Transcript Balancing Redox Equations

Balancing Redox Equations
0
analyze:
0
+2
-2
Mg + S  MgS
1. assign oxidation numbers to all elements
2. figure out change in oxidation numbers
–
what is oxidized?
Mg: 0 to +2 = oxidation
–
what is reduced?
S:
0 to -2 = reduction
0
0
+2 -2
Mg + S  MgS
half-reactions:
Mg is oxidized: 0 to +2
Mg  Mg+2 + 2eS is reduced: 0 to -2
S + 2e-  S-2
electrons lost = electrons gained
add half-reactions:
Mg  Mg+2 + 2e-  S-2
S
+
2e
____+______________________________
Mg + S + 2e-  Mg+2 +2e- + S-2
0
+1 -1
0
+2 -1
Zn + 2HCl  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
Zn +2HCl  H2 + ZnCl2
Zn  Zn+2 + 2e+1 + 2e-  H
2H
2
______________________________________
Zn +
+1
2H
+2e

+2
Zn
+2e
+ H2
Balancing Redox Equations
1. assign oxidation # to all elements
2. determine elements that changed oxidation number
3. identify element oxidized & element reduced
4. write half-reactions (diatomics must stay as is)
5. # electrons lost/gained must be equal;
(multiply half-reactions if necessary)
6. add half-reactions;
transfer co-efficients to skeleton equation
7. balance rest of equation by counting atoms
0
+1 +5 -2
+2 +5 -2
0
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
Half-Reactions
multiply by 2
Cu  Cu+2 + 2eAg+1 + 1e-  Ag
Cu  Cu+2 + 2e+______________________
2Ag+1 + 2e-  2Ag
Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e-
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
0
+1 +5 -2
+2 +5 -2
+4 -2
+1 -2
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
Cu: 0 to +2 = oxidized
can’t just say N!
H: +1 to +1; no change
+5
say:N
in
HNO
or
N
3
O: -2 to -2; no change
* N: starts as +5
• ends as +5 [no change – Cu(NO3)2]
• ends as +4 = reduction [NO2]
half-reactions:
Cu  Cu+2 + 2eN+5 + 1e-  N+4
half-reactions
Cu  Cu+2 + 2eN+5 + 1e-  N+4
multiply half-reactions as necessary:
# of electrons lost = # gained
Cu  Cu+2 + 2e2 (N+5 + e-  N+4)
balance remaining atoms by inspection
Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O
balanced half-reaction co-efficients might not completely balance the equation