中国海洋大学：《无机化学 Concise Inorganic Chemistry》课程教学资源（PPT课件，英文）Chapter 6 Redox

Chapter 6 Oxidation and Reduction(RedoxReaction)Fourissues:1.Basicknowledge2.StandardReductionPotentials3.Introduction of Voltaic Cells4.Electrolysis and Electrolytic Cells

Chapter 6 Oxidation and Reduction (Redox Reaction) Four issues： 1. Basic knowledge 2. Standard Reduction Potentials 3. Introduction of Voltaic Cells 4. Electrolysis and Electrolytic Cells

1Basicknowledge1-1 Concepts1-2BalancingRedox Equations1-3Voltaiccells1-3-1 What kinds of redox reactionsbecanharnessed to produce electrical energy?1-3-2 Electrode Types and Cell Notation1-3-3CellPotentials

1 Basic knowledge 1-1 Concepts 1-2 Balancing Redox Equations 1-3 Voltaic cells 1-3-1 What kinds of redox reactions can be harnessed to produce electrical energy? 1-3-2 Electrode Types and Cell Notation 1-3-3 Cell Potentials

1-1ConceptsZn+Cu2+=Zn2++CuC,Hi20,+6 0,=6 CO, +6H,0(1)OxidationNumber假设分子中成键的电子都归电负性较大的原子，而得到的某元素的一个原子所带的形式电荷数（表观电荷数）

1-1 Concepts Zn + Cu2+ = Zn2+ + Cu C6H12O6 + 6 O2 = 6 CO2 + 6 H2O (1) Oxidation Number 假设分子中成键的电子都归电负性较大的原子，而得到 的某元素的一个原子所带的形式电荷数（表观电荷数）

Rules for assigning oxidation number:The oxidation number of an element in an elementarysubstanceisO.Generally, in compounds, the oxidation number of His+1 , the oxidation number of O is-2,the oxidation numberof halide F is -l,the oxidation number of Group 1 metals(IA)is +1,theoxidation number of Group2 elements(IlA)is +2.The sum of the oxidation numbers of all the atoms in aneutral species is O; in an ion, it is equal to the charge of thation

Rules for assigning oxidation number： u The oxidation number of an element in an elementary substance is 0. u Generally, in compounds, the oxidation number of H is +1 ，the oxidation number of O is -2, the oxidation number of halide F is -1, the oxidation number of Group 1 metals(IA) is +1, the oxidation number of Group 2 elements( IIA ) is +2. u The sum of the oxidation numbers of all the atoms in a neutral species is 0; in an ion, it is equal to the charge of that ion

Eg.S4O.2-, 0xid.no.S× 4 +2x6 =-2, 0xid.no.S=2.5CrO,0xid.no.Cr+ 2x5=0;oxid.no.Cr=-10Thus, the oxidation number of can be integers or fractions

Eg. S4O6 2- , oxid. no. S  4 + 26 = -2, oxid. no. S = 2.5 CrO5 oxid. no. Cr + 25 = 0 ; oxid. no. Cr = -10 Thus, the oxidation number of can be integers or fractions

(2)Definition of Oxidation andReduction:Oxidation is defined as an increase in oxidation number, andreduction as a decrease in oxidation number. A redoxreaction can be defined as a reaction that has oxidationnumber changes. In a redox reaction, there are usually atleast two reactants. One with oxidation number decrease isreferred to as the oxidizing agent, the other with oxidationnumber increase is referred to as the reducing agentZn -2e → Zn2+ (Oxidation, Zn is reducing agent)Cu2++2e-→Cu (reduction, Cu is oxidizing agent)These two reaction equations are called half equationsObviously, a redox reaction equations is composed of twohalfequations

(2) Definition of Oxidation and Reduction： Oxidation is defined as an increase in oxidation number, and reduction as a decrease in oxidation number. A redox reaction can be defined as a reaction that has oxidation number changes. In a redox reaction, there are usually at least two reactants. One with oxidation number decrease is referred to as the oxidizing agent, the other with oxidation number increase is referred to as the reducing agent. Zn – 2e → Zn2+ （Oxidation, Zn is reducing agent） Cu2+ + 2e→ Cu (reduction, Cu is oxidizing agent) These two reaction equations are called half equations. Obviously, a redox reaction equations is composed of two half equations

ISelf Redox reactions : the oxidizing and reducing agentisthesamecompound.歧化反应：the oxidizing and reducing agent is the sameelementina substance.2KCI0,=2KCI+30CI, + H,O=HCI+HCIO

lSelf Redox reactions : the oxidizing and reducing agent is the same compound. 歧化反应: the oxidizing and reducing agent is the same element in a substance. 2KClO3 =2 KCl +3 O2 Cl2 + H2O = HCl +HClO

Redoxelectronicpair：同一元素的两种不同的氧化态构成氧化还原电对。Ox/Redeg.Cu2+/Cu, Zn2+/Zn, H+/H2, Cl,/CIOx1 + Red2 → Ox2 + Redl

Redox electronic pair：同一元素的两种不同的氧化态， 构成氧化还原电对。Ox / Red eg. Cu2+ /Cu, Zn2+ /Zn, H+ /H2 , Cl2 /Cl￾Ox1 + Red2 → Ox2 + Red1

1-2BalancingRedoxEquationshalf equations method or ion-electron method

1-2 Balancing Redox Equations half equations method or ion-electron method

Procedures:Write the net ionic equations instead of completeionic equationsSplit the equation into two half-equations, one oxidation and onereduction.Balance the atoms of the two half-equations, first with respect toatomsthat have oxidation number changes and then with respect to H and0.In acidic solution,addH+ You cannot add OH-!)and add H,O on the sideoflessOatoms.In basic solution, add OH-and H,OYou cannot add Ht!).Add OH-on theside of less O atoms. and add OH-on the side of more H atoms when Oatomsareequal.Balancethechargeofthetwohalf-equationsFind theleastcommon multipleof the charges of the two half-eguations.combinethemsoas to makethe numberofelectronsarinedin reductionequal to the number lostin oxidation

Procedures： u Write the net ionic equations instead of complete ionic equations. Split the equation into two half-equations, one oxidation and one reduction. u Balance the atoms of the two half-equations, first with respect to atomsthat have oxidation number changes and then with respect to H and O. In acidic solution, add H+ (You cannot add OH- !) and add H2O on the side of less O atoms. In basic solution, add OH- and H2O(You cannot add H+ !). Add OH- on the side of less O atoms, and add OH- on the side of more H atoms when O atoms are equal. u Balance the charge of the two half-equations Find the least common multiple of the charges of the two half-equations, combine them so as to make the number of electrons arined in reduction equal to the number lost in oxidation