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 3.Write "electronic" equations of atoms that changed oxidation number

Electron Transfer Method (Change in Oxidation Number Method) Steps: 4.Adjust coefficients to get equal number of electrons lost (ox.) and gained (red.) 5.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

H 2 SO 3 + I 2 + H 2 O  H 2 SO 4 + HI oxidation rxn: S  S + 2 e reduction rxn: I e -1  2 I

H 2 SO 3 + I 2 + H 2 O  H 2 SO 4 + HI oxidation rxn: S  S + 2 e reduction rxn: I e -1  2 I put in coefficients from above H 2 SO 3 + I 2 + H 2 O  H 2 SO HI (balanced)

KClO 3 + FeSO 4 + H 2 SO 4  KCl + Fe 2 (SO 4 ) 3 + H 2 O ox. rxn: 2 Fe +2  Fe e Fe +2  3 Fe e red. rxn: Cl + 6 e -1  Cl -1 x 3

KClO 3 + FeSO 4 + H 2 SO 4  KCl + Fe 2 (SO 4 ) 3 + H 2 O ox. rxn: 2 Fe +2  Fe e -1 x Fe +2  3 Fe e red. rxn: Cl + 6 e -1  Cl -1 put in the coefficients from above KClO FeSO 4 + H 2 SO 4  KCl + 3 Fe 2 (SO 4 ) 3 + H 2 O

KClO 3 + FeSO 4 + H 2 SO 4  KCl + Fe 2 (SO 4 ) 3 + H 2 O ox. rxn: 2 Fe +2  Fe e -1 x Fe +2  3 Fe e red. rxn: Cl + 6 e -1  Cl -1 put in the coefficients from above KClO FeSO 4 + H 2 SO 4  KCl + 3 Fe 2 (SO 4 ) 3 + H 2 O Finish (O not balanced)

KClO 3 + FeSO 4 + H 2 SO 4  KCl + Fe 2 (SO 4 ) 3 + H 2 O ox. rxn: 2 Fe +2  Fe e -1 x Fe +2  3 Fe e red. rxn: Cl + 6 e -1  Cl -1 put in the coefficients from above KClO FeSO 4 + H 2 SO 4  KCl + 3 Fe 2 (SO 4 ) 3 + H 2 O Finish (O not balanced) KClO FeSO H 2 SO 4  KCl + 3 Fe 2 (SO 4 ) H 2 O

Ag 2 S + HNO 3  AgNO 3 + NO + S + H 2 O

Ag 2 S + HNO 3  AgNO 3 + NO + S + H 2 O ox. rxn:S -2  S + 2 e red. rxn: N + 3 e -1  N

Ag 2 S + HNO 3  AgNO 3 + NO + S + H 2 O ox. rxn:S -2  S + 2 e -1 x S -2  3 S + 6 e red. rxn: N + 3 e -1  N x N + 6 e -1  2 N

Ag 2 S + HNO 3  AgNO 3 + NO + S + H 2 O ox. rxn:S -2  S + 2 e -1 x S -2  3 S + 6 e red. rxn: N + 3 e -1  N x N + 6 e -1  2 N put in the coefficients from above 3 Ag 2 S + 2 HNO 3  AgNO NO + 3 S + H 2 O Finish (Ag not balanced, fixing the Ag changes other coefficients)

Ag 2 S + HNO 3  AgNO 3 + NO + S + H 2 O ox. rxn:S -2  S + 2 e -1 x S -2  3 S + 6 e red. rxn: N + 3 e -1  N x N + 6 e -1  2 N put in the coefficients from above 3 Ag 2 S + 2 HNO 3  AgNO NO + 3 S + H 2 O Finish (Ag not balanced, fixing the Ag changes other coefficients) 3 Ag 2 S + 8 HNO 3  6 AgNO NO + 3 S + 4 H 2 O

Pt + HCl + KNO 3 + KCl  K 2 PtCl 6 + NO + H 2 O ox. rxn:Pt  Pt + 4 e Pt  3 Pt + 12 e red. rxn: N + 3 e -1  N N + 12 e -1  4 N put in the coefficients 3 Pt + HCl + 4 KNO 3 + KCl  3 K 2 PtCl NO + H 2 O Finish 3 Pt + 16 HCl + 4 KNO KCl  3 K 2 PtCl NO + 8 H 2 O X 3 X 4

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 H 2 O to the side with less O Balance both reactions for H by adding H +1 to the side with less H

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

Cr 2 O 7 -2 (aq) + Cl -1 (aq)  Cr +3 (aq) + Cl 2 (aq) in acid solution Step 1 – write half reactions ox. rxn:Cl -1 (aq)  Cl 2 (aq) (check ox #’s if necessary) red. rxn:Cr 2 O 7 -2 (aq)  Cr +3 (aq) Step 2 - balance except for O and H ox2 Cl -1 (aq)  Cl 2 (aq) redCr 2 O 7 -2 (aq)  2 Cr +3 (aq)

Step 3 - balance for O by adding H 2 O, balance H by adding H +1 ox2 Cl -1 (aq)  Cl 2 (aq) red Cr 2 O 7 -2 (aq) + 14 H +1 (aq)  2 Cr +3 (aq) + 7 H 2 O(l) Step 4 - balance for charge by adding electrons ox2 Cl -1 (aq)  Cl 2 (aq) + 2 e -1 red Cr 2 O 7 -2 (aq) + 14 H +1 (aq) + 6 e -1  2 Cr +3 (aq) + 7 H 2 O(l)

Step 5 - make TOTAL electrons lost and gained equal ox3 [ 2 Cl -1 (aq)  Cl 2 (aq) + 2 e -1 ] 6 Cl -1 (aq)  3 Cl 2 (aq) + 6 e -1 redCr 2 O 7 -2 (aq) + 14 H +1 (aq) + 6 e -1  2 Cr +3 (aq) + 7 H 2 O(l) Step 6 - add half reactions and cancel species that appear on both sides ox 6 Cl -1 (aq)  3 Cl 2 (aq) + 6 e -1 redCr 2 O 7 -2 (aq) + 14 H +1 (aq) + 6 e -1  2 Cr +3 (aq) + 7 H 2 O(l) Cr 2 O 7 -2 (aq) + 14 H +1 (aq) + 6 Cl -1 (aq)  3 Cl 2 (aq) + 2 Cr +3 (aq) + 7 H 2 O(l)

ClO -1 (aq) + Cr(OH) 4 -1 (aq)  CrO 4 -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 oxCr(OH) 4 -1 (aq)  CrO 4 -2 (aq) redClO -1 (aq)  Cl -1 (aq) Step 2 – they are already balanced for atoms other than O and H

Step 3 – balance O (add H 2 O) and THEN H (add H + ) oxCr(OH) 4 -1 (aq)  CrO 4 -2 (aq) + 4 H +1 (aq) redClO -1 (aq) + 2 H +1 (aq)  Cl -1 (aq) + H 2 O(l) Step 4 – balance charge oxCr(OH) 4 -1 (aq)  CrO 4 -2 (aq) + 4 H +1 (aq) + 3 e -1 redClO -1 (aq) + 2 H +1 (aq) + 2 e -1  Cl -1 (aq) + H 2 O(l) Step 5 – balance TOTAL electrons ox2 x [Cr(OH) 4 -1 (aq)  CrO 4 -2 (aq) + 4 H +1 (aq) + 3 e -1 ] red3 x [ClO -1 (aq) + 2 H +1 (aq) + 2 e -1  Cl -1 (aq) + H 2 O(l)] ox2 Cr(OH) 4 -1 (aq)  2 CrO 4 -2 (aq) + 8 H +1 (aq) + 6 e -1 red3 ClO -1 (aq) + 6 H +1 (aq) + 6 e -1  3 Cl -1 (aq) + 3 H 2 O(l)

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

1.Fe +2 + MnO 4 -1  Fe +3 + Mn +2 (in acidic solution) 1&2)oxFe +2  Fe +3 redMnO 4 -1  Mn +2 3)oxFe +2  Fe +3 redMnO H +1  Mn H 2 O 4)oxFe +2  Fe +3 + e -1 redMnO H e -1  Mn H 2 O 5)ox5 [ Fe +2  Fe +3 + e -1 ]  5 Fe +2  5 Fe e -1 redMnO H e -1  Mn H 2 O 6)5 Fe +2 + MnO H +1  5 Fe +3 + Mn H 2 O

6. MnO NO 2 -1  MnO 2 (s) + NO 3 -1 (in basic solution) 1&2)NO 2 -1  NO 3 -1 MnO 4 -1  MnO 2 3) NO H 2 O  NO H +1 MnO H +1  MnO H 2 O 4) NO H 2 O  NO H e -1 MnO H e -1  MnO H 2 O 5)ox3 [ NO H 2 O  NO H e -1 ] 3 NO H 2 O  3 NO H e -1 red2 [ MnO H e -1  MnO H 2 O ] 2 MnO H e -1  2 MnO H 2 O 6)3 NO MnO H +1  3 NO MnO 2 + H 2 O basic solution – have to add 2 OH -1 3 NO MnO H 2 O  3 NO MnO 2 + H 2 O + 2 OH -1 3 NO MnO H 2 O  3 NO MnO OH -1 ox red

2.Cr 2 O I -1  Cr +3 + I 2 (s)(in acidic solution) 1)oxI -1  I 2 redCr 2 O 7 -2  Cr +3 2)ox2 I -1  I 2 redCr 2 O 7 -2  2 Cr +3 3)ox2 I -1  I 2 redCr 2 O H +1  2 Cr H 2 O 4)ox2 I -1  I e -1 redCr 2 O H e -1  2 Cr H 2 O 5)ox3 [ 2 I -1  I e -1 ]  6 I -1  3 I e -1 redCr 2 O H e -1  2 Cr H 2 O 6)Cr 2 O H I -1  2 Cr H 2 O + 3 I 2