1. Balancing Redox Equations in Acidic Conditions
Take the reaction between potassium permanganate （KMnO4) and sodium sulfite (NaSO3)
7 steps are required to balance the full ionic equation from 2 separate half equations (oxidation & reduction)

2. Step 1
Write 2 half-equations for the reaction.
MnO4-→Mn2+
SO32-→SO42-

3. Step 2 Balance oxygen atoms using H2O MnO4- → Mn2+ + 4H2O
SO32- + H2O → SO42-

4. Step 3 Balance H with H+ 8H+ + MnO4- → Mn3+ + 4H2O
SO32- + H2O → SO H+

5. Step 4 Balance the charges with electrons
8H+ + MnO4- + 5e- → Mn2+ 4H2O
SO32- + H2O → SO H+ + 2e-

6. Step 5
Multiply the 2 half-equations by whole numbers (the lowest common multiple of the 2 stoichiometric coefficients in front of the electrons) so that electrons gained in the reduction reaction equals electrons given out by the oxidation reaction
In this case the reduction reaction needs to be multiplied by 2 while the oxidation reaction needs to be multiplied by 5 to get a common number of electrons 10
16H+ + 2MnO e- → 2Mn2+ 8H2O
5SO H2O → 5SO H+ + 10e-

7. Step 6 Add the 2 half-reactions
16H+ + 2MnO e- + 5SO H2O → 2Mn2+ 8H2O + 5SO H+ + 10e-
The electrons cancel on the both sides to give:
16H+ + 2MnO SO H2O → 2Mn2+ 8H2O + 5SO H+

8. Step 7 Subtract H+ & H2O which occur on both sides of the equation
The consumption of 16H+ & the production of 10H+ is equal to a net consumption of 6H+
The consumption of 5 H2O molecules & the production of 8 H2O molecules is equal to the net production of 3 H2O molecules
6H+ + 2MnO SO32- → 2Mn2+ 3H2O + 5SO42-

9. Points to note
The H2O molecules always appear on the right hand-side (RHS) of the reduction (8H+ + MnO4- + 5e- → Mn2+ 4H2O)
reaction but on the left-hand side (LHS) of the oxidation (SO32- + H2O → SO H+ + 2e- reaction
The H+ ions always appear on the opposite side of H2O molecules in both of the half-equations and the net ionic equation.
6H+ + 2MnO SO32- → 2Mn2+ 3H2O + 5SO42-

10. Balancing redox equations for neutral or alkaline conditions
The reaction taking place is between potassium permanganate and sodium sulfite to form manganese dioxide (MnO2)
The method used for balancing equations in acidic conditions is used; then 1 OH- is added for every H+ in the equation

11. Step 1
Write the 2 half-reactions
MnO4- → MnO2
SO32-SO42-

12. Steps 2-7
Follow the same steps as steps 2-4 for the reaction in acidic media to get the following 2 half-equations:
4H+ + MnO4- + 3e- → MnO2 + 2H2O (x2)
SO32- + H2O → SO H+ + 2e- (x3)
Multiplying the reduction reaction by 2, the oxidation by 3, adding together and simplifying gives:
2H+ + 2MnO4- + 3SO32- → 2MnO2 + H2O + 3SO42-

13. Step 8
Add OH- to convert any H+ to H2O. Any OH- added to 1 side of the equation must also be added to the other side.
2H+ + 2OH- 2MnO4- + 3SO32- → 2MnO2 + H2O + 3SO OH-
Which simplifies to:
2H2O + 2MnO4- + 3SO32- → 2MnO2 + H2O + 3SO OH-
Which simplifies further to:
H2O + 2MnO4- + 3SO32- → 2MnO2 + 3SO OH-

14. Balancing Redox equations in strongly alkaline conditions
Take the reaction between potassium permanganate & sodium sulfite in strongly alkaline media
MnO4- → MnO42-
SO32-SO42-
Following the same steps of balancing the equation in acidic media, then adding OH- for every H+ results in:
2MnO4- + SO OH-→ 2MnO42- + SO42- + H2O

15. Redox Titrations Similar to acid-base titrations
Acid-base titration: transfer of 1 or more hydrogen ions (protons) from the acid to the base
Redox Titration: transfer of one/more electrons from a reducing agent to an oxidizing agent

16. Oxidizing agents for redox titrations
Acidified manganate (VII) ions (permanagate)
8H+ + MnO4- + 5e- → Mn2+ 4H2O
MnO4- is purple but Mn2+ is almost colourless

17. Oxidizing agent for redox titrations 2
Acidified dichromate (VI) ions
14H+ + Cr2O e-  2Cr3+ + 7H2O
Cr2O72- are orange in colour
Cr3+ is green
Can be used as primary standards (a reagent which is very pure, & can be used to prepare a solution of known concentration)

18. Some more oxidizing agents
Iron (III) ions/salts
Fe3+ + e-  Fe2+
Iodine:
I2 + 2e-  2I-
I2 is red brown while I- is colourless
Acidified hydrogen peroxide
H2O2 + 2H+ + 2e-  2H2O

19. Reducing agents for redox titrations
Iron(II) salts/ions
Fe2+  Fe3+ + e-
Hydrogen peroxide if a more powerful oxidizing agent e.g. dichromate (VI) or manganate (VII) is present
H202  2H+ + O2 + 2e-
Iodide ions: 2I-  I2 + 2e-
Sodium thiosulfate (VI): 2S2O32-  S4O e-