Chemistry

Redox Equations

Reduction vs Oxidation

Reduction:
- Gain of electrons
- Decrease in oxidation state

Oxidation:
- Loss of electrons
- Increase in oxidation state

For covalent compound, the bonds are assumed to be ionic, with the more electronegative atom assuming the role of the negative charge.

Fluorine is the most electronegative atom, with electronegativity increasing across the period and down the group.

Example: H2O
Due to the fact that oxygen is more electronegative, the oxidation state of O = -2 while that of H = +1

Common oxidation states

Group 1 → +1
Group 2 → +2
Group 3 → +3
Hydrogen → +1
Hydrogen in metal hydrides → -1
Oxygen → -2
Oxygen in peroxides → -1

Disproportionation reactions

A disproportionation reaction is a redox reaction where atoms of an element in a single substance undergoes oxidation and reduction simultaneously. The oxidation state of the atom increases and decreases at the same time.

2H2O2 → 2H2O + O2
The oxidation state of oxygen decreases from -1 in hydrogen peroxide to -2 in water, and increased from -1 in hydrogen peroxide to 0 in oxygen.

Comproportionation reactions

A comproportionation reaction is a redox reaction where two reactants each containing the same element with different oxidation numbers, form a product in which the element involved achieves the same oxidation number.

5I^- + IO3^- + 6H^+ → 3I2 + 3H20

The oxidation state of iodine increases from -1 in iodide to 0 in iodine, and decreases from +5 in iodate (V) to 0 in iodine.

Balancing redox equations

A - Balance all half-equations
O - Balance oxygen by adding 1H20/ 2OH^-
H - Balance hydrogen by adding 2H^+ for every H2O/ 1 H2O for every 2OH^-
E - Balance charges with electrons

Balancing redox titrations -Ion electron half equation method

Example 1: Acidic medium

Step 1: Write down the balanced equation.

MnO4^- + Fe^2+ → Mn^2+ + Fe^3+ (done in acidic medium)

Step 2: write down the half equations.

MnO4^- → Mn^2+
Fe^2+ → Fe^3+

Step 3: Balance the equations for elements that have undergone a change in oxidation state.

MnO4^- → Mn^2+ (already balanced)
Fe^2+ → Fe^3+ (already balanced)

Step 4: Balance oxygen atoms by adding 1 H2O for every deficient oxygen to the side with less oxygen.

MnO4^- → Mn^2+ + 4H2O
Fe^2+ → Fe^3+ (no oxygen)

Step 5: Balance hydrogen atoms by adding 2H^+ for every H2O added to the side with less hydrogen.

MnO4^- + 8H^+ → Mn^2+ + 4H2O
Fe^2+ → Fe^3+ (no hydrogen)

Step 6: Balance charges by adding electrons

MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O
Fe^2+ → Fe^3+ + e-

Step 7: Combine half equations. (No. of moles of e^- lost in oxidation = No. of moles of e^- gained in reduction)

MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O (x1)
Fe^2+ → Fe^3+ + e- (x5)

∴ Overall balanced equation:
8H^+ + MnO4^- + 5Fe^2+ → Mn^2+ + 4H2O +5Fe^3+