Balancing Redox Equations

1. Balancing Redox Equations

2. 0 0 Analyze Mg + S  MgS +2 -2 • What is oxidized? • What is reduced? • Assign Oxidation Numbers. • Figure out change in oxidation numbers. Mg goes from 0 to +2: Oxidation S goes from 0 to -2: Reduction

3. 2 electrons Mg + S  MgS lost • Identify what species is oxidized & what species is reduced. • Figure out the 2 half-reactions. gained 2 electrons

4. Half-Reactions Mg + S  MgS 0 0 +2 -2 Mg is oxidized: Mg  Mg+2 + 2e- S is reduced: S + 2e- S-2

5. Adjust half-reactions so electrons lost = electrons gained. Mg  Mg+2 + 2e- S + 2e- S-2 • Add half-reactions. __________________________________ Mg + S + 2e-  Mg+2 +2e- + S-2 • Balance everything else by counting atoms.

6. 0 Zn + AgCl  ZnCl2 + Ag +1 -1 +2 -1 0 Zn + AgCl  ZnCl2 + Ag Lost 2 electrons 2 2 X 2 Gained 1 electron

7. Half-reactions Oxidation: Zn  Zn2+ + 2e- Reduction: Ag+ + 1e-  Ag Reduction: 2 Ag+ + 2e-  2 Ag X 2 Zn + 2 Ag+ + 2e- Zn2+ + 2 Ag + 2 e-

8. Those 7 pesky diatomics … • A little quirky…

9. Zn + HCl  H2 + ZnCl2 0 0 +1 -1 +2 -1 • Zn goes from 0 to +2: oxidation. • H goes from +1 to 0: reduction. • Cl goes from -1 to -1: No change.

10. 2 electrons lost Zn + HCl  H2 + ZnCl2 X 2 * Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 Gains 1 electron per H ______________________________________ Zn + 2H+1 +2e- Zn+2 +2e- + H2

11. Transfer the coefficients! Zn + 2H+1 H2 + Zn+2 • This is what you’ve got from adding the 2 half-reactions. Zn + HCl  H2 + ZnCl2 • This is the skeleton equation you started with. • Transfer the coefficients. Zn + 2HCl  H2 + ZnCl2

12. Balancing Redox Equations • Assign oxidation numbers to all atoms in equation. • See which elements have changes in oxidation number. • Identify atoms that are oxidized & atoms that are reduced. • Write the half-reactions. Diatomics have to be written as diatomics. • Make the number of electrons lost & gained equal in magnitude by multiplying half-reactions as needed. • Add the half-reactions. Transfer coefficients to skeleton equation. • Balanceremainder of equation by counting up atoms.

13. Cu + AgNO3 Cu(NO3)2 + Ag 0 +1 +5 -2 +2 +5 -2 0 • Cu goes from 0 to +2: oxidation. • Ag goes from +1 to 0: reduction. • N goes from +5 to +5: No change. • O goes from -2 to -2: No change.

14. Half-Reactions Cu  Cu+2 + 2e- Ag+1 + 1e-  Ag Cu  Cu+2 + 2e- 2Ag+1 + 2e-  2Ag Multiply by 2 ______________________ Cu + 2Ag+1 + 2e- 2Ag + Cu+2 + 2e-

15. Transfer Coefficients • Compare skeleton equation & sum of half-reactions: Cu + AgNO3 Ag + Cu(NO3)2 Vs. Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e- • Transfer the coefficients! Cu + 2AgNO3  2Ag + Cu(NO3)2

16. Cu + HNO3 Cu(NO3)2 + NO2 + H2O -2 +2 +1 -2 0 +1 +5 +5 -2 +4 -2 • Assign Oxidation Numbers. • Identify which species are oxidized & which reduced. Cu: 0 to +2 = oxidized H: +1 to +1 so no change N: all starts as +5. Some ends as +5, some as +4 = reduction O: -2 to -2 so no change

17. Change of +2 Cu + HNO3 Cu(NO3)2 + NO2 + H2O 2 2 • Find change in oxidation number. • Write half-reactions. Cu  Cu+2 + 2e- N+5 + 1e-  N+4 2( ) = -2 Change of -1

18. What’s oxidized? What’s reduced? • What is oxidized? • What is reduced? • What is the oxidizing agent? • What is the reducing agent? Cu Can’t just say N. It’s the N in the HNO3. Or the HNO3 or N+5. N+5 Cu

19. Multiply half-reactions as necessary. Cu  Cu+2 + 2e- 2N+5 + 2e-  2N+4 • Now the # of electrons lost = # gained. Cu + 2HNO3 Cu(NO3)2 + 2NO2 + H2O • Add half-reactions. Transfer coefficients. • Balance remaining atoms by inspection. Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O