1. To balance a redox reaction, assign oxidation numbers and identify which elements are oxidized and reduced. 2. Write separate half-reactions for the oxidation and reduction processes. 3. Adjust the half-reactions so the number of electrons lost equals the number gained, then add the half-reactions together. 4. Transfer coefficients to balance the full reaction equation.

1. Balancing Redox Equations

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

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

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

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

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
• Zn goes from 0 to +2: oxidation.
• H goes from +1 to 0: reduction.
• Cl goes from -1 to -1: No change.
0 +1 -1 0 +2 -1

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

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
1. Assign oxidation numbers to all atoms in equation.
2. See which elements have changes in oxidation
number.
3. Identify atoms that are oxidized & atoms that are
reduced.
4. Write the half-reactions. Diatomics have to be
written as diatomics.
5. Make the number of electrons lost & gained equal in
magnitude by multiplying half-reactions as needed.
6. Add the half-reactions. Transfer coefficients to
skeleton equation.
7. Balance remainder of equation by counting up
atoms.

13. Cu + AgNO3  Cu(NO3)2 + Ag
• 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.
0 +1 +5 -2 +2 +5 -2 0

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
1. Assign Oxidation Numbers.
2. Identify which species are oxidized &
which reduced.
0 +1+5 -2 +2 +5 -2 +4 -2 +1 -2
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. Cu + HNO3  Cu(NO3)2 + NO2 + H2O
3. Find change in oxidation number.
4. Write half-reactions.
Cu  Cu+2 + 2e-
N+5 + 1e-  N+4
Change of +2
Change of -1
2( ) = -2
2 2

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