Solution:

(a) KMnO₄ (Potassium permanganate)

Apply oxidation state rules:

K: Group 1 metal

• K = +1 (rule 6)

O: Oxygen in compound

• O = -2 (rule 3, no exceptions here)

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

| Compound | Formula | Oxidation States | Sum Check | |----------|---------|------------------|-----------| | (a) | KMnO₄ | K: +1, Mn: +7, O: -2 | 1 + 7 + 4(-2) = 0 ✓ | | (b) | H₂SO₄ | H: +1, S: +6, O: -2 | 2(+1) + 6 + 4(-2) = 0 ✓ | | (c) | NH₄⁺ | N: -3, H: +1 | -3 + 4(+1) = +1 ✓ | | (d) | Cr₂O₇²⁻ | Cr: +6, O: -2 | 2(+6) + 7(-2) = -2 ✓ |

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)

Solution:

Given reaction:

$\ce{Sn^{2+}(aq) + 2Fe^{3+}(aq) -> Sn^{4+}(aq) + 2Fe^{2+}(aq)}$

Task: Identify redox components and write half-reactions

(a) What is oxidized?

Oxidation = increase in oxidation state

Examine each species:

Sn²⁺ → Sn⁴⁺:

• Oxidation state: +2 → +4
• Increase from +2 to +4
• Change: +2

Fe³⁺ → Fe²⁺:

• Oxidation state: +3 → +2
• Decrease from +3 to +2
• Change: -1

Answer (a):

$\boxed{\text{Sn}^{2+} \text{ is oxidized}}$

Explanation:

• Sn²⁺ loses 2 electrons to become Sn⁴⁺
• Oxidation state increases: +2 → +4
• Loss of electrons = oxidation (OIL)

(b) What is reduced?

Reduction = decrease in oxidation state

From our analysis above:

Fe³⁺ → Fe²⁺:

• Oxidation state: +3 → +2
• Decrease from +3 to +2

Answer (b):

$\boxed{\text{Fe}^{3+} \text{ is reduced}}$

Explanation:

• Fe³⁺ gains 1 electron to become Fe²⁺
• Oxidation state decreases: +3 → +2
• Gain of electrons = reduction (RIG)
• Note: 2 Fe³⁺ ions reduced (coefficient of 2)

(c) The oxidizing agent

Oxidizing agent:

• Species that causes oxidation
• Gets reduced itself
• Accepts electrons

What gets reduced? Fe³⁺

Answer (c):

$\boxed{\text{Fe}^{3+} \text{ is the oxidizing agent}}$

Explanation:

• Fe³⁺ accepts electrons (gets reduced)
• By accepting electrons, it causes Sn²⁺ to lose electrons (oxidize)
• Oxidizing agent = gets reduced

(d) The reducing agent

Reducing agent:

• Species that causes reduction
• Gets oxidized itself
• Donates electrons

What gets oxidized? Sn²⁺

Answer (d):

$\boxed{\text{Sn}^{2+} \text{ is the reducing agent}}$

Explanation:

• Sn²⁺ donates electrons (gets oxidized)
• By donating electrons, it causes Fe³⁺ to gain electrons (reduce)
• Reducing agent = gets oxidized

(e) Balanced half-reactions

Half-reactions show electron transfer explicitly

Oxidation half-reaction:

Sn²⁺ is oxidized (loses electrons)

Change: Sn²⁺ → Sn⁴⁺

Balance atoms: Already balanced (1 Sn on each side)

Balance charges:

• Left: +2
• Right: +4
• Need to add 2e⁻ to right side

Oxidation half-reaction:

$\boxed{\ce{Sn^{2+}(aq) -> Sn^{4+}(aq) + 2e-}}$

Check:

• Atoms: 1 Sn = 1 Sn ✓
• Charge: +2 = +4 + 2(-1) = +2 ✓

Interpretation:

• Sn²⁺ loses 2 electrons
• Electrons shown as products (lost)
• This is oxidation

Reduction half-reaction:

Fe³⁺ is reduced (gains electrons)

Change: Fe³⁺ → Fe²⁺

Balance atoms: Already balanced (1 Fe on each side)

Balance charges:

• Left: +3
• Right: +2
• Need to add 1e⁻ to left side

Reduction half-reaction:

$\boxed{\ce{Fe^{3+}(aq) + e- -> Fe^{2+}(aq)}}$

Check:

• Atoms: 1 Fe = 1 Fe ✓
• Charge: +3 + (-1) = +2 ✓

Interpretation:

• Fe³⁺ gains 1 electron
• Electrons shown as reactant (gained)
• This is reduction

Verify overall reaction:

To reconstruct original equation from half-reactions:

Oxidation: Sn²⁺ → Sn⁴⁺ + 2e⁻

Reduction: Fe³⁺ + e⁻ → Fe²⁺

Equalize electrons:

• Oxidation produces 2e⁻
• Reduction consumes 1e⁻
• Need to multiply reduction by 2:

Oxidation (×1): Sn²⁺ → Sn⁴⁺ + 2e⁻

Reduction (×2): 2Fe³⁺ + 2e⁻ → 2Fe²⁺

Add half-reactions:

$\ce{Sn^{2+} + 2Fe^{3+} + 2e- -> Sn^{4+} + 2Fe^{2+} + 2e-}$

Cancel electrons:

$\ce{Sn^{2+} + 2Fe^{3+} -> Sn^{4+} + 2Fe^{2+}}$

This matches the original equation! ✓

Summary Table

| Question | Answer | Reasoning | |----------|---------|-----------| | (a) Oxidized | Sn²⁺ | Oxidation state: +2 → +4 (increase) | | (b) Reduced | Fe³⁺ | Oxidation state: +3 → +2 (decrease) | | (c) Oxidizing agent | Fe³⁺ | Gets reduced, accepts electrons | | (d) Reducing agent | Sn²⁺ | Gets oxidized, donates electrons | | (e) Oxidation half | Sn²⁺ → Sn⁴⁺ + 2e⁻ | Shows electron loss | | (e) Reduction half | Fe³⁺ + e⁻ → Fe²⁺ | Shows electron gain |

Electron flow visualization:

Electron transfer:

$\ce{Sn^{2+} ->[\text{loses 2e⁻}] Sn^{4+}}$

$\ce{2Fe^{3+} ->[\text{gain 2e⁻ total}] 2Fe^{2+}}$

Electrons transferred:

• Source: Sn²⁺ (2 electrons released)
• Destination: 2 Fe³⁺ (2 electrons accepted, 1 per Fe³⁺)

Net effect:

• 2 electrons transferred from Sn²⁺ to 2 Fe³⁺
• Sn²⁺ oxidized, Fe³⁺ reduced
• Complementary processes

Memory aids:

OIL RIG:

• Oxidation Is Loss (Sn²⁺ loses e⁻)
• Reduction Is Gain (Fe³⁺ gains e⁻)

Oxidizing vs Reducing agents:

• Oxidizing agent does opposite (gets reduced)
• Reducing agent does opposite (gets oxidized)

Think of it as:

• Oxidizing agent "causes" oxidation by "accepting" electrons
• Reducing agent "causes" reduction by "donating" electrons

Real-world context:

This reaction is an example of:

1. Ion-electron transfer in solution

• Common in analytical chemistry
• Used in redox titrations

2. Tin chemistry

• Sn can exist as Sn²⁺ (stannous) or Sn⁴⁺ (stannic)
• Sn²⁺ is good reducing agent (readily oxidized)

3. Iron chemistry

• Fe commonly exists as Fe²⁺ or Fe³⁺
• Fe³⁺/Fe²⁺ couple used in many biological systems

Applications:

• Analytical determination of Sn²⁺ or Fe³⁺ concentrations
• Understanding corrosion (iron oxidation)
• Biological electron transport chains (similar Fe³⁺/Fe²⁺ reactions)

Practice tip:

To identify redox components:

1. ✓ Assign oxidation states to all species
2. ✓ Find what increases (oxidized) and decreases (reduced)
3. ✓ Remember: oxidizing agent gets reduced, reducing agent gets oxidized
4. ✓ Write half-reactions showing explicit electron transfer
5. ✓ Verify: electrons lost = electrons gained

Solution:

Given unbalanced equation:

$\ce{MnO4-(aq) + C2O4^{2-}(aq) -> Mn^{2+}(aq) + CO2(g)}$

Condition: Acidic solution

Method: Half-reaction method

Step 1: Assign oxidation states

Identify what's oxidized and reduced

MnO₄⁻ → Mn²⁺

MnO₄⁻:

• O: -2 (4 oxygen)
• Mn: ?

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

Mn²⁺:

• Mn: +2

Manganese change: +7 → +2

• Decrease (gains 5 electrons)
• Reduction

C₂O₄²⁻ → CO₂

C₂O₄²⁻ (oxalate ion):

• O: -2 (4 oxygen)
• C: ?

Sum = -2: 2(C) + 4(-2) = -2 2C = +6 C = +3

CO₂:

• O: -2 (2 oxygen)
• C: ?

Sum = 0: C + 2(-2) = 0 C = +4

Carbon change: +3 → +4

• Increase (loses 1 electron per C, 2 total)
• Oxidation

Summary:

• Reduction: MnO₄⁻ → Mn²⁺ (Mn: +7 → +2)
• Oxidation: C₂O₄²⁻ → CO₂ (C: +3 → +4)

Step 2: Write two half-reactions

Reduction half-reaction:

$\ce{MnO4- -> Mn^{2+}}$

Oxidation half-reaction:

$\ce{C2O4^{2-} -> CO2}$

Step 3: Balance reduction half-reaction

$\ce{MnO4- -> Mn^{2+}}$

Balance atoms (except H and O):

Mn: 1 on left, 1 on right ✓

Balance oxygen:

Left: 4 O in MnO₄⁻ Right: 0 O

Add 4 H₂O to right:

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

Now: 4 O on both sides ✓

Balance hydrogen:

In acidic solution: use H⁺

Left: 0 H Right: 8 H (in 4 H₂O)

Add 8 H⁺ to left:

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

Now: 8 H on both sides ✓

Balance charge:

Left: -1 + 8(+1) = +7 Right: +2

Add electrons to more positive side (left):

Need to go from +7 to +2 → add 5e⁻ to left

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

Check charge:

• Left: -1 + 8 - 5 = +2
• Right: +2 ✓

Balanced reduction half-reaction:

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

Step 4: Balance oxidation half-reaction

$\ce{C2O4^{2-} -> CO2}$

Balance atoms (except H and O):

C: 2 on left, 1 on right

Add coefficient 2 to CO₂:

$\ce{C2O4^{2-} -> 2CO2}$

Now: 2 C on both sides ✓

Balance oxygen:

Left: 4 O Right: 4 O (in 2 CO₂) ✓

Already balanced!

Balance hydrogen:

Left: 0 H Right: 0 H ✓

No H to balance!

Balance charge:

Left: -2 Right: 0

Add electrons to more positive side (right):

Need to go from -2 to 0 → add 2e⁻ to right

$\ce{C2O4^{2-} -> 2CO2 + 2e-}$

Check charge:

• Left: -2
• Right: 0 - 2 = -2 ✓

Balanced oxidation half-reaction:

$\boxed{\ce{C2O4^{2-} -> 2CO2 + 2e-}}$

Step 5: Equalize electrons

Reduction: 5e⁻ consumed Oxidation: 2e⁻ produced

Find LCM of 5 and 2: LCM = 10

Multiply reduction by 2:

$\ce{2MnO4- + 16H+ + 10e- -> 2Mn^{2+} + 8H2O}$

Multiply oxidation by 5:

$\ce{5C2O4^{2-} -> 10CO2 + 10e-}$

Now both involve 10 electrons ✓

Step 6: Add half-reactions

Reduction (×2): $\ce{2MnO4- + 16H+ + 10e- -> 2Mn^{2+} + 8H2O}$

Oxidation (×5): $\ce{5C2O4^{2-} -> 10CO2 + 10e-}$

Add them:

$\ce{2MnO4- + 16H+ + 10e- + 5C2O4^{2-} -> 2Mn^{2+} + 8H2O + 10CO2 + 10e-}$

Cancel 10e⁻ from both sides:

$\boxed{\ce{2MnO4- + 16H+ + 5C2O4^{2-} -> 2Mn^{2+} + 8H2O + 10CO2}}$

Step 7: Verify the balanced equation

Check atoms:

Manganese:

• Left: 2 (in 2 MnO₄⁻)
• Right: 2 (in 2 Mn²⁺) ✓

Carbon:

• Left: 10 (in 5 C₂O₄²⁻, each has 2 C)
• Right: 10 (in 10 CO₂) ✓

Oxygen:

• Left: 8 (in 2 MnO₄⁻) + 20 (in 5 C₂O₄²⁻) = 28
• Right: 8 (in 8 H₂O) + 20 (in 10 CO₂) = 28 ✓

Hydrogen:

• Left: 16 (in 16 H⁺)
• Right: 16 (in 8 H₂O) ✓

All atoms balanced! ✓

Check charge:

Left side:

• 2 MnO₄⁻: 2(-1) = -2
• 16 H⁺: 16(+1) = +16
• 5 C₂O₄²⁻: 5(-2) = -10
• Total: -2 + 16 - 10 = +4

Right side:

• 2 Mn²⁺: 2(+2) = +4
• 8 H₂O: 0
• 10 CO₂: 0
• Total: +4

Charges balanced! ✓

Final Answer:

$\boxed{\ce{2MnO4-(aq) + 16H+(aq) + 5C2O4^{2-}(aq) -> 2Mn^{2+}(aq) + 8H2O(l) + 10CO2(g)}}$

Summary of the reaction:

What happens:

• MnO₄⁻ (purple) reduced to Mn²⁺ (pale pink/colorless)
• C₂O₄²⁻ (oxalate) oxidized to CO₂ (gas)

Observations:

• Purple color fades (MnO₄⁻ consumed)
• Bubbles form (CO₂ gas produced)
• Solution becomes colorless or pale pink

Electron transfer:

• Each MnO₄⁻ gains 5e⁻
• Each C₂O₄²⁻ loses 2e⁻
• Need 2 MnO₄⁻ and 5 C₂O₄²⁻ for electron balance

Step-by-step summary:

| Step | Action | Result | |------|--------|--------| | 1 | Assign oxidation states | Mn: +7→+2 (reduced), C: +3→+4 (oxidized) | | 2 | Write half-reactions | MnO₄⁻ → Mn²⁺, C₂O₄²⁻ → CO₂ | | 3 | Balance reduction half | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | | 4 | Balance oxidation half | C₂O₄²⁻ → 2CO₂ + 2e⁻ | | 5 | Equalize electrons | ×2 and ×5 to get 10e⁻ each | | 6 | Add and cancel e⁻ | 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂ | | 7 | Verify | Atoms and charges balanced ✓ |

Applications:

This reaction is used in:

1. Analytical chemistry:

• Redox titration to determine oxalate concentration
• MnO₄⁻ is self-indicating (purple → colorless at endpoint)
• Standard method for analyzing calcium oxalate

2. Quantitative analysis:

• Can determine C₂O₄²⁻ by titrating with standard KMnO₄
• Can determine Ca²⁺ by precipitating as CaC₂O₄, then titrating

3. Teaching tool:

• Classic example of complex redox balancing
• Shows importance of half-reaction method
• Demonstrates color change in redox reactions

Key insights:

Why this method works:

1. Separates complex reaction into manageable parts
2. Balances atoms systematically (metals, O with H₂O, H with H⁺)
3. Balances charge with electrons explicitly
4. Equalizes electron transfer ensures conservation
5. Combines half-reactions gives complete balanced equation

Common student mistakes to avoid:

• ✗ Forgetting to multiply coefficients when equalizing electrons
• ✗ Not checking final atom and charge balance
• ✗ Adding electrons to wrong side
• ✗ Forgetting to balance oxygen before hydrogen

This is a challenging problem that demonstrates:

• Complete half-reaction method
• Multi-step balancing procedure
• Verification importance
• Real-world application in analytical chemistry