Balancing Redox Equations in
Acidic Conditions


Take the reaction between potassium permanganate
（ KMnO
4
) and sodium sulfite (NaSO
3
)
7 steps are required to balance the full ionic equation from 2 separate half equations ( oxidation & reduction )

 Write 2 half-equations for the reaction.


MnO
4
→ Mn 2+
SO
3
2→ SO
4
2-




Balance oxygen atoms using H
2
O
MnO
4
→ Mn 2+ + 4H
2
O
SO
3
2+ H
2
O → SO
4
2-

 Balance H with H +


8H + + MnO
4
-
SO
3
2-
→ Mn 3+ + 4H
2
O
+ H
2
O → SO
4
2+ 2H +

 Balance the charges with electrons


8H + + MnO
4
-
SO
3
2-
+ 5e → Mn 2+ 4H
2
O
+ H
2
O → SO
4
2+ 2H + + 2e -

 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
4
-
5SO
3
2-
+ 10e → 2Mn 2+ 8H
2
O
+ 5H
2
O → 5SO
4
2+ 10H + + 10e -

 Add the 2 half-reactions


16H + + 2MnO
+ 5SO
4
2-
4
+ 10e + 5SO
3
2-
+ 10H + + 10e -
+ 5H
2
O → 2Mn 2+ 8H
2
O
The electrons cancel on the both sides to give:
 16H + + 2MnO
5SO
4
2+ 10H +
4
+ 5SO
3
2+ 5H
2
O → 2Mn 2+ 8H
2
O +

 Subtract H + & H
2
O 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 H
2
O molecules & the production of 8 H
2
O molecules is equal to the net production of 3 H
2
O molecules
6H + + 2MnO
4
+ 5SO
3
2→ 2Mn 2+ 3H
2
O + 5SO
4
2-

•
•
•
•
The H
2
O molecules always appear on the right handside (RHS) of the reduction (8H + + MnO
4
+ 5e →
Mn 2+ 4H
2
O) reaction but on the left-hand side (LHS) of the oxidation (SO
3
2+ H
2
O → SO
4
2+ 2H + + 2e reaction
The H + ions always appear on the opposite side of H
2
O molecules in both of the half-equations and the net ionic equation.
6H + + 2MnO
4
+ 5SO
3
2→ 2Mn 2+ 3H
2
O + 5SO
4
2-

Balancing redox equations for neutral or alkaline conditions
 The reaction taking place is between potassium permanganate and sodium sulfite to form manganese dioxide (MnO
2
)
 The method used for balancing equations in acidic conditions is used; then 1 OH is added for every H + in the equation

 Write the 2 half-reactions


MnO
4
→ MnO
2
SO
3
2-

SO
4
2-

 Follow the same steps as steps 2-4 for the reaction in acidic media to get the following 2 halfequations:


4H + + MnO
4
-
SO
3
2-
+ 3e → MnO
2
+ H
2
O → SO
4
2+ 2H +
+ 2H
+ 2e -
2
O (x2)
(x3)
 Multiplying the reduction reaction by 2, the oxidation by 3, adding together and simplifying gives:
 2H + + 2MnO
4
+ 3SO
3
2→ 2MnO
2
+ H
2
O + 3SO
4
2-





Add OH to convert any H + to H
2
O. Any OH added to 1 side of the equation must also be added to the other side.
2H + + 2OH 2MnO
3SO
4
2+ 2OH -
4
-
Which simplifies to:
+ 3SO
3
2→ 2MnO
2
+ H
2
O +
2H
2
O + 2MnO
+2 OH -
4
+ 3SO
3
2→ 2MnO
2
+ H
2
O + 3SO
4
2-


Which simplifies further to:
H
2
O + 2MnO
4
+ 3SO
3
2→ 2MnO
2
+ 3SO
4
2+2 OH -

Balancing Redox equations in strongly alkaline conditions





Take the reaction between potassium permanganate & sodium sulfite in strongly alkaline media
MnO
SO
3
2-
4
→

SO
MnO
4
2-
4
2-
Following the same steps of balancing the equation in acidic media, then adding OH- for every H+ results in:
2MnO
4
+ SO
3
2+ 2OH → 2MnO
4
2+ SO
4
2+ H
2
O

 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

Oxidizing agents for redox titrations
 Acidified manganate (VII) ions (permanagate)


8H + + MnO
4
-
MnO
4
-
+ 5e → Mn 2+ 4H
2
O is purple but Mn 2+ is almost colourless

Oxidizing agent for redox titrations
2




Acidified dichromate (VI) ions
14H + + Cr
2
O
7
2+ 6e -

2 Cr 3+ + 7H
2
O
Cr
2
O
7
2are orange in colour
Cr 3+ is green
 Can be used as primary standards (a reagent which is very pure, & can be used to prepare a solution of known concentration)



Iron (III) ions/salts
Fe 3+ + e -

Fe 2+


 I
I
Iodine:
2
+ 2e -

2I -
2 is red brown while I is colourless


Acidified hydrogen peroxide
H
2
O
2
+ 2H + + 2e -

2H
2
O

Reducing agents for redox titrations


Iron(II) salts/ions
Fe 2+

Fe 3+ + e -




Hydrogen peroxide if a more powerful oxidizing agent e.g. dichromate (VI) or manganate (VII) is present
H
2
0
2

2H + + O
2
+ 2e
Iodide ions: 2I -

I
2
-
+ 2e -
Sodium thiosulfate (VI): 2S
2
O
3
2-

S
4
O
6
2+ 2e-