1. Chapter 19.2

2. 1. Explain what must be conserved in redox
equations.
2. Balance redox equations by using the half-
reaction method.

3. • The half-reaction method for balancing redox
equations consists of seven steps:
1. Write the formula equation if it is not given in the
problem. Then write the ionic equation.
2. Assign oxidation numbers. Delete substances
containing only elements that do not change oxidation
state.
H2S + HNO3 H2SO4 + NO2 + H2OFormula Equation:
H2S + H+ + NO3
- 2H+ + SO4
-2 + NO2 + H2OIonic Equation:
H2S + H+ + NO3
- 2H+ + SO4
-2 + NO2 + H2O
+1 -2 +1 +5 -2 +1 +1-2 -2 -2+6 +4

4. 3. Write the half-reaction for oxidation.
• Balance the atoms.
• Balance the charge.
4. Write the half-reaction for reduction.
• Balance the atoms.
• Balance the charge.
H2S SO4
-2
-2 +6
H2S SO4
-2
-2 +6
+ 4 H2O + 10 H+
H2S SO4
-2
-2 +6
+ 4 H2O + 10 H+ + 8 e-
NO3
- NO2
+5 +4
NO3
- NO2
+5 +4
+ H2O+ 2 H+
NO3
- NO2
+5 +4
+ H2O+ 2 H+ + e-

5. 5. Conserve charge by adjusting the coefficients in front
of the electrons so that the number lost in oxidation
equals the number gained in reduction.
6. Combine the half-reactions, and cancel out anything
common to both sides of the equation.
H2S SO4
-2
-2 +6
+ 4 H2O + 10 H+ + 8 e-
NO3
- NO2
+5 +4
+ H2O+ 2 H+ + e-
(
(
)
)
1
8
H2S SO4
-2
-2 +6
+ 4 H2O + 10 H+ + 8 e-
8 NO3
- 8 NO2
+4
+ 8 H2O+ 16 H++ 8 e-
+5 6 4
8 NO3
- + H2S + 6 H+ 8 NO2 + 4 H2O + SO4
-2
+6+5 +4-2

6. 7. Combine ions to form the compounds shown in the
original formula equation. Check to ensure that all other
ions balance.
8 HNO3 + H2S 8 NO2 + 4 H2O + SO4
-2 + 2H+
8 NO3
- + H2S + 6 H+ 8 NO2 + 4 H2O + SO4
-2
+6+5 +4-2
8 HNO3 + H2S 8 NO2 + 4 H2O + H2SO4
-2

9. A deep purple solution of potassium permanganate is
titrated with a colorless solution of iron(II) sulfate and
sulfuric acid. The products are iron(III) sulfate,
manganese(II) sulfate, potassium sulfate, and water—all
of which are colorless. Write a balanced equation for this
reaction.
4 4 2 4
2 4 3 4 2 4 2
KMnO + FeSO + H SO
Fe (SO ) + MnSO + K SO + H O

+ 2+ 2 + 2
4 4 4
3+ 2 2+ 2 + 2
4 4 4 2
K + MnO +Fe + SO + 2H + SO
2Fe + 3SO + Mn + SO + 2K + SO + H O
– – –
– – –

10. +1 +7 2 +2 +6 2 +1 +6 2
+3 +6
+ 2+ 2 + 2–
4 4 4
3+ 2– 2+ 2– + 2
4 4
2 +2 +6 2 +1 +6 2 +
4 2
1 2
K + MnO +Fe + SO + 2H + SO
2Fe + 3SO + Mn + SO + 2K + SO + H O

– – –
– –
–
– –
– –
+7 2 +2 +3 +2
2+ 3+ 2+
4MnO + Fe Fe + Mn
–
–
+2 +
2 3+
3
+
Fe Fe
2
+2
+ 3+
+3
Fe Fe + e –
+7
+ 2
4 2
2
+
+
MnO + 8H + 5 Mn + 4H Oe – –
+7
4
+2
2+
MnO Mn–
+
+7 +2
2+
4 2MnO + 8H Mn + 4H O–

11. + 2+
4 21(MnO + 8H + 5 Mn + 4H O)e – –
2+ 3+
5(Fe Fe + )e –
lost in oxidation 1
gained in reduction 5
e
e

–
–
+ 2+
4 2MnO + 8H + 5 Mn + 4H Oe – –
2+ 3+
Fe Fe + e –
2+ + 2+ 3+
4 2MnO + 5Fe + 8H + 5 Mn + 5Fe + 4H O + 5e e– – –
5 5 5

12. 2+ + 3+ 2+
4 22(5Fe + MnO + 8H 5Fe + Mn + 4H O)–
2+ + 3+ 2+
4 210Fe + 2MnO + 16H 10Fe + 2Mn + 8H O
4 4 2 4
2 4 3 4 2 4 2
10FeSO + 2KMnO + 8H SO
5Fe (SO ) + 2MnSO + K SO + 8H O