复旦大学：《普通化学 General Chemistry》课程教学资源（课堂讲义）Chap 7 - Oxidation & Reduction

PART 7-Oxidation Reduction Reference: Chapter 11 in the textbook

PART 7 – Oxidation & Reduction Oxidation & Reduction Reference: Chapter 11 in the textbook Reference: Chapter 11 in the textbook 1

Oxidation number Oxidation number or oxidation state Definition In a compound the charge of each element would have if all shared e- pairs in the lewis structure were transferred to the more electronegative atom e.g. HCL, Co, CO2, H2O, Cl2, P4, Cao, Distinguish between the Oxidation Number and the Formal charge e.g. HCl, CO, cO2

Oxidation Numbe r z O da o u be o O da o S a e xidati on N u mber or Oxidati on S t a t e: „ Definition: In a compound, the charge of each element would have if all shared e- pairs in the Lewis Structure were transferred to the more electrone gative atom. „ e.g. HCl, CO, CO 2, H 2O, Cl 2, P 4, CaO, … z Distinguish between the Oxidation Number and the Formal Charge „ e g HCl CO CO e.g. HCl, CO, CO 2, … 2

Rules for Assigning Oxidation (Ref: textbook, pg. 393): Be applied in order o Sum of all atoms' oxidation #s in a compound must equal to its net charge 2 Elementary substance (all the same atom):=0 3 In a compound: Group IA=+1; Group lIA=+2 B Al are +3: Fis-1 4 In a compound: H is +1; except in MH where H is-1 ⑤ In a compound:Ois-2 3

Rules for Assig g nin Oxidation # z ( pg ) pp Ref: textbook, pg. 393): Be applied in order: ① Sum of all atoms Sum of all atoms’ oxidation #s in a compound must oxidation #s in a compound must equal to its net charge. ② Elementary substance (all the same atom): = 0 ③ I d G IA 1 G IIA 2 In a compound: Group IA = +1; Group IIA = +2; B, Al are +3; F is -1. ④ In a compound: H is +1; except in MH where H is -1. ⑤ In a compound: O is -2. 3

Oxidation number Periodic Table Oxidation number: related to an atoms valence electrons Highest possible oxidation number of its valence electrons LoWest possible oxidation number #f of electrons needed to gain for an octet Example: C(Group IVA:+4, -4): N(Group VA: +5, -3) s( GroupⅥA:+6,-2);Cl( GroupⅦA:+7,-1) 4

Oxidation Number & Periodic Table z Oxidation number: related to an atom’s valence electrons „ Highest possible oxidation number: Highest possible oxidation number: = # of its valence electrons „ Lowest possible oxidation number: = # of electrons needed to gain for an octet z Example: „ C (Group IVA: +4, -4); N (Group VA: +5, -3); S (Group VIA: +6, -2); Cl (Group VIIA: +7, -1). 4

Practice Assigning the oxidation number of each elements in the following compounds (a) CaCao (b)H2S46 (c)V2 d)OF (e)Ncl3 f) sbH (g) socl2

Practice z Ass g g e o da o u be o eac igning the oxidation number of each elements in the following compounds: „ ( )C CO (a) CaCrO4 „ (b) H2S4O6 ( ) 2 „ (c) VO2+ „ (d) OF2 „ (e) NCl (e) NCl3 „ (f) SbH3 „ (g) SOCl2 5

Oxidation Reduction reaction Oxidation Reduction(Red-OX) a reaction in which the oxidation of any element changes, (indicating the transfer of electrons), and comprises 2 half-reactions Oxidation( Reducing agent Oxidation number t, lose electrons, being oxidized Reduction(Oxidizing agent): Oxidation number 1, gain electrons, being reduced 6

Oxidation & Reduction Reaction z O idation & Red ction ( O xidation & Red uction (Red - o x ): „ A reaction in which the oxidation # of any element changes, (indicating the transfer of electrons), and compri 2 h lf ises 2 h alf-reactions. „ Oxidation (Reducing agent): Oxidation number ↑, lose electrons, being oxidized; „ Reduction (Oxidizing agent): Oxidation number ↓, gain electrons, bein g reduced. 6

Write Red-ox Half reactions (a)Identifying each oxidation/reduction half reactions; (b )oxidizing/reducing agents; and(c)write each half reactions 1. H2(g)+ Cuo(s))Cu(s)+ H2o(g) Il.Fe2O3()+C(s)>Fe(+co 2(g) IL.HNO3(aq)+ Cu2O(s)>Cu(NO3)2(ag)+No( 9)+ H2o( Ⅳ.NaCl(aq)+AgNo3(aq)→Agc|(s)+NaNO3(aq)

Write Red-ox Half Reactions z (a) Identif ing each o idation/red ction (a) Identifying each o xidation/red uction half reactions; (b) oxidizing/reducing agents; and (c) write each half reactions: I. H 2 (g) + CuO (s) Æ Cu (s) + H 2O (g) II. Fe 2 O 3 (l) + C (s) Æ Fe (l) + CO 2 II. Fe (g) 2 O 3 (l) C (s) Æ Fe (l) CO 2 (g) III. HNO 3 (aq) + Cu 2O (s) Æ Cu(NO 3 ) 2 (aq) + NO (g) + H 2O (l) IV. NaCl (aq) + AgNO NaCl (aq) + AgNO (aq) Æ AgCl (s) + NaNO (aq) 3 (aq) Æ AgCl (s) + NaNO 3 (aq) 7

Balance the red-ox Reactions Half-Reaction Method (ref: Textbook, pg 401) o Divide the whole reaction into an oxidation half-reaction and a reduction half-reaction Balance elements other than h and o first ③UseH2 o to balance c Use H+ to balance h 6 Use e to balance charge 6 In basic solution add oh- to each side to neutralize h+ shown. Combine OH- and Ht to form H2o o Combine the two half-reactions together and simplify 8

Balance the Red-ox Reactions z Half-Reaction Method (ref: Textbook, pg.401 ) ① Divide the whole reaction into an oxidation half-reaction and a reduction half and a reduction half-reaction; reaction; ② Balance elements other than H and O first; ③ Use H 2O to balance O; ④ Use H + Use H to balance H; to balance H; ⑤ Use e- to balance charge; ⑥ In basic solution, add OH- to each side to neutralize H + shown. Combine OH- and H + to form H 2 O; ⑦ Combine the two half-reactions together and simplify. 8

Practice 1: Balance red-ox Reaction In an acidic solution: H2O2 +I>2+H2o ( Oxidation)2l-→ +2e (Reduction) H2O2 2 H++ 2e>2H2O Combined:(Ox)+(Red) H2O2+2H+2|→l2+2H2O

Practice 1: Balance Red-ox Reaction z In an acidic solution: H 2 O 2 + I - → I2 + H 2 O (Oxidation) 2 I - → I2 + 2 e - (Reduction) H 2 O 2 + 2 H + + 2 e - → 2 H 2O z Combined: (Ox) + (Red) : H 2 O 2 + 2 H + + 2 I - → I2 + 2 H 2 O 9

Practice 2 Balance red-ox Reaction In an acidic solution. CHSCHO Cr2 072-> C6H5COOH Cr3+ (Ox)CBHSCHO H2O-CBH5COOH 2 H+2e Red)Cr2O72-+ 14 H++ 6e>2 Cr3++ 7 H2O Combined: 3 X(OX)+(Red 3CBH5CHO +3 H2O+ cr2 072+ 14H -3CcH-COOH +6 H++ 2 Cr3++7H,O Simplify: 3 C6H5CHO 8 H*+ Cr2o ->3 C._COOH 2 Cr3++4H.O 10

Practice 2: Balance Red-ox Reaction z In an acidic solution: In an acidic solution: C 6 H 5CHO + Cr 2 O 7 2- → C 6 H 5COOH + Cr3+ (Ox) C 6 H 5CHO + H 2O → C 6 H 5COOH + 2 H + + 2 e- (Red) Cr 2 O 7 2- + 14 H + + 6 e- → 2 Cr3+ (Red) Cr 2 O 7 + 14 H + 6 e → 2 Cr + 7H 2 O Combined: 3 x (Ox) + (Red) : 3 C 6 H 5CHO + 3 H 2 O + Cr 2 O 7 2- + 14 H + → 3 C 6 H 5COOH + 6 H + + 2 Cr3+ + 7 H 2 O Simplify： 3 C 6 H 5CHO + 8 H + + Cr 2 O 7 2- 3 C H COOH 2 C 3 4 H O 10 → 3 C 6 H 5COOH + 2 C r 3 + + 4 H 2 O