Q.Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Concept and intuition

In any redox reaction, the total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent. Here, MnO₄⁻ is reduced to Mn²⁺ (Mn changes oxidation state from +7 to +2, a gain of 5 electrons), and SO₂ is oxidized to HSO₄⁻ (S changes from +4 to +6, a loss of 2 electrons). To make electrons equal, we need the smallest whole‑number ratio that balances the electron transfer: 2 MnO₄⁻ (gaining 10 e⁻) for every 5 SO₂ (losing 10 e⁻). That gives X:Y = 2:5.

Step‑by‑step reasoning

1. Assign oxidation numbers

   • In MnO₄⁻: O is –2 (×4 = –8), overall charge –1 ⇒ Mn = +7.
   • In Mn²⁺: Mn = +2.
   • In SO₂: O is –2 (×2 = –4), molecule neutral ⇒ S = +4.
   • In HSO₄⁻: O is –2 (×4 = –8), H is +1, overall charge –1 ⇒ S = +6.

2. Write the half‑reactions

   • Reduction: MnO₄⁻ → Mn²⁺ Change: Mn from +7 to +2 ⇒ gain of 5 e⁻. In acidic medium: balance O with H₂O, then H⁺:

• Oxidation: SO₂ → HSO₄⁻ Change: S from +4 to +6 ⇒ loss of 2 e⁻. Balance O with H₂O, then H⁺:

3. Equalize electrons transferred
   • Reduction half‑reaction involves 5 e⁻; oxidation involves 2 e⁻.
   • LCM of 5 and 2 is 10. Multiply reduction by 2, oxidation by 5:

4. Add the half‑reactions and simplify
   • Combine: $$2\mathrm{MnO_4^- + 16H^+ + 5SO_2 + 10H_2O \rightarrow 2Mn^{2+} + 8H_2O + 5HSO_4^- + 15H^+} …$$

We start by recognizing that ferrous oxalate is a mixed reducing agent: it contains both Fe²⁺ (which oxidises to Fe³⁺) and oxalate ion C₂O₄²⁻ (which oxidises to CO₂). In acidic medium, K₂Cr₂O₇ is a strong oxidising agent, with Cr⁶⁺ being reduced to Cr³⁺. The stoichiometry is found by equating the total number of electrons transferred.

1. Write the half-reactions for the reducing agent (ferrous oxalate)

   • Fe²⁺ → Fe³⁺ + e⁻ (1 electron lost per Fe²⁺)
   • C₂O₄²⁻ → 2 CO₂ + 2 e⁻ (2 electrons lost per oxalate ion) Since one mole of ferrous oxalate (FeC₂O₄) contains 1 mole of Fe²⁺ and 1 mole of C₂O₄²⁻, the total electrons lost by 1 mole of ferrous oxalate = $1 + 2 = 3$ moles of electrons.

2. Write the half-reaction for the oxidising agent (dichromate in acid)

This shows that 1 mole of K₂Cr₂O₇ accepts 6 moles of electrons.

3. Set up the electron balance Let $x$ be the moles of K₂Cr₂O₇ required to oxidise 1 mole of ferrous oxalate.
   • Electrons gained by $x$ moles of dichromate = $x \times 6$
   • Electrons lost by 1 mole of ferrous oxalate = $3$ For a complete redox reaction: $6x = 3 \quad \Rightarrow \quad x = \frac{3}{6} = 0.5$ …

Step 1 — Oxidation half-reaction

$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$: carbon goes from +3 to +4 (two carbons), so each oxalate ion loses 2 electrons.

Step 2 — Reduction half-reaction

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$: manganese goes from +7 to +2, so each permanganate ion gains 5 electrons.

Step 3 — Equalise electrons

To balance electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2, so that $5 \times 2 = 10$ electrons lost equal $2 \times 5 = 10$ electrons gained:

• $5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$
• $2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$

Step 4 — Combine and read off coefficients …

We have two separate redox equations to balance. The first involves permanganate with sulfite in acidic medium; the second involves thiosulfate with permanganate in basic medium. The letters (a),

(b),

(c),

(d) are placeholders for the coefficients we must find. The trick is that the same letter appears in both equations, so the coefficients must be consistent across both.

Concept & Intuition

Redox balancing works by splitting the reaction into oxidation and reduction halves. In acidic solution we add H⁺ and H₂O; in basic solution we add OH⁻ and H₂O. Once each half is balanced for atoms and charge, we multiply them so electrons cancel. The final coefficients give us the values of (a),

(b),

(c),

(d).

Step‑by‑step balancing

Equation (i):

1. Identify half‑reactions

   • Reduction: $\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+}$ (Mn goes from +7 to +2, gain of 5 e⁻)
   • Oxidation: $\mathrm{SO}_3^{2-} \to \mathrm{SO}_4^{2-}$ (S goes from +4 to +6, loss of 2 e⁻)

2. Balance the reduction half (acidic)

(Check: atoms and charge balanced.)

3. Balance the oxidation half (acidic)

(Check: atoms and charge balanced.)

4. Equalize electrons Reduction gives 5 e⁻, oxidation gives 2 e⁻. LCM = 10. Multiply reduction by 2:

Multiply oxidation by 5:

5. Add and cancel Add the two half‑reactions:

Cancel 10 e⁻, cancel 10 H⁺ from both sides, and cancel 5 H₂O from left with 5 of the 8 H₂O on right:

So from equation (i): $x = 2$, $(a) = 5$, $(b) = 6$, $(c) = 3$.

Equation (ii):

We already know from (i): $b = 6$, $c = 3$. So we need to find $d$ and check consistency.

6. Identify half‑reactions

   • Reduction: $\mathrm{MnO}_4^- \to \mathrm{MnO}_2$ (Mn from +7 to +4, gain of 3 e⁻)
   • Oxidation: $\mathrm{S}_2\mathrm{O}_3^{2-} \to \mathrm{SO}_4^{2-}$ (S in thiosulfate: average oxidation state +2; in sulfate +6; each S loses 4 e⁻, but there are 2 S atoms per thiosulfate, so total loss = 8 e⁻ per $\mathrm{S}_2\mathrm{O}_3^{2-}$)

7. Balance the reduction half (basic)

   In basic medium:

(Check: atoms and charge balanced.)

8. Balance the oxidation half (basic) Start with $\mathrm{S}_2\mathrm{O}_3^{2-} \to 2\mathrm{SO}_4^{2-}$. Balance S: already 2 on each side. Balance O: left 3 O, right 8 O → add 5 H₂O to left:

But we are in basic medium, so add 10 OH⁻ to both sides to neutralise H⁺:

Cancel 5 H₂O from both sides:

Concept and Intuition

In any redox reaction, the number of electrons transferred is directly tied to the change in oxidation state of the species being oxidised or reduced. Here, the question asks for the number of moles of electrons involved in producing 3.0 moles of the oxidised product. That means we focus on the oxidation half-reaction: sulphite ion ($\mathrm{SO}_3^{2-}$) is oxidised to sulphate ion ($\mathrm{SO}_4^{2-}$). The dichromate ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$) is the oxidising agent, but we don’t need its full balancing — we only need the electron count per mole of product.

Step-by-step reasoning

1. Identify the oxidation half-reaction Sulphite ion ($\mathrm{SO}_3^{2-}$) is oxidised to sulphate ion ($\mathrm{SO}_4^{2-}$). In acidic medium, the half-reaction is:

This shows that each mole of sulphite that becomes sulphate releases 2 moles of electrons.

2. Confirm the oxidation state change

   In $\mathrm{SO}_3^{2-}$, sulfur has oxidation state +4 (since oxygen is -2, total -6, so S = +4).

   In $\mathrm{SO}_4^{2-}$, sulfur has oxidation state +6.

   The change is from +4 to +6, which is a loss of 2 electrons per sulfur atom. This matches the half-reaction.

3. Relate to the question

   The “oxidised product” is sulphate ($\mathrm{SO}_4^{2-}$). We are producing 3.0 moles of it.

   Since each mole of sulphate formed comes from one mole of sulphite and releases 2 moles of electrons, the total moles of electrons is:

   $$$$

Oxidation states: In $\mathrm{ClO_3^-}$, Cl is $+5$; it is reduced to $0$ in $\mathrm{Cl_2}$ — gains $5\,e^-$. Each $\mathrm{Cl^-}$ ($-1$) is oxidised to $0$ — loses $1\,e^-$.

Electron balance: to supply the 5 electrons gained by one chlorate, we need 5 $\mathrm{Cl^-}$, so $w=5$.

Atom balance:

• Cl: LHS $=1+5=6$; RHS $=2z \Rightarrow z=3$. …

Concept: Redox stoichiometry - H2O2 acts as a REDUCING agent towards acidified KMnO4, decolourising the purple MnO4- to colourless Mn2+.

Balanced equation (acidic medium):

2 MnO4- + 5 H2O2 + 6 H+ -> 2 Mn2+ + 5 O2 + 8 H2O

Equivalently by n-factors: MnO4- (Mn +7 -> +2) gains 5 electrons; H2O2 (O -1 -> 0) loses 2 electrons. For electron balance: …