1. Redox Basics #68. Assign OX#
a. UO22+ Net charge is 2+ and OX#(O) = -2. Therefore 2+ = OX#(U) +2(-2) = 2+. Or OX#(U) = +6
c. NaBiO3 Net charge = 0. OX#(O) = -2, OX#(Na) = +1. Therefore 0 = +1 + OX#(Bi) + 3(-2). Or OX#(Bi) = +5
d. As4 Net charge = 0, As4 is an element. Therefore OX#(As) = 0
g. Na2S2O3 OX#(Na) & OX#(O) known. OX#(S) =

2. Redox Basics #72.Redox?, oxidized?, reduced? OA?, RA?
CH4(g) + H2O(g)  CO(g) + 3H2(g)
-4, , ,
OX# of carbon increases from -4 to +2; C loses e-s and is oxidized; methane = RA
OX# of hydrogen decreases from +1 to 0; H gains e-s and is reduced; water = OA

3. Redox Basics #72.Redox?, oxidized?, reduced? OA?, RA?
c. Zn(s) +2HCl(aq)  ZnCl2(aq) + H2(g)
, ,
Zn is oxidized; Zn loses e-s; Zn = RA
H is reduced; H gains e-s; HCl = OA
2H+(aq)+2CrO42-(aq)Cr2O72(aq)+H2O(l)
, , , -2

4. Balance Redox Rxn in Acid
#74a. Cu(s) +NO3-(aq)  Cu2+(aq)+ NO(g)
, , -2
Cu is oxidized; N is reduced
Cu(s)  Cu2+(aq) oxidation ½ rxn
No O or H to balance in oxid ½ rxn
NO3-(aq)  NO(g) reduction ½ rxn
NO3-(aq)  NO(g) +2H2O(l) Balance O
4H+(aq) + NO3-(aq)  NO(g) +2H2O(l)
Balance H

5. Balance Redox Rxn in Acid(2)
#74a. Cu(s) +NO3-(aq)  Cu2+(aq)+ NO(g)
Cu(s)  Cu2+(aq) + 2e- Balance e
3e- + 4H+(aq) + NO3-(aq)  NO(g) +2H2O(l) Balance e
Note that 2 e are lost and 3 e are gained.
Multiple oxidation ½ rxn by 3 and reduction ½ rxn by 2 to balance e-s

6. Balance Redox Rxn in Acid(3)
#74a. Cu(s) +NO3-(aq)  Cu2+(aq)+ NO(g)
3Cu(s)  3Cu2+(aq) + 6e-
6e- + 8H+ + 2NO3-  2NO(g) + 4H2O(l)
Add two half-rxns
3Cu + 6e- + 8H+ + 2NO3-  3Cu e- + 2NO(g) +4H2O(l) Cancel as needed
3Cu + 8H+ + 2NO3-  3Cu NO(g) +4H2O(l)

7. Balance Redox Rxn in Acid(4)
Note that all electrons cancel
3Cu + 8H+ + 2NO3-  3Cu2+ + 2NO(g) + 4H2O(l)
Check R P 3 Cu 8 H 2 N 6 O +6
Charge

8. Balance Redox Rxn in Base
MnO NO2-  MnO2 + NO3-
+7, -2 +3, -2 +4, -2 +5, -2
N is oxidized; Mn is reduced
Work on oxidation ½ rxn
NO2-  NO3- balance O
NO H2O  NO3- balance H
NO2- + H2O  NO3- + 2H+ balance e
NO2- + H2O  NO3- + 2H+ + 2e-

9. Balance Redox Rxn in Base (2)
Work on reduction ½ reaction
MnO4-  MnO2 Balance O
MnO4-  MnO2 + 2H2O Balance H
MnO H+  MnO2 + 2H2O Balance e-
MnO H e-  MnO2 + 2H2O
Notice that in oxid ½ rxn, 2 e- are lost, but in red ½ rxn, 3 e- are gained.
#e- lost must equal #e- gained

10. Balance Redox Rxn in Base (3)
Multiply oxid ½ rxn by 3
3NO H2O  3NO3- + 6H+ + 6e-
Multiply red ½ rxn by 2
2MnO H e-  2MnO2 + 4H2O
Add ½ rxns together
3NO H2O + 2MnO H e-  3NO3- + 6H+ + 6e- + 2MnO2 + 4H2O

11. Balance Redox Rxn in Base (3)
3NO H2O + 2MnO H e-  3NO3- + 6H+ + 6e- + 2MnO2 + 4H2O
Cancel
3NO2- + 2MnO H+  3NO MnO2 + H2O Now add steps to balance in base.
Add 2 OH- to each side to make soln basic. Note that 2H+ + 2OH-  2H2O

12. Balance Redox Rxn in Base (4)
3NO2- + 2MnO H+ + 2OH-  3NO MnO2 + H2O + 2OH-
3NO2- + 2MnO4- + 2H2O  3NO MnO2 + H2O + 2OH-
Cancel waters
3NO2- + 2MnO4- + H2O  3NO MnO2 + 2OH-

13. Balance Redox Rxn in Base (5)P 3 N 15 O 2 Mn H -5 Ch
3NO2- + 2MnO4- + H2O  3NO MnO2 + 2OH-
A redox rxn in base must have OH- left over NOT H+ (which means acid)
Check