1. Balancing Redox Equations

2. analyze: Mg + S  MgS Mg: 0 to +2 = oxidation S: 0 to -2 = reduction-2 +2
analyze: Mg + S  MgS
1. assign oxidation numbers to all elements
2. figure out change in oxidation numbers
what is oxidized?
what is reduced?
Mg: 0 to +2 = oxidation
S: to -2 = reduction

3. Mg + S  MgS +2 -2 half-reactions: Mg is oxidized: 0 to +2
+2 -2
half-reactions:
Mg is oxidized: 0 to +2
Mg  Mg+2 + 2e-
S is reduced: 0 to -2
S + 2e-  S-2

4. electrons lost = electrons gained
add half-reactions:
Mg  Mg+2 + 2e-
S + 2e-  S-2
____+______________________________
Mg + S + 2e-  Mg+2 +2e- + S-2

5. Zn + 2HCl  H2 + ZnCl2 +1 -1 +2 -1 Zn goes from 0 to +2 = oxidation+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

6. Zn +2HCl  H2 + ZnCl2 Zn  Zn+2 + 2e- 2H+1 + 2e-  H2
Zn  Zn+2 + 2e-
2H+1 + 2e-  H2
______________________________________
Zn + 2H+1 +2e-  Zn+2 +2e- + H2

7. Balancing Redox Equations
assign oxidation # to all elements
determine elements that changed oxidation number
identify element oxidized & element reduced
write half-reactions (diatomics must stay as is)
# electrons lost/gained must be equal; (multiply half-reactions if necessary)
add half-reactions; transfer co-efficients to skeleton equation
balance rest of equation by counting atoms

8. Cu + AgNO3  Cu(NO3)2 + Ag +1 +5 -2 +2 +5 -2+1 +5 -2 +2 +5 -2
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

9. Half-Reactions +______________________ Cu  Cu+2 + 2e- Ag+1 + 1e-  Ag
multiply by 2
Cu  Cu+2 + 2e-
2Ag+1 + 2e-  2Ag
+______________________
Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e-

10. Transfer Co-efficients
compare skeleton equation & sum of ½ rxns:
Cu + AgNO3  Ag + Cu(NO3)2
vs.
Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e-
transfer co-efficients!
Cu + 2AgNO3  2Ag + Cu(NO3)2

11. Cu + HNO3  Cu(NO3)2 + NO2 + H2O Cu: 0 to +2 = oxidized+1 +5 -2 +2 +5 -2 +1 -2 +4 -2
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
Cu: 0 to +2 = oxidized
H: to +1; no change
O: to -2; no change
* N: starts as +5
ends as +5 [no change – Cu(NO3)2]
ends as +4 = reduction [NO2]
can’t just say N!
say:N in HNO3 or N+5
half-reactions:
Cu  Cu+2 + 2e-
N+5 + 1e-  N+4

12. multiply half-reactions as necessary: # of electrons lost = # gained
Cu  Cu+2 + 2e-
N+5 + 1e-  N+4
multiply half-reactions as necessary:
# of electrons lost = # gained
Cu  Cu+2 + 2e-
2 (N+5 + e-  N+4)
balance remaining atoms by inspection
balanced half-reaction co-efficients might not completely balance the equation
Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O