Balancing Redox Reactions

Half Reaction Method 1.Write the formula equation if it is not given. Then write the ionic equation. Formula eq: H 2 S + HNO 3  H 2 SO 4 + NO 2 + H 2 O Ionic eq: H 2 S + H + + NO 3 -  2H + + SO NO 2 + H 2 O 2.Assign oxidation numbers. Delete substances containing only elements that do not change oxidation state. H 2 S + NO 3 -  SO NO 2

3. Write the half reaction of oxidation H 2 S  SO Balance the atoms. Add H2O to balance for extra oxygens, then add H+ to balance for hydrogens H 2 S + 4 H 2 O  SO H + -In basic solutions add OH- to cancel out the positive charge from the H+

5. Balance the charge by adding e - -Look at the change in oxidation state H 2 S + 4 H 2 O  SO H + + 8e - 6. Repeat steps 1-5 for the half-reaction for reduction NO 3 -  NO 2 NO H + + e -  NO 2 + H 2 O

7. Conserve the charge by adding coefficients in front of the half-reactions so that they cancel out 1(H 2 S + 4 H 2 O  SO H + + 8e - ) 8(NO H + + e -  NO 2 + H 2 O). 8. Combine the half-reactions, and cancel out anything common to both sides 8NO H 2 S + 6H +  8NO 2 + 4H 2 O + SO 4 2-

9. Combine ions to form the compounds shown in the original formula equation. 8HNO H 2 S  8NO 2 + 4H 2 O + SO H + 10.Check to ensure that all other ions balance out 8HNO H 2 S  8NO 2 + 4H 2 O + H 2 SO 4

Balancing Redox Full Reactions in Acid Solution MnO C 2 O 4 2- →MnO 2 + CO 3 2- MnO 4 - → MnO 2 C 2 O 4 2- → CO MnO C 2 O H 2 O + 8H + → 2MnO 2 + 6CO H + + 4H 2 O 2MnO C 2 O H 2 O → 2MnO 2 + 6CO H + 1.Separate into half reactions2.Balance each half reaction 3.Add the two half reactions4.Simplify 5.Check + 4H + + 3e - + 2H 2 O2(2( 3(3( + 2e - + 4H + ) 2 ) 24

Balancing Redox Full Reactions in Basic Solution 2MnO C 2 O H 2 O → 2MnO 2 + 6CO H + 1.Balance in acid solution2.Add OH - to neutralize the H + 4OH - 4H 2 O2

Balancing Redox Full Reactions in Basic Solution 2MnO C 2 O → 2MnO 2 + 6CO OH - 2H 2 O

Balancing Redox Full Reactions in Acid Solution Fe+O 2 → H 2 O+Fe(OH) 3 Fe +→+ Fe(OH) 3 O 2 → H 2 0 4Fe + 12H 2 O + 3O 2 +12H + → 12H + + 4Fe(OH) 3 + 6H 2 O 4Fe + 6H 2 O + 3O 2 → 4Fe(OH) 3 3H 2 O3H + + 3e - ) + H 2 O)+4H + 4e - + 3( 4(4( 6

Balance the redox reaction As→H 2 AsO 4 - +AsH 3 (alkaline) You must separate into two half reactions! 3As + 12H 2 O + 15H + + 5As → 3H 2 AsO H + + 5AsH 3 8As + 12H 2 O → 3H 2 AsO H + + 5AsH 3 3OH - 3OH - + 8As + 9H 2 O → 3H 2 AsO AsH 3 As H 2 AsO 4 - AsH 3 + 4H 2 O+6H + + 5e - 3H e - 3(3( ) 5(5( ) → → 3