(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 /ClOx1 + 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