Chapter 17 Electrochemistry Electrochemistry deals with the relationships between electricity and chemical reactions. Oxidation-reduction (redox) reactions were introduced in Chapter 4 Can be simple displacement reactions: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)

Zn + Sn2+  Zn2+ + Sn

Zn + Cu2+  Zn2+ + Cu

Oxidation-Reduction Reactions Redox reactions can also be more complex, with structural and composition changes as well as an exchange of electrons. 5VO2+(aq) + MnO4-(aq) + H2O(l)  5VO2+(aq) + Mn2+(aq) + 2H+(aq) 5Fe2+(aq) + MnO4-(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) + 4H2O(l) 21m03vd1

Oxidation and Reduction You can’t have one without the other! Decrease in number of electrons (loss of electrons) Increase in oxidation number Oxidation number: -3 -2 -1 0 +1 +2 +3 Increase in number of electrons (gain of electrons) Decrease in oxidation number Reduction

Half-Reactions Oxidation-reduction reactions can be written as separate oxidation and reduction reactions, called half-reactions. Oxidation: VO2+(aq) + H2O(l)  VO2+(aq) + 2H+(aq) + e- VO2+ is called the reducing agent (or reductant), because it causes the reduction of another substance; the reducing agent is oxidized in the process

Half-Reactions Reduction: MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l) MnO4- is called the oxidizing agent (or oxidant), because it causes the oxidation of another substance; the oxidizing agent is reduced in the process Oxidation-reduction can be considered to be the competition between two substances for electrons. The one with the greater attraction for additional electrons becomes the oxidizing agent; the one with the least is the reducing agent.

Spontaneous Oxidation-Reduction Reactions If we place a metal in a solution of another metal ion, sometimes we get metal deposition, sometimes not. How do we decide? Zn + Sn2+  Zn2+ + Sn Sn + Zn2+  no rxn

Spontaneous Redox How do we know which substances will act as oxidizing agents or as reducing agents? An activity series gives the relative activity of substances as oxidizing or reducing agents. Review from Chapter 4. Determine an activity series in several ways: activity in displacing H2 from water activity in displacing metals from metal ion solutions (more active metal displaces a less active metal from solution) generation of an electrochemical potential or voltage

Spontaneous Redox Na displaces H2 from H2O Zn displaces Pb from a solution of Pb2+

Spontaneous Redox Pb + 2Ag+  Pb2+ + 2Ag Ag + Pb2+  no rxn

Spontaneous Redox The reverse reactions can be made to occur with electric current, but they are not spontaneous. General rule, using an activity series: stronger stronger weaker weaker oxidizing + reducing  reducing + oxidizing agent agent agent agent Depending on the relative strengths, the reaction can go to completion, or reach a state of equilibrium.

Activity Series

Group Work Predict which of the following combinations will undergo an oxidation-reduction reaction: Mg + K+ Mg + Zn2+ H2 + Ni2+ H2 + Cu2+

Group Quiz Predict which of the following combinations will undergo an oxidation-reduction reaction: Sn + Pb2+ Na + H2O Pb + H2O Co + H+(aq) Ag + H+(aq)

20.2 Balancing Oxidation-Reduction Equations Some redox equations can be balanced by inspection, just like other types of reactions. Zn + CuCl2  ZnCl2 + Cu Net ionic equation: Zn + Cu2+  Zn2+ + Cu Need to balance atoms and charge What is wrong with the following? Al + Cu2+  Al3+ + Cu (Al-can demo)

Balancing Equations 5Cr3+ + 3MnO4- + 8H2O  5CrO42- + 3Mn2+ + 16H+ How do we balance an equation as complex as this? Two methods: half-reaction method oxidation number change method Will focus on the half-reaction method since it is useful in more circumstances.

Half-Reaction Method Write an unbalanced half-reaction for either oxidation or reduction. Balance the half-reaction: a. Balance all atoms other than H and O. b. Balance O by adding H2O to the equation. c. Balance H by adding H+ to the equation. d. Balance ionic charges by adding electrons to the equation. e. If in basic solution, add OH- to each side of the equation to neutralize all the H+. f. Cancel any H2O occurring on both sides.

Half-Reaction Method continued Write an unbalanced half-reaction for the other process. Balance by the same procedure. Equalize the number of electrons lost and gained by multiplying each coefficient in each half-reaction by the appropriate constant. Add the two half-reactions and cancel equal amounts of anything occurring on both sides. Make a final check of atom and charge balances.

Balancing Equations Cr3+ + MnO4-  CrO42- + Mn2+ How do we balance this equation in acidic solution? 1. Cr3+  CrO42- 2a. Cr already balanced 2b. Cr3+ + 4H2O  CrO42- 2c. Cr3+ + 4H2O  CrO42- + 8H+ 2d. Cr3+ + 4H2O  CrO42- + 8H+ + 3e-

Balancing Equations 3. MnO4-  Mn2+ 4a. Mn already balanced 4b. MnO4-  Mn2+ + 4H2O 4c. MnO4- + 8H+  Mn2+ + 4H2O 4d. MnO4- + 8H+ + 5e-  Mn2+ + 4H2O 5. 3 e- lost, but 5 e- gained. Equalize by multiplying half-reactions: 5(Cr3+ + 4H2O  CrO42- + 8H+ + 3e-) 3(MnO4- + 8H+ + 5e-  Mn2+ + 4H2O)

Balancing Equations 5Cr3+ + 20H2O  5CrO42- + 40H+ + 15e- 3MnO4- + 24H+ + 15e-  3Mn2+ + 12H2O 6. Add the equations: 5Cr3+ + 3MnO4- + 20H2O + 24H+ + 15e-  5CrO42- + 3Mn2+ + 12H2O + 40H+ + 15e- Cancel any substances on both sides of the equation: 5Cr3+ + 3MnO4- + 8H2O  5CrO42- + 3Mn2+ + 16H+ 7. Check atom and charge balance: 5 Cr, 3 Mn, 20 O, 16 H, 12 + charges on each side

Balancing Equations in Basic Solution Neutralize any H+ with OH- to form H2O and add the same number of OH- to the other side of the equation. Cr3+ + 4H2O  CrO42- + 8H+ + 3e- Add 8OH- to each side of the equation: Cr3+ + 4H2O + 8OH-  CrO42- + 8H+ + 8OH- + 3e- Form water by neutralization: Cr3+ + 4H2O + 8OH-  CrO42- + 8H2O + 3e- Cancel any water occurring on both sides: Cr3+ + 8OH-  CrO42- + 4H2O + 3e-

Group Work Balance the following equation in acidic solution: MnO4- + Cl-  Mn2+ + ClO3- Answer: MnO4- + 8H+ + 5e-  Mn2+ + 4H2O Cl- + 3H2O  ClO3- + 6H+ + 6e- 6MnO4- + 5Cl- + 18H+  6Mn2+ + 5ClO3- + 9H2O

Group Quiz 3V, 1Cr, 7O, 2H, 6 + charges on each side Balance the following chemical equation in acidic solution: VO2+ + CrO42-  VO2+ + Cr3+ 3VO2+ + CrO42- + 2H+  3VO2+ + Cr3+ + H2O 3V, 1Cr, 7O, 2H, 6 + charges on each side