Redox Reactions OxidationReduction Reactions Redox Chemical reactions where
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Redox Reactions
Oxidation-Reduction Reactions (Redox) Chemical reactions where reactants exchange electrons. Eg: rusting and corrosion; all types of batteries, alkaline, Ni. Cad, 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.
l Redox reactions are chemical reactions in which 2 or more atoms undergo a change in oxidation number. l Some redox reactions are spontaneous and result in a decrease in potential energy They liberate energy which can be converted into electrochemical energy battery. l l l 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 Decreased +2 oxidation +1 number 0 -1 -2 -3 increased -4 oxidation number
Typical redox reactions Magnesium burns in air to produce a bright light 2 Mg(s) + O 2 (g) 2 Mg. O(s) Mg is oxidized by O 2 ; O 2 is called the oxidizing agent. 2 Mg 2+ + 4 e. Each Mg loses 2 electrons (4 e- lost) O 2 is reduced by Mg; Mg is called the reducing agent. O 2 + 4 e 2 O 2 Each O in O 2 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) + 2 Ag. NO 3 (aq) Net ionic equation: Cu(s) + 2 Ag+ (aq) 2 Ag(s) + Cu(NO 3)2(aq) 2 Ag(s) + Cu 2+(aq) Note: spectator ions NO 3 - are not shown in equation because they do not react chemically. Separately: 2 Ag+(aq) + 2 e 2 Ag(s) gain of electrons reduction copper wire is the reducing agent Cu(s) Cu 2+(aq) + 2 eloss of electrons oxidation silver ion is the oxidizing agent
Copper penny in nitric acid Cu(s) + 4 HNO 3 (aq) 2 NO 2 (g) + Cu(NO 3)2(aq) + 2 H 2 O(l) Net ionic equation: Cu(s) + 2 NO 3 - (aq) + 4 H+ (aq) 2 NO 2 (g) + Cu 2+(aq) + 2 H 2 O(l) Separately: Cu(s) Cu 2+(aq) + 2 e. Cu loses of electrons oxidation 2 NO 3 - (aq) + 4 H+ (aq) + 2 e 2 NO 2 (g) + 2 H 2 O(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). 1 The oxidation state of elements in their standard states is zero. Cu(s), Na(s), O 2(g) 2 Oxidation state for monoatomic ions are the same as their charge. Cu 2+, I-, N 3 - 3 Oxygen is assigned an oxidation state of -2 in its covalent compounds except as a peroxide. H 2 O, NO 2 (ox. # -2) H 2 O 2, Na 2 O 2 (ox. # -1) 4 In compounds with nonmetals hydrogen is assigned the oxidation state +1. H 2 SO 4, HCl, Na. HCO 3
5 In its compounds fluorine is always – 1. SF 6 6 The sum of the oxidation states must be zero in compounds. Pb. Cl 4 +4 + 4(-1) = 0 Na. Br. O 3 +1 + +5 + 3(-2) = 0 7. The sum of the oxidation states must be equal the charge of the ion. NO 3 Cr 2 O 7 2+5 + 3(-2) = -1 2(+6) + 7(-2) = -2 8. Group IA elements are always +1 Cs. F, Na. Cl 9. Hydrogen is always +1 with the exception of hydrides. (H is -1). Li. H 10. Group IIA elements are always +2. Ca. F 2 , Ba. Cl 2
Oxidation States 1. 2. 3. 4. 5. Assign the oxidation states to each element in the following. CO 2 C = +4 O = -2 NO 3 O = -2 N = +5 H 2 SO 4 S = +6 O = -2 H = +1 Fe 2 O 3 Fe = +3 O = -2 Na 2 Cr 2 O 7 Na = +1 Cr = +6 O = -2
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 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 a) Fe (s) + O 2(g) ® Fe 2 O 3(s) b) Fe 2 O 3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO 2(g) c) SO 3 -2 + H+ + Mn. O 4 - ® SO 4 -2 + H 2 O + 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 + Cl 2 ® Na+ + Cl SO 3 -2 + H+ + Mn. O 4 - ® SO 4 -2 + H 2 O + 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 1 Write separate half reactions 2 For each half reaction balance all reactants except H and O 3 Balance O using H 2 O 4 Balance H using H+ 5 Balance charge using e 6 Multiply equations to make electrons equal 7 Add equations and cancel identical species 8 Check that charges and elements are balanced.
Br. O 3 - + I- Br- in an acidic solution + I 2 1 Write separate half reactions Br. O 3 - Br- I- I 2 Br. O 3 - Br- 2 I- I 2 2 For each half reaction balance all reactants except H and O done 3 Balance O using H 2 O Br. O 3 - Br- + 3 H 2 O done 4 Balance H using H+ Br. O 3 - + 6 H+ Br- + 3 H 2 O 2 Idone I 2
5. Balance charge using e- Br. O 3 - + 6 H+ + 6 e 6. I 2 + 2 e- Br- + 3 H 2 O 3[2 I- I 2 + 2 e- ] Add equations and cancel identical species Br. O 3 - + 6 H+ + 6 I 8. 2 I- Multiply equations to make electrons equal Br. O 3 - + 6 H+ + 6 e 7. Br- + 3 H 2 O Br- + 3 I 2 + Check that charges and elements are balanced. 3 H 2 O
Practice The following reactions occur in acidic solutions. Balance them. a) Mn. O 4 - + H 2 O 2 ® Mn+2 + O 2 b) I- + NO 3 - ® I 2 + NO(g) c) Cr 2 O 72 - + Fe 2+ ® Cr 3+ + Fe 3+ d) Mn+2 + Bi. O 3 - ® Bi+3 + Mn. O 4 -
Basic Solution Do everything you would with acid, but add one more step. Add enough OH- to both sides to neutralize the H+
Al + NO 3 - Al+3 + NH 3 in a basic solution 1 Write separate half reactions Al Al 3+ NO 3 - NH 3 2 For each half reaction balance all reactants except H and O done 3 Balance O using H 2 O Al Al 3+ done NO 3 - done NH 3 + 3 H 2 O 4 Balance H using H+ Al Al 3+ done NO 3 - + 9 H+ NH 3 + 3 H 2 O
5. Balance charge using e. Al Al 3+ + 3 e. NO 3 - + 7 H+ + 6 e. NH 3 + H 2 O 6. Add OH- to both sides to balance H+. Create H 2 O. Al Al 3+ + 3 e. NO 3 - + 7 H+ + 6 e- + 7 OH- NH 3 + 3 H 2 O + 7 OH Al Al 3+ + 3 e- NO 3 - + 7 H 2 O + 6 e. NH 3 + 3 H 2 O + 7 OH 7. Multiply equations to make electrons equal 2[Al Al 3+ + 3 e- ] NO 3 - + 7 H 2 O + 6 e. NH 3+ 3 H 2 O + 7 OH 8. Add equations and cancel identical species 2 Al + NO 3 - + 4 H 2 O 2 Al 3+ + NH 3 + 7 OH 9. Check that charges and elements are balanced.
Practice The following reactions occur in basic solutions. Balance them. a) Cr(OH)3 + OCl- + ® Cr. O 4 -2 + Cl- + H 2 O b) Mn. O 4 - + Fe+2 ® Mn+2 + Fe+3 c) Fe(OH)2 + H 2 O 2 ® Fe(OH)4 - d) S 2 O 32 - + OCl- ® SO 42 - + Cl-
Homework: p. 671 # 5 p 673 #6 -8, 3, 4, 7,
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