Balancing Redox Reactions

A)     Oxidation number method

1.   Write correct formula for each reactant & product.

Ex: K2Cr2O7 + HCl → KCl + CrCl3 + H2O + Cl2

2.   Write oxidation numbers for all elements

K+12Cr+62O-27 + H+1Cl-1 → K+1Cl-1 + Cr+3Cl-13 + H+12O-2 + Cl02

3.   Calculate the increase or decrease in oxidation number per atom

• Oxidation number of Cr decreased from +6 to +3
  K2Cr+62O7 → 2Cr+32Cl3

• Oxidation number of Cl increased from -1 to 0
  HCl-1 → Cl02

4.   The decrease in oxidation number should be equal to increase in oxidation number

• Decrease in Oxid. num of Cr = 6 units per molecule of K2Cr2O7. Increase in oxid. num of Cl = 1 unit per molecule of HCl
  We need 6 HCl for 6 units increase in oxid num. so the reaction becomes

K2Cr2O7 + 6 HCl → 2CrCl3 + 3Cl2

5.   Balance the remaining atoms. Potassium, Hydrogen & oxygen in this case.

K2Cr2O7 + 14 HCl → +2KCl + 2CrCl3 + 7 H2O + 3Cl2

The reaction is balanced.

6.   If the reacting is taking place in water, add H+ or OH- ions on the appropriate side so that the total ionic charges of reactants & products are equal.

7.   If reaction is carried out in acidic solution, use H+ ions in the equation.

If in basic solution, use OH- ions.

B)     Half reaction method

The 2 half reactions are balanced separately and then added together
Let’s balance the following reaction
H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)

Step 1

• Write half-cell reactions
  

1. Oxidation reaction Fe2+ (aq) → Fe3+ (aq) → eq (i)
2. Reduction reaction H2O2 (aq) → H2O (l) → eq (ii)

Step 2

• Balance the atoms other than O & H in each half reaction.
  Here Fe atoms are already balanced.

Step 3

• For acidic medium, add H2O to balance O atoms and H+ to balance H atoms
• For basic medium, add OH- to balance H atoms and H2O to balance O atoms.
  The given reaction is taking place in acidic medium so to balance O atoms we add H2O & to balance H atoms we add H+

.·. Eq (ii) becomes

H2O2 + 2H+ → H2O + H2O

=> H2O2 + 2H+ → 2H2O

Step 4

• To balance the charges, add electrons to the side deficient of electrons

Fe2+ → Fe3+ + e-             → (iii)

H2O2 + 2H+ + 2e- → 2H2O         → (iv)

Step 5

• Make the number of electrons in the two half reactions equal by multiplying the reactions by appropriate coefficients. Here, multiply equation (iii) by 2

2Fe2+ → 2Fe3+ + e-

Step 6

• Add the two half-cell reactions

2Fe2+ + H2O2 + 2H+ → 2Fe3+ + 2H2O

Step 7

• Verify that the equation contains same type and number of atoms and same charges on both equations. If not, the equation is not balanced so check the steps again.
  Thus, the balanced equation is

2Fe2+ (aq) + H2O2 (aq) + 2H+ → 2Fe3+ (aq) + 2H2O (l)