Oxidation-Reduction Reactions

Overview

Oxidation-reduction (redox) reactions involve the transfer of electrons between species. One species loses electrons (oxidation) while another gains electrons (reduction).

Key Definitions

Term	Definition	Memory Aid
Oxidation	Loss of electrons	OIL = Oxidation Is Loss
Reduction	Gain of electrons	RIG = Reduction Is Gain
Oxidizing agent	Causes oxidation; gets reduced
Reducing agent	Causes reduction; gets oxidized

Oxidation States (Numbers)

Rules for Assigning Oxidation States

Rule	Oxidation State
Elements in free state	0
Monatomic ions	Ion charge
F in compounds	-1
H in compounds	+1 (usually)
H in metal hydrides	-1
O in compounds	-2 (usually)
O in peroxides	-1
Sum in neutral compound	0
Sum in ion	Ion charge

Examples

H₂SO₄:

$$\text{H: } +1 \times 2 = +2$$ $$\text{O: } -2 \times 4 = -8$$ $$\text{S: } +6 \quad \text{(to balance)}$$ $$\text{Total: } +2 + 6 + (-8) = 0 \checkmark$$

MnO₄⁻:

$$\text{O: } -2 \times 4 = -8$$ $$\text{Mn: } +7 \quad \text{(to give -1 charge)}$$ $$\text{Total: } +7 + (-8) = -1 \checkmark$$

Identifying Redox Reactions

Oxidation

• Oxidation state increases
• Electrons lost (appear as products)

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad \text{(oxidation)}$$

Reduction

• Oxidation state decreases
• Electrons gained (appear as reactants)

$$\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \quad \text{(reduction)}$$

Balancing Redox Equations

Half-Reaction Method (Acidic Solution)

1. Separate into oxidation and reduction half-reactions
2. Balance atoms other than O and H
3. Balance O by adding H₂O
4. Balance H by adding H⁺
5. Balance charge by adding e⁻
6. Equalize electrons in both half-reactions
7. Add half-reactions and simplify

Example: Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺ (acidic)

Oxidation half-reaction:

$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$

Reduction half-reaction:

$$\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$$ $$\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \text{(add H}_2\text{O for O)}$$ $$\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \text{(add H}^+ \text{ for H)}$$ $$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad \text{(add } e^- \text{ for charge)}$$

Combine:

$$5(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-)$$ $$1(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O})$$

Final balanced equation:

$$5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

Half-Reaction Method (Basic Solution)

Follow acidic method, then:

• Add OH⁻ to both sides equal to H⁺
• Combine H⁺ + OH⁻ → H₂O
• Cancel water molecules

Common Oxidizing Agents

Agent	Product	Half-Reaction
MnO₄⁻ (acidic)	Mn²⁺	$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
MnO₄⁻ (basic)	MnO₂	$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$
Cr₂O₇²⁻ (acidic)	Cr³⁺	$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
Cl₂	Cl⁻	$\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$
O₂	O²⁻	$\text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-}$

Common Reducing Agents

Agent	Product	Half-Reaction
Metals	Cations	$\text{M} \rightarrow \text{M}^{n+} + ne^-$
H₂	H⁺	$\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$
Fe²⁺	Fe³⁺	$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$
I⁻	I₂	$2\text{I}^- \rightarrow \text{I}_2 + 2e^-$

Disproportionation

A species is both oxidized and reduced.

Example: Cl₂ in base

$$\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}$$

• $\text{Cl}_2 (0) \rightarrow \text{Cl}^- (-1)$ reduction
• $\text{Cl}_2 (0) \rightarrow \text{ClO}^- (+1)$ oxidation

Activity Series

Metals are listed by their tendency to lose electrons:

$$\text{Li} > \text{K} > \text{Ba} > \text{Ca} > \text{Na} > \text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Ni} > \text{Sn} > \text{Pb} > \text{H} > \text{Cu} > \text{Hg} > \text{Ag} > \text{Pt} > \text{Au}$$

• Metals above H can displace H from acids
• Higher metals can reduce lower metal ions

Example

$$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \checkmark \quad \text{(Zn above Cu)}$$ $$\text{Cu} + \text{Zn}^{2+} \rightarrow \text{No reaction} \quad \text{(Cu below Zn)}$$

Identifying Redox vs Non-Redox

Redox	Non-Redox
Oxidation states change	No oxidation state change
Electron transfer occurs	No electron transfer
Synthesis, decomposition, single replacement, combustion	Acid-base, precipitation (usually)