Redox Reactions

Oxidation-Reduction Reactions (Redox) Chemical reactions where reactants exchange electrons. Eg: rusting and corrosion; all types of batteries, alkaline, NiCad, car; metabolism OXIDATION: a process in which a substance loses electrons resulting in an increase in oxidation number. REDUCTION: a process in which a substance gains electrons resulting in a lower oxidation number. OXIDATION NUMBER: a positive or negative # assigned to an atom, ion or element. It indicates how electrons are shared. See rules for determining numbers.

Redox reactions are chemical reactions in which 2 or more atoms undergo a change in oxidation number. Some redox reactions are spontaneous and result in a decrease in potential energy They liberate energy which can be converted into electrochemical energy battery. other redox reactions are not spontaneous and require energy to proceed. Electrolytic cells hydrolysis of water recharging a battery

LEO says GER Loss of electrons; oxidation / Gain of electrons; reduction) +4 +3 +2 +1 0 -1 -2 -3 -4 Decreased oxidation number REDUCTION OXIDATION increased oxidation number

Typical redox reactions Magnesium burns in air to produce a bright light 2Mg(s) + O2 (g) 2MgO(s) Mg is oxidized by O2 ;O2 is called the oxidizing agent. 2Mg 2Mg2+ + 4e- Each Mg loses 2 electrons (4 e- lost) O2 is reduced by Mg; Mg is called the reducing agent. O2 + 4e- 2O2- Each O in O2 gains 2 electrons (4 e- gained) IF ONE STUBSTANCE IS OXIDIZED, ANOTHER IN THE SAME REACTION MUST BE REDUCED.

Copper wire in silver nitrate solution Cu(s) + 2AgNO3 (aq) 2Ag(s) + Cu(NO3)2(aq) Net ionic equation: Cu(s) + 2Ag+ (aq) 2Ag(s) + Cu2+(aq) Note: spectator ions NO3- are not shown in equation because they do not react chemically. Separately: 2Ag+(aq) + 2e- 2Ag(s) gain of electrons  reduction copper wire is the reducing agent Cu(s) Cu2+(aq) + 2e- loss of electrons  oxidation silver ion is the oxidizing agent

Copper penny in nitric acid Cu(s) + 4 HNO3 (aq) 2NO2 (g) + Cu(NO3)2(aq) + 2H2O(l) Net ionic equation: Cu(s) + 2NO3- (aq) + 4H+ (aq) 2NO2 (g) + Cu2+(aq) + 2H2O(l) Separately: Cu(s) Cu2+(aq) + 2e- Cu loses of electrons  oxidation 2NO3- (aq) + 4H+ (aq) + 2e- 2NO2 (g) + 2H2O(l) How do you know N is reduced? N gains of electrons  reduction

Oxidation States A way of keeping track of the electrons. need the rules for assigning (memorize). The oxidation state of elements in their standard states is zero. Cu(s), Na(s), O2(g) Oxidation state for monoatomic ions are the same as their charge. Cu2+, I-, N3- Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide. H2O, NO2 (ox. # -2) H2O2, Na2O2 (ox. # -1) In compounds with nonmetals hydrogen is assigned the oxidation state +1. H2SO4, HCl, NaHCO3

In its compounds fluorine is always –1. SF6 The sum of the oxidation states must be zero in compounds. PbCl4 NaBrO3 +4 + 4(-1) = 0 +1 + +5 + 3(-2) = 0 7. The sum of the oxidation states must be equal the charge of the ion. NO3- Cr2O7 2- +5 + 3(-2) = -1 2(+6) + 7(-2) = -2 8. Group IA elements are always +1 CsF, NaCl Hydrogen is always +1 with the exception of hydrides. (H is -1). LiH 10.Group IIA elements are always +2. CaF2 , BaCl2

Oxidation States Assign the oxidation states to each element in the following. CO2 NO3- H2SO4 Fe2O3 Na2Cr2O7 C = +4 O = -2 O = -2 N = +5 S = +6 O = -2 H = +1 Fe = +3 O = -2 Cr = +6 O = -2 Na = +1

Oxidation-Reduction summary Oxidation means an increase in oxidation state - lose electrons. Reduction means a decrease in oxidation state - gain electrons. The substance that is oxidized is called the reducing agent. The substance that is reduced is called the oxidizing agent.

Agents Oxidizing agents Reducing agents gets reduced gains electrons. More negative oxidation state. Reducing agents gets oxidized. Loses electrons. More positive oxidation state.

Identify the… Oxidizing agent; Reducing agent Substance oxidized; Substance reduced in the following reactions Fe (s) + O2(g) ® Fe2O3(s) Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g) c) SO3-2 + H+ + MnO4- ® SO4-2 + H2O + Mn+2

homework P. 653 #2 P.657 # 9-11 P. 659 #12, 13 P.662 #19

Half-Reactions All redox reactions can be thought of as happening in two halves. One produces electrons - Oxidation half. The other requires electrons - Reduction half. Write the half reactions for the following. Na + Cl2 ® Na+ + Cl- SO3-2 + H+ + MnO4- ® SO4-2 + H2O + Mn+2

Steps to Balancing Redox Equations In aqueous solutions the key is the number of electrons produced must be the same as those required. For reactions in acidic solution an 8 step procedure. For reactions in basic solutions, one more step is required.

Acidic Solution Write separate half reactions For each half reaction balance all reactants except H and O Balance O using H2O Balance H using H+ Balance charge using e- Multiply equations to make electrons equal Add equations and cancel identical species Check that charges and elements are balanced.

BrO3- + I- Br- + I2 in an acidic solution Write separate half reactions BrO3- Br- I- I2 For each half reaction balance all reactants except H and O BrO3- Br- 2 I- I2 done Balance O using H2O BrO3- Br- + 3 H2O 2 I- I2 Balance H using H+ BrO3- + 6H+ Br- + 3 H2O 2 I- I2

BrO3- + 6H+ + 6e- Br- + 3 H2O 2 I- I2 + 2e- Balance charge using e- BrO3- + 6H+ + 6e- Br- + 3 H2O 2 I- I2 + 2e- Multiply equations to make electrons equal BrO3- + 6H+ + 6e- Br- + 3 H2O 3[2 I- I2 + 2e- ] Add equations and cancel identical species BrO3- + 6H+ + 6I- Br- + 3 I2 + 3 H2O 8. Check that charges and elements are balanced.

Practice The following reactions occur in acidic solutions. Balance them. a) MnO4- + H2O2 ® Mn+2 + O2 b) I- + NO3- ® I2 + NO(g) c) Cr2O72- + Fe2+ ® Cr3+ + Fe3+ d) Mn+2 + BiO3- ® Bi+3 + MnO4-

Basic Solution Do everything you would with acid, but add one more step. Add enough OH- to both sides to neutralize the H+

Al + NO3- Al+3 + NH3 in a basic solution Write separate half reactions Al Al3+ NO3- NH3 For each half reaction balance all reactants except H and O done done Balance O using H2O Al Al3+ NO3- NH3 + 3H2O done Balance H using H+ Al Al3+ NO3- + 9 H+ NH3 + 3H2O

Balance charge using e- Al Al3+ + 3e- NO3- + 7H+ + 6e- NH3 + H2O Add OH- to both sides to balance H+ . Create H2O. Al Al3+ + 3e- NO3- + 7 H+ + 6e- + 7OH- NH3 + 3H2O + 7OH- Al Al3+ + 3e- NO3- + 7H2O + 6e- NH3 + 3H2O + 7OH- 7. Multiply equations to make electrons equal 2[Al Al3+ + 3e- ] NO3- + 7H2O + 6e- NH3+ 3H2O + 7OH- 8. Add equations and cancel identical species 2Al + NO3- + 4H2O 2Al3+ + NH3 + 7OH- 9. Check that charges and elements are balanced.

Practice The following reactions occur in basic solutions. Balance them. Cr(OH)3 + OCl- + ® CrO4-2 + Cl- + H2O MnO4- + Fe+2 ® Mn+2 + Fe+3 Fe(OH)2 + H2O2 ® Fe(OH)4 - d) S2O32- + OCl- ® SO42- + Cl-

Homework:p. 671 # 5 p673 #6-8, 3,4,7,