Balancing Redox Reactions Half Reaction Method 1 Write

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 - 2 H+ + SO 42 - + 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 42 - + NO 2

3. Write the half reaction of oxidation H 2 S SO 42 - 4. Balance the atoms. Add H 2 O to balance for extra oxygens, then add H+ to balance for hydrogens H 2 S + 4 H 2 O SO 42 - + 10 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 42 - + 10 H+ + 8 e- 6. Repeat steps 1 -5 for the half-reaction for reduction NO 3 - NO 2 NO 3 - + 2 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 42 - + 10 H+ + 8 e-) 8(NO 3 - + 2 H+ + e- NO 2 + H 2 O) . 8. Combine the half-reactions, and cancel out anything common to both sides 8 NO 3 - + H 2 S + 6 H+ 8 NO 2 + 4 H 2 O + SO 42 -

9. Combine ions to form the compounds shown in the original formula equation. 8 HNO 3 - + H 2 S 8 NO 2 + 4 H 2 O + SO 42 - + 2 H+ 10. Check to ensure that all other ions balance out 8 HNO 3 - + H 2 S 8 NO 2 + 4 H 2 O + H 2 SO 4

Balancing Redox Full Reactions in Acid Solution 4. 3. 5. 2. 1. Simplify Add the two half reactions Check Separate Balance each into half reactions Mn. O 4 - + C 2 O 42 - → 2( Mn. O 4 - + 4 H+ + 3 e- → Mn. O 2 + CO 32+ 2 H 2 O ) 3(C 2 O 42 - + 2 H 2 O → 2 CO 32 - + 4 H+ + 2 e-) 2 4 2 Mn. O 4 - + 3 C 2 O 42 - + 6 H 2 O + 8 H+ → 2 Mn. O 2 + 6 CO 32 - + 12 H+ + 4 H 2 O 2 Mn. O 4 - + 3 C 2 O 42 - + 2 H 2 O → 2 Mn. O 2 + 6 CO 32 - + 4 H+

Balancing Redox Full Reactions in Basic Solution 1. 2. - toacid Balance Add OHin neutralize solutionthe H+ 2 Mn. O 4 - + 3 C 2 O 42 - + 2 H 2 O → 2 Mn. O 2 + 6 CO 32 - + 4 OH- 4 H+ 4 OH 2 4 H 2 O

Balancing Redox Full Reactions in Basic Solution 2 Mn. O 4 - + 3 C 2 O 42 - + 4 OH- → 2 Mn. O 2 + 6 CO 32 - + 2 H 2 O -12

Balancing Redox Full Reactions in Acid Solution Fe + O 2 4( Fe + 3( 4 e- + O 2 → 3 H 2 O + H 2 O → 4 H+ + 3 H+ Fe(OH)3 + 3 e -) → H 20 + H 2 O) 6 4 Fe + 12 H 2 O + 3 O 2 +12 H+ → 12 H+ + 4 Fe(OH)3 + 6 H 2 O 4 Fe + 6 H 2 O + 3 O 2 → 4 Fe(OH)3

Balance the redox reaction As → H 2 As. O 4+ As. H 3 You must separate into two half reactions! 3(As + 5( 3 H+ 4 H 2 O + As → H 2 As. O 4 - + 3 e- → + (alkaline) 6 H+ + As. H 3 ) 3 As + 12 H 2 O + 15 H+ + 5 As → 3 H 2 As. O 4 - 5 e-) 3 + 18 H+ + 5 As. H 3 8 As + 12 H 2 O → 3 H 2 As. O 4 - + 3 H+ + 5 As. H 3 3 OH 3 OH- + 8 As + 9 H 2 O → 3 H 2 As. O 4 - + 5 As. H 3