Chapter 19.2
1. Explain what must be conserved in redox
equations.
2. Balance redox equations by using the half-
reaction method.
• The half-reaction method for balancing redox
equations consists of seven steps:
1. Write the formula equation if it is n...
3. Write the half-reaction for oxidation.
• Balance the atoms.
• Balance the charge.
4. Write the half-reaction for reduct...
5. Conserve charge by adjusting the coefficients in front
of the electrons so that the number lost in oxidation
equals the...
7. Combine ions to form the compounds shown in the
original formula equation. Check to ensure that all other
ions balance....
A deep purple solution of potassium permanganate is
titrated with a colorless solution of iron(II) sulfate and
sulfuric ac...
+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...
+ 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 2...
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 ...
Chapter 19.2 : Balancing Redox Equations
Chapter 19.2 : Balancing Redox Equations
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Chapter 19.2 : Balancing Redox Equations

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Chapter 19.2 : Balancing Redox Equations

  1. 1. Chapter 19.2
  2. 2. 1. Explain what must be conserved in redox equations. 2. Balance redox equations by using the half- reaction method.
  3. 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. 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. 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. 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
  7. 7. 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 – – – – – –
  8. 8. +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–
  9. 9. + 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
  10. 10. 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
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