1. 4-6 Balancing Redox Reactions (Section 18.4)

2. Steps for balancing: 1) Assign oxidation numbers to the elements.
Steps for balancing:
1) Assign oxidation numbers to the elements.
2) Write the 2 half-reactions, one for oxidation and the other for reduction.
3) Balance all elements except H and O.
4) Balance electrons and then add half-reactions.
5) In ACID: add H2O to balance O’s, then add H+ to balance H’s.
In BASE: do the same as acid, then add OH- to both sides to eliminate H+.
6) Check final equation to ensure that # of atoms and charges balance.

3. Cr2O7-2(aq) + I-1(aq) → Cr+3(aq) + I2(aq)+6 -2 -1 +3 e-
Reduction ½ reaction: Gain e- Reduced
Cr2O ?e-  Cr+3
Oxidation ½ reaction: Loss e- Oxidized
I-1  I2 + ?e- +6 +3 -1

4. Balance all but H and O Cr2O7 + e-  Cr+3 I-1  I2 + e- 6 2 2 2 +6 +32 2

5. Balance e- s and combine ½ rxns
Cr2O e-  2 Cr+3
(2I-1  I e-) 3
= 6I-1  3I2 + 6e-
Cr2O7 + 6e- + 6I-1  2 Cr+3 + 3I2 + 6e-

6. Cr2O7 + 6e- + 6I-1  2 Cr+3 + 3I2 + 6e- If in Acidic solution:
Cr2O I  2 Cr+3 + 3I2
+ 14 H+
+ 7 H2O

7. Cr2O7 + 6e- + 6I-1  2 Cr+3 + 3I2 + 6e- If in basic solution:
Cr2O I  2 Cr+3 + 3I2
Cr2O I H2O  2 Cr+3 + 3I2 + 7 H2O + 14 OH-
Cr2O I H2O  2 Cr+3 + 3I OH-
+14 H+
+ 7 H2O
+ 14 OH-
+ 14 OH- 7

8. MnO4-1(aq) + Fe+2(aq) → Fe+3(aq) + Mn+2(aq) +7 -2 +2 +3 +2
MnO4-1(aq) + Fe+2(aq) → Fe+3(aq) + Mn+2(aq)   e-
Reduction: (ger)
MnO e-  Mn
Oxidation: (Leo)
Fe+2  Fe e- +2 +7 5 1

9. MnO4-1 + 5 e- + 5 Fe+2  Mn+2 + 5 Fe+3 + 5 e-
MnO e-  Mn+2
(Fe+2  Fe e- )
= Fe+2  Fe e-
MnO e- + 5 Fe+2  Mn Fe e- 5

10. MnO4-1 + 5 Fe+2  Mn+2 + 5 Fe+3 In acidic solution:
+ 8 H+
+ 4 H2O

11. MnO4-1 + 5Fe+2 + 8 H+  Mn+2 + 5Fe+3 + 4 H2O
In Basic solution:
MnO Fe H+  Mn+2 + 5Fe H2O
MnO Fe H2O  Mn Fe H2O + 8 OH-
MnO Fe+2 + 4H2O  Mn+2 + 5Fe+3 + 8OH-
+ 8 OH-
+ 8 OH- 4

12. Great Job!!!!!!