Balancing Redox Equations in Acidic Conditions Take the reaction between potassium permanganate KMnO 4 ) and sodium sulfite (NaSO 3 ) 7 steps are required.

Balancing Redox Equations in Acidic ConditionsTake 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)

Step 1Write 2 half-equations for the reaction.MnO4

-→Mn2+

SO32-→SO4

2-

Step 2Balance oxygen atoms using H2O

MnO4- → Mn2+ + 4H2O

SO32- + H2O → SO4

2-

Step 3Balance H with H+

8H+ + MnO4- → Mn3+ + 4H2O

SO32- + H2O → SO4

2- + 2H+

Step 4Balance the charges with electrons8H+ + MnO4

- + 5e- → Mn2+ 4H2O

SO32- + H2O → SO4

2- + 2H+ + 2e-

Step 5Multiply 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+ + 2MnO4- + 10e- → 2Mn2+ 8H2O

5SO32- + 5H2O → 5SO4

2- + 10H+ + 10e-

Step 6Add the 2 half-reactions16H+ + 2MnO4

- + 10e- + 5SO32- + 5H2O →

2Mn2+ 8H2O + 5SO42- + 10H+ + 10e-

The electrons cancel on the both sides to give:

16H+ + 2MnO4- + 5SO3

2- + 5H2O → 2Mn2+

8H2O + 5SO42- + 10H+

Step 7Subtract H+ & H2O which occur on both sides

of the equationThe 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+ + 2MnO4- + 5SO3

2- → 2Mn2+ 3H2O + 5SO4

2-

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 (SO3

2- + H2O → SO42- + 2H+ + 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+ + 2MnO4- + 5SO3

2- → 2Mn2+ 3H2O + 5SO42-

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

Step 1Write the 2 half-reactionsMnO4

- → MnO2

SO32-SO4

2-

Steps 2-7Follow 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 → SO4

2- + 2H+ + 2e- (x3)Multiplying the reduction reaction by 2, the

oxidation by 3, adding together and simplifying gives:

2H+ + 2MnO4- + 3SO3

2- → 2MnO2 + H2O + 3SO4

2-

Step 8Add 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- + 3SO3

2- → 2MnO2 + H2O + 3SO4

2- + 2OH-

Which simplifies to:2H2O + 2MnO4

- + 3SO32- → 2MnO2 + H2O +

3SO42- +2 OH-

Which simplifies further to:H2O + 2MnO4

- + 3SO32- → 2MnO2 + 3SO4

2- +2 OH-

Balancing Redox equations in strongly alkaline conditionsTake the reaction between potassium

permanganate & sodium sulfite in strongly alkaline media

MnO4- → MnO4

2-

SO32-SO4

2-

Following the same steps of balancing the equation in acidic media, then adding OH- for every H+ results in:

2MnO4- + SO3

2- + 2OH-→ 2MnO42- + SO4

2- + H2O

Redox TitrationsSimilar to acid-base titrationsAcid-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

Oxidizing agents for redox titrationsAcidified manganate (VII) ions (permanagate)8H+ + MnO4

- + 5e- → Mn2+ 4H2O

MnO4- is purple but Mn2+ is almost colourless

Oxidizing agent for redox titrations 2Acidified dichromate (VI) ions14H+ + Cr2O7

2- + 6e- 2Cr3+ + 7H2O

Cr2O72- are orange in colour

Cr3+ is greenCan be used as primary standards (a reagent

which is very pure, & can be used to prepare a solution of known concentration)

Some more oxidizing agentsIron (III) ions/saltsFe3+ + e- Fe2+

Iodine:I2 + 2e- 2I-

I2 is red brown while I- is colourlessAcidified hydrogen peroxideH2O2 + 2H+ + 2e- 2H2O

Reducing agents for redox titrationsIron(II) salts/ionsFe2+ 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- S4O6

2- + 2e-