Ch. 20: Electrochemistry Lecture 1: Redox Review

Electrochemistry Terminology #1 Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na+ + e- Reduction – A process in which an element attains a more negative oxidation state Cl2 + 2e-  2Cl-

Electrochemistry Terminology #2 An old memory device for oxidation and reduction goes like this… OIL RIG Oxidation Is Losing (electrons) Reduction Is Gaining (electrons)

Electrochemistry Terminology #3 You’ll need to know this for the SAT test and for college chemistry, but not for AP Chem. Oxidizing agent The substance that is reduced is the oxidizing agent Reducing agent The substance that is oxidized is the reducing agent

Oxidation Number Rules (p. 137 in book) Zero (0) An atom in its elemental form Cu(s) and each H atom in H2 has an oxidation # of zero. Its charge A monatomic metal ion e.g., K+ has an oxidation # of +1. -2 Oxygen (usually) Except in peroxides (i.e., H2O2): each oxygen is -1. +1 Hydrogen bonded to nonmetals -1 Hydrogen bonded to metals -1 Fluorine (always, except F2(g)) -1 Other halogens (except with N, O, or other halogens)

Oxidation Number Rules (p. 137 in book) (continued) The SUM of oxidation numbers of all atoms: Is zero in a neutral compound Is the charge of the ion in a polyatomic ion

Oxidation Number Rules EXAMPLE What is the oxidation number of sulfur in each compound? H2S S8 (s) SCl2 Na2SO3 SO42- -2 +2 +4 +6

RedOx Reactions Two half reactions: Cu(s)  Cu2+ + 2e- Cu(s) + 2 AgNO3 (aq)  Cu(NO3)2 (aq) + 2 Ag (s) Net ionic: Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s) Two half reactions: Cu(s)  Cu2+ + 2e- 2Ag+ + 2e-  2Ag(s) Cu is oxidized (loses electrons); Ag is reduced (gains electrons)

Zn(s) + 2 HNO3 (aq)  Zn(NO3)2(aq) + H2(g) RedOx Reactions Zn(s) + 2 HNO3 (aq)  Zn(NO3)2(aq) + H2(g) net ionic: Zn(s) + 2 H+ (aq)  Zn2+ (aq) + H2(g) Two half reactions: Zn(s)  Zn2+ + 2e- 2 H+ + 2e-  H2(g) Zinc is oxidized (loses electrons); H is reduced (gains electrons)

RedOx Reactions Write the overall reaction if Mn is oxidized (lose electrons so reverse rxn) and Cr is reduced. Reduction Rxns: Mn2+ + 2e-  Mn(s) 3x(Mn(s)Mn2++2e-) Cr3++ 3e-  Cr(s) 2x(Cr3++3e- Cr(s)) Add the two together: 3 Mn(s) + 2 Cr3+ (aq) +6e-  3 Mn2+ (aq) + 2 Cr (s)+6e- net ionic: 3 Mn(s) + 2 Cr3+ (aq)  3 Mn2+ (aq) + 2 Cr (s)