Oxidation-Reduction Chapter 20

Oxidation states & balancing redox equations [20.1-20.2] Cu+2 + Al → Cu + Al+3

Redox reaction: transfer of electrons Reduction: process of gaining electrons; becoming more negative LEO the lion Oxidation: process of losing electrons; becoming more positive goes GER

Oxidation and Reduction Reactant reduced is the oxidizing agent. H+ oxidizes Zn by taking electrons from it. (e- acceptor) Reactant oxidized is the reducing agent. Zn reduces H+ by giving it electrons. (e- donor)

Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

Oxidation numbers = charge of an atom Ca+2 F-1 Mn+7 -2 -2 x 4 = -8 K2SO4 CO3-2 ___ + 3(-2) = -2 Fe3O4 3(__) + 4(-2) = 0 Rules: Elements = 0 Hydrogen +1; except in metal hydrides (NaH), then it is -1. Placement is the key! Oxygen = -2; except in peroxides (H2O2; O = -1) In a compound, oxidation numbers must balance out to net charge of zero. 2(+1) + ___ + (-8) = 0

BALANCING SIMPLE REDOX REACTIONS Cu+2 + Al → Cu + Al+3 1. Cu+2 → Cu Al → Al+3 2. no changes 3. 2e- + Cu+2 → Cu Al → Al+3 + 3e- 4. (2e- + Cu+2 → Cu)3 (Al → Al+3 + 3e-)2 5. 6e- + 3Cu+2 → 3Cu 2Al → 2Al+3 + 6e- 2Al + 3Cu+2 → 3Cu + 2Al+3 2(0) + 3(+2) = 3(0) + 2(+3) Half reaction method: Write ½ reactions. Balance atoms, if necessary (except H & O) Balance charges by adding electrons. Multiply (if needed) to cancel electrons. Check work: charges & atoms BALANCING SIMPLE REDOX REACTIONS

MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq) Half-Reaction Method Consider the reaction between MnO4− and C2O42− : MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)

MnO4− + C2O42-  Mn2+ + CO2 Half-Reaction Method First, we assign oxidation numbers. MnO4− + C2O42-  Mn2+ + CO2 +7 +3 +4 +2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

BALANCING REDOX REACTIONS IN ACIDIC SOLUTIONS MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq) 1. C2O42−  CO2 MnO4−  Mn2+ 2. C2O42−  2 CO2 3. MnO4−  Mn2+ + 4 H2O 4. 8 H+ + MnO4−  Mn2+ + 4 H2O 5. C2O42−  2 CO2 + 2 e− 5 e− + 8 H+ + MnO4−  Mn2+ +4 H2O 6. 5 C2O42−  10 CO2 + 10 e− 10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2 Write ½ reactions Balance all atoms except H & O, if needed. Balance oxygen atoms by adding H2O to one side since aqueous. Balance hydrogen atoms by adding H+ Balance electric charge by adding electrons Make electrons equal, to cancel Check atoms & charge. BALANCING REDOX REACTIONS IN ACIDIC SOLUTIONS

Balancing redox reactions in basic solutions Zn + VO3-1  VO+2 + Zn+2 Zn  Zn+2 VO3-1  VO+2 3. VO3-1  VO+2 + 2H2O 4 H+ + VO3-1  VO+2 + 2H2O 2 (e- + 4 H+ + VO3-1  VO+2 + 2H2O) Zn  Zn+2 + 2 e- 6. 2e- + 8 H+ + 2VO3-1 2VO+2 +4H2O Zn + 2 VO3-1 + 8H+  2VO+2 + 4H2O + Zn+2 8. + 8 OH- + 8 OH- 8 H2O Zn + 2 VO3-1 + 8H2O  2VO+2 + 4H2O + Zn+2 + 8 OH- Zn + 2 VO3-1 + 4H2O  2VO+2 + Zn+2 + 8 OH- Write ½ reactions Balance all atoms except H & O, if needed. Balance oxygen atoms by adding H2O to one side since aqueous. Balance hydrogen atoms by adding H+ Balance electric charge by adding electrons Make electrons equal, to cancel Check atoms & charge. Now add OH- to both sides to eliminate H+ Balancing redox reactions in basic solutions