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Balancing Redox Equations

Balancing Redox Equations. Electron Transfer Method (Change in Oxidation Number Method) works best for formula equations (no ions present) Steps: 1. Write formulas for reactants and products (skeletal equation) – usually given 2. Assign oxidation numbers for every atom

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Balancing Redox Equations

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  1. Balancing Redox Equations

  2. Electron Transfer Method (Change in Oxidation Number Method) works best for formula equations (no ions present) Steps: 1. Write formulas for reactants and products (skeletal equation) – usually given 2. Assign oxidation numbers for every atom 3. Write "electronic" equations of atoms that changed oxidation number

  3. Electron Transfer Method (Change in Oxidation Number Method) Steps: 4. Adjust coefficients to get equal number of electrons lost (ox.) and gained (red.) • Insert coefficients into skeletal equation. CAUTION: the coefficients in the electronic equations reflect ONLY those atoms oxidized and reduced – sometimes the OVERALL equation is not balanced yet

  4. H2SO3 + I2 + H2O  H2SO4 + HI +1 +4-2 0 +1 -2 +1 +6 -2 +1-1 +4 +6 oxidation rxn: S  S + 2 e-1 0 -1 reduction rxn: I2 + 2 e-1 2 I

  5. H2SO3 + I2 + H2O  H2SO4 + HI +1 +4-2 0 +1 -2 +1 +6 -2 +1-1 +4 +6 oxidation rxn: S  S + 2 e-1 0 -1 reduction rxn: I2 + 2 e-12 I put in coefficients from above H2SO3 + I2 + H2O  H2SO4 + 2 HI (balanced)

  6. x 3 KClO3 + FeSO4 + H2SO4 KCl + Fe2(SO4)3 + H2O +1+5-2 +2+6-2 +1+6-2 +1-1 +3 +6-2 +1 -2 +2 +3 ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 +2 +3 6 Fe+23 Fe2+3 + 6 e-1 +5 -1 red. rxn: Cl + 6 e-1 Cl-1

  7. KClO3 + FeSO4 + H2SO4 KCl + Fe2(SO4)3 + H2O +1+5-2 +2+6-2 +1+6-2 +1-1 +3 +6-2 +1 -2 +2 +3 ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 x 3 +2 +3 6 Fe+23 Fe2+3 + 6 e-1 +5 -1 red. rxn: Cl + 6 e-1 Cl-1 put in the coefficients from above KClO3 + 6 FeSO4 + H2SO4 KCl + 3 Fe2(SO4)3 + H2O

  8. KClO3 + FeSO4 + H2SO4 KCl + Fe2(SO4)3 + H2O +1+5-2 +2+6-2 +1+6-2 +1-1 +3 +6-2 +1 -2 +2 +3 ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 x 3 +2 +3 6Fe+2 3 Fe2+3 + 6 e-1 +5 -1 red. rxn: Cl + 6 e-1 Cl-1 put in the coefficients from above KClO3 + 6 FeSO4 + H2SO4 KCl + 3 Fe2(SO4)3 + H2O Finish (O not balanced)

  9. KClO3 + FeSO4 + H2SO4 KCl + Fe2(SO4)3 + H2O +1+5-2 +2+6-2 +1+6-2 +1-1 +3 +6-2 +1 -2 +2 +3 ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 x 3 +2 +3 6Fe+2 3 Fe2+3 + 6 e-1 +5 -1 red. rxn: Cl + 6 e-1 Cl-1 put in the coefficients from above KClO3 + 6 FeSO4 + H2SO4 KCl + 3 Fe2(SO4)3 + H2O Finish (O not balanced) KClO3 + 6 FeSO4 + 3 H2SO4 KCl + 3 Fe2(SO4)3 + 3 H2O

  10. Ag2S + HNO3 AgNO3 + NO + S + H2O +1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2

  11. Ag2S + HNO3 AgNO3 + NO + S + H2O +1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2 -2 0 ox. rxn: S-2 S + 2 e-1 +5 +2 red. rxn: N + 3 e-1 N

  12. Ag2S + HNO3 AgNO3 + NO + S + H2O +1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2 -2 0 ox. rxn: S-2 S + 2 e-1x 3 -2 0 3 S-23 S + 6 e-1 +5 +2 red. rxn: N + 3 e-1 N x 2 +5 +2 2 N + 6 e-12 N

  13. Ag2S + HNO3 AgNO3 + NO + S + H2O +1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2 -2 0 ox. rxn: S-2 S + 2 e-1 x 3 -2 0 3 S-23 S + 6 e-1 +5 +2 red. rxn: N + 3 e-1 N x 2 +5 +2 2 N + 6 e-12 N put in the coefficients from above 3 Ag2S + 2 HNO3 AgNO3 + 2 NO + 3 S + H2O Finish (Ag not balanced, fixing the Ag changes other coefficients)

  14. Ag2S + HNO3 AgNO3 + NO + S + H2O +1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2 -2 0 ox. rxn: S-2 S + 2 e-1 x 3 -2 0 3 S-2 3 S + 6 e-1 +5 +2 red. rxn: N + 3 e-1 N x 2 +5 +2 2 N + 6 e-1 2 N put in the coefficients from above 3 Ag2S + 2 HNO3 AgNO3 + 2 NO + 3 S + H2O Finish (Ag not balanced, fixing the Ag changes other coefficients) 3 Ag2S + 8 HNO36 AgNO3 + 2 NO + 3 S + 4 H2O

  15. X 3 Pt + HCl + KNO3 + KCl  K2PtCl6 + NO + H2O 0 +1-1 +1+5-2 +1-1 +1+4-1 +2-2 +1-2 0 +4 ox. rxn: Pt  Pt + 4 e-1 0 +4 3 Pt  3 Pt + 12 e-1 +5 +2 red. rxn: N + 3 e-1 N +5 +2 4 N + 12 e-1 4 N put in the coefficients 3 Pt + HCl + 4 KNO3 + KCl  3 K2PtCl6 + 4 NO + H2O Finish 3 Pt + 16 HCl + 4 KNO3 + 2 KCl  3 K2PtCl6 + 4 NO + 8 H2O X 4

  16. Half Reaction Method (see Chapter 20) works best for ionic reactions (ions present) reactions occur in acidic or basic solution Steps: 1. Break skeletal ionic equation into an oxidation equation and a reduction equation (called half reactions) 2. Balance both reactions for all atoms except O and H 3. Balance both reactions for O by adding H2O to the side with less O Balance both reactions for H by adding H+1 to the side with less H

  17. Half Reaction Method (continued) Steps: 4. Balance both reactions for charge by adding electrons to the side with the greater positive charge 5. Adjust coefficients in the balanced half reactions to get the number of electrons lost equal to the number of electrons gained 6. Add the two half reactions and cancel electrons and other species that appear on both sides of the equation

  18. Cr2O7-2(aq) + Cl-1(aq)  Cr+3(aq) + Cl2(aq) in acid solution Step 1 – write half reactions ox. rxn: Cl-1(aq)  Cl2(aq) (check ox #’s if necessary) red. rxn: Cr2O7-2(aq)  Cr+3(aq) Step 2 - balance except for O and H ox 2 Cl-1(aq)  Cl2(aq) red Cr2O7-2(aq)  2 Cr+3(aq)

  19. Step 3 - balance for O by adding H2O, balance H by adding H+1 ox 2 Cl-1(aq)  Cl2(aq) red Cr2O7-2(aq) + 14 H+1(aq)  2 Cr+3(aq) + 7 H2O(l) Step 4 - balance for charge by adding electrons ox 2 Cl-1(aq)  Cl2(aq) + 2 e-1 red Cr2O7-2(aq) + 14 H+1(aq) + 6 e-1 2 Cr+3(aq) + 7 H2O(l)

  20. Step 5 - make TOTAL electrons lost and gained equal ox 3 [ 2 Cl-1(aq)  Cl2(aq) + 2 e-1 ] 6 Cl-1(aq)  3 Cl2(aq) + 6 e-1 red Cr2O7-2(aq) + 14 H+1(aq) + 6 e-1 2 Cr+3(aq) + 7 H2O(l) Step 6 - add half reactions and cancel species that appear on both sides ox 6 Cl-1(aq)  3 Cl2(aq) + 6 e-1 red Cr2O7-2(aq) + 14 H+1(aq) + 6 e-1 2 Cr+3(aq) + 7 H2O(l) ------------------------------------------------------------------------------------- Cr2O7-2(aq) + 14 H+1(aq) + 6 Cl-1(aq) 3 Cl2(aq) + 2 Cr+3(aq) + 7 H2O(l)

  21. ClO-1(aq) + Cr(OH)4-1(aq)  CrO4-2(aq) + Cl-1(aq) basic solution Balance like an acidic reaction, then add OH-1 ions to neutralize the H+1 Step 1 – half reactions ox Cr(OH)4-1(aq)  CrO4-2(aq) red ClO-1(aq)  Cl-1(aq) Step 2 – they are already balanced for atoms other than O and H

  22. Step 3 – balance O (add H2O) and THEN H (add H+) ox Cr(OH)4-1(aq)  CrO4-2(aq) + 4 H+1(aq) red ClO-1(aq) + 2 H+1(aq)  Cl-1(aq) + H2O(l) Step 4 – balance charge ox Cr(OH)4-1(aq)  CrO4-2(aq) + 4 H+1(aq) + 3 e-1 red ClO-1(aq) + 2 H+1(aq) + 2 e-1 Cl-1(aq) + H2O(l) Step 5 – balance TOTAL electrons ox 2 x [Cr(OH)4-1(aq)  CrO4-2(aq) + 4 H+1(aq) + 3 e-1] red 3 x [ClO-1(aq) + 2 H+1(aq) + 2 e-1 Cl-1(aq) + H2O(l)] ox 2 Cr(OH)4-1(aq)  2 CrO4-2(aq) + 8 H+1(aq) + 6 e-1 red 3 ClO-1(aq) + 6 H+1(aq) + 6 e-1 3 Cl-1(aq) + 3 H2O(l)

  23. Step 6 – add and cancel 2 Cr(OH)4-1(aq)  2 CrO4-2(aq) + 8 H+1(aq) + 6 e-1 3 ClO-1(aq) + 6 H+1(aq) + 6 e-1 3 Cl-1(aq) + 3 H2O(l) ------------------------------------------------------------------------------------ 2Cr(OH)4-1(aq) + 3ClO-1(aq)  2CrO4-2(aq) + 2H+1(aq) + 3Cl-1(aq) + 3H2O(l) because this is in basic solution, we add 2 OH-1 ions to each side to convert the 2 H+1on the right side into H2O final equation balanced for atoms and charge is 2Cr(OH)4-1(aq) + 3ClO-1(aq) + 2OH-1(aq)2CrO4-2(aq) + 3Cl-1(aq) + 5H2O(l) the 2 H+1 (on the right) combine with the 2 OH-1 to make 2 H2O and then the 2 H2O combine with the 3 H2O already there to make 5 H2O total on the right Becomes 2 H2O

  24. 1. Fe+2 + MnO4-1 Fe+3 + Mn+2 (in acidic solution) 1&2) ox Fe+2 Fe+3 red MnO4-1 Mn+2 3) ox Fe+2 Fe+3 red MnO4-1 + 8 H+1 Mn+2 + 4 H2O 4) ox Fe+2 Fe+3 + e-1 red MnO4-1 + 8 H+1 + 5 e-1 Mn+2 + 4 H2O 5) ox 5 [ Fe+2 Fe+3 + e-1 ] 5 Fe+2 5 Fe+3 + 5 e-1 red MnO4-1 + 8 H+1 + 5 e-1 Mn+2 + 4 H2O 6) 5 Fe+2 + MnO4-1 + 8 H+1 5 Fe+3 + Mn+2 + 4 H2O

  25. 6. MnO4-1 + NO2-1 MnO2 (s) + NO3-1 (in basic solution) 1&2) NO2-1 NO3-1 MnO4-1 MnO2 3) NO2-1 + H2O  NO3-1 + 2 H+1 MnO4-1 + 4 H+1 MnO2 + 2 H2O 4) NO2-1 + H2O  NO3-1 + 2 H+1 + 2 e-1 MnO4-1 + 4 H+1 + 3 e-1 MnO2 + 2 H2O 5) ox 3 [ NO2-1 + H2O  NO3-1 + 2 H+1 + 2 e-1 ] 3 NO2-1 + 3 H2O  3 NO3-1 + 6 H+1 + 6 e-1 red 2 [ MnO4-1 + 4 H+1 + 3 e-1 MnO2 + 2 H2O ] 2 MnO4-1 + 8 H+1 + 6 e-1 2 MnO2 + 4 H2O 6) 3 NO2-1 + 2 MnO4-1 + 2 H+1 3 NO3-1 + 2 MnO2 + H2O basic solution – have to add 2 OH-1 3 NO2-1 + 2 MnO4-1 + 2 H2O  3 NO3-1 + 2 MnO2 + H2O + 2 OH-1 3 NO2-1 + 2 MnO4-1 + H2O  3 NO3-1 + 2 MnO2 + 2 OH-1 ox red

  26. 2. Cr2O7-2 + I-1 Cr+3 + I2 (s) (in acidic solution) 1) ox I-1 I2 red Cr2O7-2 Cr+3 2) ox 2 I-1 I2 red Cr2O7-2 2 Cr+3 3) ox 2 I-1 I2 red Cr2O7-2 + 14 H+1 2 Cr+3 + 7 H2O 4) ox 2 I-1 I2 + 2 e-1 redCr2O7-2 + 14 H+1 + 6 e-1 2 Cr+3 + 7 H2O 5) ox 3 [ 2 I-1 I2 + 2 e-1 ]  6 I-1 3 I2 + 6 e-1 red Cr2O7-2 + 14 H+1 + 6 e-1 2 Cr+3 + 7 H2O 6) Cr2O7-2 + 14 H+1 + 6 I-1  2 Cr+3 + 7 H2O + 3 I2

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