Oxidation States (Oxidation Numbers)

Oxidation state: Charge an atom would have if compound were ionic

Used to:

• Track electron transfer
• Identify what's oxidized/reduced
• Balance redox equations

Rules for Assigning Oxidation States

Apply in order (earlier rules take precedence):

1. Free elements: Oxidation state = 0

• Examples: Na(s), O₂(g), Cl₂(g), S₈(s), Fe(s)
• All atoms in elemental form = 0

2. Monatomic ions: Oxidation state = ionic charge

• Na⁺ = +1
• Cl⁻ = -1
• Ca²⁺ = +2
• O²⁻ = -2
• Al³⁺ = +3

3. Oxygen in compounds: Usually -2

• Exceptions:
  • Peroxides (H₂O₂, Na₂O₂): O = -1
  • Superoxides (KO₂): O = -½
  • Bonded to F (OF₂): O = +2

4. Hydrogen in compounds: Usually +1

• Exception: Metal hydrides (NaH, CaH₂): H = -1

5. Fluorine: Always -1 (most electronegative)

6. Group 1 metals (Li, Na, K, etc.): Always +1

7. Group 2 metals (Mg, Ca, Ba, etc.): Always +2

8. Aluminum: Usually +3

9. Sum of oxidation states:

• Neutral molecule: Sum = 0
• Polyatomic ion: Sum = ion charge

Examples of Assigning Oxidation States

Example 1: H₂O

• H: +1 (rule 4)
• O: -2 (rule 3)
• Check: 2(+1) + (-2) = 0 ✓

Example 2: H₂SO₄

• H: +1
• O: -2
• S: ?

Sum must = 0: 2(+1) + S + 4(-2) = 0 2 + S - 8 = 0 S = +6

Example 3: MnO₄⁻ (permanganate ion)

• O: -2
• Mn: ?

Sum must = -1: Mn + 4(-2) = -1 Mn - 8 = -1 Mn = +7

Example 4: Cr₂O₇²⁻ (dichromate ion)

• O: -2
• Cr: ?

Sum must = -2: 2(Cr) + 7(-2) = -2 2Cr - 14 = -2 2Cr = 12 Cr = +6

Example 5: NH₃

• H: +1
• N: ?

Sum must = 0: N + 3(+1) = 0 N = -3

Example 6: Fe₃O₄ (magnetite)

• O: -2
• Fe: ?

Sum must = 0: 3(Fe) + 4(-2) = 0 3Fe - 8 = 0 3Fe = 8 Fe = +8/3 (average)

Actually: Mixed oxidation states

• 1 Fe³⁺ and 2 Fe²⁺
• Average: (1×3 + 2×2)/3 = 7/3... wait
• Correct: FeO·Fe₂O₃ has Fe²⁺ and Fe³⁺

Identifying Oxidation and Reduction

Oxidation: Increase in oxidation state (becomes more positive)

• Atom loses electrons
• Example: Fe → Fe²⁺ (0 → +2, oxidized)

Reduction: Decrease in oxidation state (becomes more negative)

• Atom gains electrons
• Example: Cl₂ → 2Cl⁻ (0 → -1, reduced)

Example: Zn + Cu²⁺ → Zn²⁺ + Cu

Assign oxidation states:

Reactants:

• Zn(s): 0 (free element)
• Cu²⁺: +2 (ion charge)

Products:

• Zn²⁺: +2
• Cu(s): 0

What changed?

Zn: 0 → +2

• Increase in oxidation state
• Lost 2 electrons
• Zn is oxidized

Cu: +2 → 0

• Decrease in oxidation state
• Gained 2 electrons
• Cu²⁺ is reduced

Oxidizing and Reducing Agents

Oxidizing agent (oxidant): Species that causes oxidation

• Gets reduced itself
• Gains electrons
• Example: Cu²⁺ (causes Zn to be oxidized)

Reducing agent (reductant): Species that causes reduction

• Gets oxidized itself
• Loses electrons
• Example: Zn (causes Cu²⁺ to be reduced)

Summary for Zn + Cu²⁺ → Zn²⁺ + Cu:

• Oxidized: Zn
• Reduced: Cu²⁺
• Oxidizing agent: Cu²⁺
• Reducing agent: Zn

Memory aid:

• Oxidizing agent gets reduced (gains electrons)
• Reducing agent gets oxidized (loses electrons)
• They do the opposite of what their name suggests!

Types of Redox Reactions

1. Combination (Synthesis)

Element + element → compound

$2Mg\text{(s)} + O2\text{(g)} \rightarrow 2MgO\text{(s)}$

• Mg: 0 → +2 (oxidized)
• O: 0 → -2 (reduced)

2. Decomposition

Compound → elements

$2H2O\text{(l)} \rightarrow 2H2\text{(g)} + O2\text{(g)}$

• H: +1 → 0 (reduced)
• O: -2 → 0 (oxidized)

3. Single Replacement

Metal displacing metal:

$Zn\text{(s)} + CuSO4\text{(aq)} \rightarrow ZnSO4\text{(aq)} + Cu\text{(s)}$

• Zn: 0 → +2 (oxidized)
• Cu: +2 → 0 (reduced)

Halogen displacing halogen:

$Cl2\text{(g)} + 2NaBr\text{(aq)} \rightarrow 2NaCl\text{(aq)} + Br2\text{(l)}$

• Cl: 0 → -1 (reduced)
• Br: -1 → 0 (oxidized)

4. Combustion

Always redox reactions

$CH4\text{(g)} + 2O2\text{(g)} \rightarrow CO2\text{(g)} + 2H2O\text{(l)}$

• C: -4 → +4 (oxidized)
• O: 0 → -2 (reduced)

5. Disproportionation

Same element both oxidized and reduced

$Cl2\text{(g)} + 2OH-\text{(aq)} \rightarrow Cl-\text{(aq)} + ClO-\text{(aq)} + H2O\text{(l)}$

• One Cl: 0 → -1 (reduced)
• Other Cl: 0 → +1 (oxidized)

Balancing Redox Equations

Half-Reaction Method

Most reliable method for complex redox equations

Steps:

1. Assign oxidation states

• Identify what's oxidized and reduced

2. Write two half-reactions

• Oxidation half-reaction (electrons as product)
• Reduction half-reaction (electrons as reactant)

3. Balance atoms in each half-reaction

• Balance all atoms except H and O
• Balance O by adding H₂O
• Balance H by adding H⁺ (acidic) or OH⁻ (basic)

4. Balance charges

• Add electrons to balance charge

5. Equalize electrons

• Multiply half-reactions so electrons cancel

6. Add half-reactions

• Cancel electrons and any common terms

7. Check balance

• Atoms and charges must balance

Example: Acidic Solution

Balance: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic solution)

Step 1: Oxidation states

• Mn: +7 → +2 (reduced)
• Fe: +2 → +3 (oxidized)

Step 2: Half-reactions

Reduction: $MnO4- \rightarrow Mn^{2+}$

Oxidation: $Fe^{2+ -> Fe^{3+}}$

Step 3: Balance atoms

Reduction half:

• Mn balanced ✓
• Add 4 H₂O to right (for 4 O):

$MnO4- \rightarrow Mn^{2+ + 4H2O}$

• Add 8 H⁺ to left (for 8 H):

$MnO4- + 8H+ \rightarrow Mn^{2+ + 4H2O}$

Oxidation half:

• Fe balanced ✓ (already done)

Step 4: Balance charges

Reduction:

• Left: -1 + 8(+1) = +7
• Right: +2
• Add 5e⁻ to left:

$MnO4- + 8H+ + 5e- \rightarrow Mn^{2+ + 4H2O}$

Check: -1 + 8 - 5 = +2 ✓

Oxidation:

• Left: +2
• Right: +3
• Add 1e⁻ to right:

$Fe^{2+ -> Fe^{3+} + e-}$

Check: +2 = +3 - 1 ✓

Step 5: Equalize electrons

Reduction uses 5e⁻, oxidation produces 1e⁻

Multiply oxidation by 5:

$5Fe^{2+ -> 5Fe^{3+} + 5e-}$

Step 6: Add half-reactions

$MnO4- + 8H+ + 5e- \rightarrow Mn^{2+ + 4H2O}$ $5Fe^{2+ -> 5Fe^{3+} + 5e-}$

Cancel 5e⁻:

$\boxed{MnO4- + 8H+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H2O}$

Step 7: Check

Atoms:

• Mn: 1, 1 ✓
• O: 4, 4 ✓
• H: 8, 8 ✓
• Fe: 5, 5 ✓

Charge:

• Left: -1 + 8 + 5(+2) = +17
• Right: +2 + 5(+3) = +17 ✓

Example: Basic Solution

Balance: Cr₂O₇²⁻ + Cl⁻ → Cr³⁺ + Cl₂ (basic solution)

Steps 1-4: Same as acidic, but use H⁺ first

Reduction: $Cr2O7^{2- + 14H+ + 6e- -> 2Cr^{3+} + 7H2O}$

Oxidation: $2Cl- \rightarrow Cl2 + 2e-$

Multiply oxidation by 3: $6Cl- \rightarrow 3Cl2 + 6e-$

Add: $Cr2O7^{2- + 14H+ + 6Cl- -> 2Cr^{3+} + 3Cl2 + 7H2O}$

Now convert to basic:

Step 5: Add OH⁻ to neutralize H⁺

Add 14 OH⁻ to both sides:

$Cr2O7^{2- + 14H+ + 14OH- + 6Cl- -> 2Cr^{3+} + 3Cl2 + 7H2O + 14OH-}$

Step 6: Combine H⁺ + OH⁻ → H₂O

14 H⁺ + 14 OH⁻ = 14 H₂O (left side)

$Cr2O7^{2- + 14H2O + 6Cl- -> 2Cr^{3+} + 3Cl2 + 7H2O + 14OH-}$

Step 7: Cancel water

14 H₂O - 7 H₂O = 7 H₂O (left side)

$\boxed{Cr2O7^{2-} + 7H2O + 6Cl- \rightarrow 2Cr^{3+} + 3Cl2 + 14OH-}$

Check charge:

• Left: -2 + 0 + 6(-1) = -8
• Right: 2(+3) + 0 + 14(-1) = 6 - 14 = -8 ✓

Common Oxidizing and Reducing Agents

Strong Oxidizing Agents

Accept electrons readily:

• MnO₄⁻ (permanganate) - strong, purple color
• Cr₂O₇²⁻ (dichromate) - orange color
• H₂O₂ (hydrogen peroxide)
• O₂ (oxygen gas)
• Halogens: F₂ > Cl₂ > Br₂ > I₂
• Concentrated H₂SO₄, HNO₃

Strong Reducing Agents

Donate electrons readily:

• Active metals: Li, K, Na, Ca, Mg
• H₂ (hydrogen gas)
• Carbon (C)
• Sn²⁺ (can be oxidized to Sn⁴⁺)
• Fe²⁺ (can be oxidized to Fe³⁺)

Spontaneity of Redox Reactions

Activity series predicts which redox reactions occur spontaneously

More active metal:

• Loses electrons more easily (better reducing agent)
• Displaces less active metal from solution

Activity series (metals, strongest to weakest):

Li > K > Ba > Ca > Na > Mg > Al > Zn > Cr > Fe > Ni > Sn > Pb > H > Cu > Ag > Au

Rule: Metal can reduce ions of any metal below it

Example:

• Zn can reduce Cu²⁺ (Zn above Cu) ✓
• Cu cannot reduce Zn²⁺ (Cu below Zn) ✗

Halogen activity series:

F₂ > Cl₂ > Br₂ > I₂

Rule: Halogen can oxidize ions of any halogen below it

Example:

• Cl₂ can oxidize Br⁻ to Br₂ ✓
• Br₂ cannot oxidize Cl⁻ ✗

Applications of Redox Reactions

Batteries and Electrochemical Cells

Convert chemical energy to electrical energy

• Based on spontaneous redox reactions
• Oxidation at anode (negative)
• Reduction at cathode (positive)

Corrosion

Unwanted oxidation of metals

• Rusting: 4Fe + 3O₂ → 2Fe₂O₃
• Fe oxidized: 0 → +3

Metallurgy

Extracting metals from ores

• Reduction of metal oxides
• Example: Fe₂O₃ + 3CO → 2Fe + 3CO₂

Bleaching

Oxidation of colored compounds

• H₂O₂, Cl₂ used as oxidizing agents

Cellular Respiration

Glucose oxidized to CO₂

• C₆H₁₂O₆ + O₂ → CO₂ + H₂O + energy
• C: -1 → +4 (oxidized)

Photosynthesis

CO₂ reduced to glucose

• Reverse of respiration
• C: +4 → -1 (reduced)

Summary

Key concepts:

1. Redox = electron transfer

   • Oxidation = loss of e⁻
   • Reduction = gain of e⁻

2. Oxidation states track electrons

   • Increase = oxidation
   • Decrease = reduction

3. Complementary processes

   • Oxidizing agent gets reduced
   • Reducing agent gets oxidized

4. Half-reaction method

   • Balance atoms (except H, O)
   • Add H₂O for O
   • Add H⁺ for H (acidic) or OH⁻ (basic)
   • Balance charge with e⁻
   • Equalize and add

5. Activity series predicts spontaneity

   • Active metals displace less active
   • Active halogens oxidize less active

Mn: Unknown, calculate from sum

Sum rule for neutral compound:

Total oxidation states = 0

Set up equation:

$\text{K} + \text{Mn} + 4(\text{O}) = 0$

$+1 + \text{Mn} + 4(-2) = 0$

$+1 + \text{Mn} - 8 = 0$

$\text{Mn} - 7 = 0$

$\text{Mn} = +7$

Answer (a):

$\boxed{\text{K} = +1, \text{Mn} = +7, \text{O} = -2}$

Verification:

• (+1) + (+7) + 4(-2) = 1 + 7 - 8 = 0 ✓

Note: Mn⁺⁷ is manganese in its highest oxidation state, making MnO₄⁻ a strong oxidizing agent (purple color).

(b) H₂SO₄ (Sulfuric acid)

Apply rules:

H: Hydrogen in compound

• H = +1 (rule 4)

O: Oxygen in compound

• O = -2 (rule 3)

S: Calculate from sum

Sum = 0 (neutral molecule):

$2(\text{H}) + \text{S} + 4(\text{O}) = 0$

$2(+1) + \text{S} + 4(-2) = 0$

$+2 + \text{S} - 8 = 0$

$\text{S} - 6 = 0$

$\text{S} = +6$

Answer (b):

$\boxed{\text{H} = +1, \text{S} = +6, \text{O} = -2}$

Verification:

• 2(+1) + (+6) + 4(-2) = 2 + 6 - 8 = 0 ✓

Note: S⁺⁶ is sulfur in its highest common oxidation state. Concentrated H₂SO₄ is an oxidizing agent.

(c) NH₄⁺ (Ammonium ion)

Apply rules:

H: +1 (rule 4)

N: Calculate

Sum = +1 (charge on ion):

$\text{N} + 4(\text{H}) = +1$

$\text{N} + 4(+1) = +1$

$\text{N} + 4 = +1$

$\text{N} = +1 - 4$

$\text{N} = -3$

Answer (c):

$\boxed{\text{N} = -3, \text{H} = +1}$

Verification:

• (-3) + 4(+1) = -3 + 4 = +1 ✓

Note: Nitrogen in NH₄⁺ has same oxidation state as in NH₃ (-3). The positive charge comes from the extra H⁺, not from N oxidation state.

(d) Cr₂O₇²⁻ (Dichromate ion)

Apply rules:

O: -2 (rule 3)

Cr: Calculate (note: 2 Cr atoms!)

Sum = -2 (charge on ion):

$2(\text{Cr}) + 7(\text{O}) = -2$

$2(\text{Cr}) + 7(-2) = -2$

$2(\text{Cr}) - 14 = -2$

$2(\text{Cr}) = -2 + 14$

$2(\text{Cr}) = +12$

$\text{Cr} = +6$

Answer (d):

$\boxed{\text{Cr} = +6, \text{O} = -2}$

Verification:

• 2(+6) + 7(-2) = 12 - 14 = -2 ✓

Note: Both Cr atoms have same oxidation state (+6). Cr₂O₇²⁻ is orange and a strong oxidizing agent.

Summary Table

Additional Insights

Common oxidation states to recognize:

Permanganate (MnO₄⁻):

• Mn: +7 (maximum for Mn)
• Strong oxidizer, purple
• Reduced to Mn²⁺ (pale pink) in acidic solution

Sulfate (SO₄²⁻) and Sulfuric acid (H₂SO₄):

• S: +6 (maximum for S)
• Concentrated H₂SO₄ is oxidizing acid

Ammonium (NH₄⁺) and Ammonia (NH₃):

• N: -3 (same in both)
• Low oxidation state, can be oxidized

Dichromate (Cr₂O₇²⁻):

• Cr: +6 (high oxidation state)
• Strong oxidizer, orange
• Reduced to Cr³⁺ (green) in acidic solution

Why these are important:

1. Identify oxidizing agents:

• High oxidation states (Mn⁺⁷, Cr⁺⁶, S⁺⁶) indicate strong oxidizers
• Can accept electrons, get reduced

2. Predict reactions:

• KMnO₄ and Cr₂O₇²⁻ commonly used in redox titrations
• Strong enough to oxidize many organic compounds

3. Color changes:

• MnO₄⁻ (purple) → Mn²⁺ (pale pink/colorless)
• Cr₂O₇²⁻ (orange) → Cr³⁺ (green)
• Useful visual indicators in titrations

4. Maximum oxidation states:

• Mn can go up to +7
• Cr and S commonly +6
• These are in highest oxidation states, can only be reduced

Practice tip:

When calculating oxidation states:

1. ✓ Start with elements that have fixed values (Group 1, 2, F, O, H)
2. ✓ Use sum rule (= 0 for compounds, = charge for ions)
3. ✓ Solve for unknown element
4. ✓ Always verify by checking sum
5. ✓ Watch for multiple atoms of same element (×2 for Cr₂, H₂, etc.)

Common mistakes to avoid:

• ✗ Forgetting to multiply by number of atoms
• ✗ Using wrong sum (0 vs ionic charge)
• ✗ Forgetting peroxide exception (O = -1 in H₂O₂)