V for the reaction:

Examples Calculate the cell potential at standard state (E o cell ) for the following reaction. Then write the half reactions I2(s) + 2 Br-(aq)  2I-(aq) + Br2(l)G o = 1.1 x 10 5 J

Standard Reduction Potentials 2H 1+ (aq) + Cu(s)H 2 (g) + Cu 2+ (aq) Cu(s)Cu 2+ (aq) + 2e  2H + (aq) + 2e  H2(g)H2(g) Overall cell reaction: Anode half-reaction: Cathode half-reaction: The standard potential of a cell is the sum of the standard half-cell potentials for oxidation at the anode and reduction at the cathode: E ° cell = E ° ox + E ° red The measured potential for this cell: E ° cell = 0.34 V

Standard Reduction Potentials E o cell is the standard cell potential when both products and reactants are at their standard states: Solutes at 1.0 M Gases at 1.0 atm Solids and liquids in pure form Temp = 25.0 o C

Standard Reduction Potentials Spotaniety of the reaction can be determined by the positive E o cell value The cell reaction is spontaneous when the half reaction with the more positive E o value is cathode Note: E o cell is an intensive property; the value is independent of how much substance is used in the reaction Ag + (aq) + e-  Ag(s)E o red = 0.80 V 2 Ag + (aq) + 2e-  2 Ag(s) E o red = 0.80V

Standard Reduction Potentials The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode.

Standard Reduction Potentials 2H + (aq, 1 M) + 2e  H 2 (g, 1 atm) 2H + (aq, 1 M) + 2e  E ° red = 0 V E ° ox = 0 V The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode.

Standard Reduction Potentials 2H + (aq) + Cu(s)H 2 (g) + Cu 2+ (aq) Cu(s)Cu 2+ (aq) + 2e  2H + (aq) + 2e  H2(g)H2(g) Overall cell reaction: Anode half-reaction: Cathode half-reaction: 0.34 V = 0 V + E ° red E ° cell = E ° ox + E ° red Cu(s) Cu 2+ (aq) + 2e  E ° = 0.34 V A standard reduction potential can be defined:

Standard Reduction Potentials

Examples Of the two standard reduction half reactions below, write the net equation and determine which would be the anode and which would be the cathode of a galvanic cell. Calculate E o cell a. Cd 2+ (aq) + 2e-  Cd(s)E o red = V Ag + (aq) + e-  Ag(s) E o red = 0.80 V b. Fe 2+ (aq) + 2e-  Fe(s) E o red = V Al 3+ (aq) + 3e-  Al(s) E o red = V

Standard Cell Potentials and Equilibrium Constants -nFE° = -RT ln K  G° = -RT ln K andUsing  G° = -nFE° log K n V E° = log K nF RT ln K nF RT =E° = in volts, at 25°C

The Nernst Equation What is the potential of a cell at 25 o C that has the following ion concentrations? Cu 2+ (aq) + 2Fe 2+ (aq)Cu(s) + 2Fe 3+ (aq) Consider a galvanic cell that uses the reaction: [Fe 2+ ] = 0.20 M[Fe 3+ ] = 1.0 × 10  4 M[Cu 2+ ] = 0.25 M

Example Use the tabulated half-cell potentials to calculate K for the following oxidation of copper by H+ 2Cu(s) + 2H + (aq)  Cu 2+ (aq) + H2(g)