Q.Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Balancing Redox Equations in Basic Medium

Concept: In redox reactions, oxidation (loss of electrons) and reduction (gain of electrons) occur simultaneously. Balancing requires equalising electron transfer and accounting for the medium (basic: add OH⁻ and H₂O).

(a) P₄(s) + OH⁻(aq) → PH₃(g) + H₂PO₂⁻(aq)

Oxidation number method:

P in P₄: 0 → P in PH₃: –3 (reduction, 3 e⁻ per P); P in H₂PO₂⁻ (hypophosphite): +1

(oxidation, 1 e⁻ per P). This is a disproportionation — for every P reduced, three

are oxidised, so one P₄ gives 1 PH₃ and 3 H₂PO₂⁻.

Balance O and H in basic medium with 3 OH⁻ + 3 H₂O on the left (6 O and 9 H, exactly what

the right side needs); charge is –3 on both sides:

(The exercise prints the product as "HPO₂⁻"; the species actually formed — used in the chapter's own eq. 7.46 — is hypophosphite, H₂PO₂⁻, and only that formula balances.)

Oxidising agent: P₄ (reduced to PH₃)

Reducing agent: P₄ (oxidised to H₂PO₂⁻)

(b) N₂H₄(l) + ClO₃⁻(aq) → NO(g) + Cl⁻(aq)

Ion-electron method (basic medium):

Oxidation half: N (–2 in N₂H₄) → N (+2 in NO)

Add 8OH⁻ to right, 4H₂O to left:

Reduction half: Cl (+5) → Cl (–1), gain 6e⁻

Add 3H₂O to left, 6OH⁻ to right:

Multiply oxidation by 3, reduction by 4 (LCM of 8 and 6 = 24):

Oxidising agent: ClO₃⁻ …

Redox reactions in solution require careful bookkeeping: atoms must balance, charge must balance, and the electrons lost in oxidation must equal those gained in reduction. The ion-electron method splits the reaction into half-reactions (oxidation and reduction), balances each in the prevailing medium (acidic or basic), then recombines them. The oxidation-number method tracks changes in oxidation states directly, assigns coefficients to equalise electron transfer, then balances the rest by inspection. Both roads lead to the same balanced equation.

In basic medium we add OH⁻ and H₂O as needed; in acidic medium we use H⁺ and H₂O. The oxidising agent is the species reduced (it accepts electrons), and the reducing agent is the species oxidised (it donates electrons).

(a) $\ce{P4(s) + OH^-(aq) -> PH3(g) + H2PO2^-(aq)}$ in basic medium

This is a disproportionation: phosphorus in P₄ (oxidation state 0) is simultaneously

reduced to $-3$ in PH₃ and oxidised to $+1$ in the hypophosphite ion, $\ce{H2PO2^-}$.

Oxidation-number method

1. Assign oxidation states. P in P₄: $0$; in PH₃: $-3$; in $\ce{H2PO2^-}$: $+1$ (from $2(+1) + x + 2(-2) = -1$).
2. Balance the electron transfer. Each P reduced ($0 \to -3$) gains 3 electrons; each P oxidised ($0 \to +1$) loses 1. So for every 1 P reduced, 3 P are oxidised — the four atoms of one P₄ split as 1 PH₃ : 3 $\ce{H2PO2^-}$:

3. Balance O and H (basic medium). The right side carries 6 O and 9 H; adding 3 OH⁻ and 3 H₂O on the left supplies exactly 6 O and 9 H:

4. Check the charge. Left: $3(-1) = -3$. Right: $3(-1) = -3$. Balanced.

Ion-electron method

• Reduction half: $\ce{P4 + 12H2O + 12e^- -> 4PH3 + 12OH^-}$ (each of 4 P gains 3 e⁻; charge $-12$ on both sides).
• Oxidation half: $\ce{P4 + 8OH^- -> 4H2PO2^- + 4e^-}$ (each of 4 P loses 1 e⁻; O: 8 = 8, H: 8 = 8, charge $-8$ on both sides).
• Equalise electrons (multiply the oxidation half by 3), add, and cancel $12e^-$, $12\,\ce{H2O}$ and $12\,\ce{OH^-}$:

Divide by 4:

Oxidising agent: P₄ (the portion reduced to PH₃) ·

Reducing agent: P₄ (the portion oxidised to $\ce{H2PO2^-}$) —

in a disproportionation the same species is both.

(b) $\ce{N2H4(l) + ClO3^-(aq) -> NO(g) + Cl^-(g)}$ in basic medium

Nitrogen in N₂H₄ (−2) is oxidised to +2 in NO; chlorine in ClO₃⁻ (+5) is reduced to −1 in Cl⁻.

Ion-electron method

1. Half-reactions.

   • Oxidation: $\ce{N2H4 -> NO}$
   • Reduction: $\ce{ClO3^- -> Cl^-}$

2. Balance atoms other than O and H.

   • Oxidation: $\ce{N2H4 -> 2 NO}$
   • Reduction: already balanced for Cl.

3. Balance O with H₂O, then H with OH⁻.

   • Oxidation: Right has 2 O, add 2 H₂O on left. Left now has 4 H (from N₂H₄) + 4 H (from 2 H₂O) = 8 H; right has 0 H. Add 8 OH⁻ on right:

 In basic medium replace 8 H⁺ by adding 8 OH⁻ to both sides:  

 Simplify:  

• Reduction: left has 3 O, right none — add 3 H₂O on the right; that leaves 6 H on the right, so (basic medium) add 3 H₂O on the left and 6 OH⁻ on the right, which simplifies to:

4. Balance charge with electrons.
   • Oxidation: Left has −8, right has 0. Add 8 e⁻ on right:

• Reduction: Left has −1, right has −7. Add 6 e⁻ on left:

5. Equalise electrons (LCM = 24). Multiply oxidation by 3, reduction by 4:

Add and cancel 24 e⁻, 24 OH⁻, 12 H₂O:

Oxidation-number method

1. Oxidation states.

   N in N₂H₄: −2; in NO: +2 (change +4 per N, so +8 per N₂H₄).

   Cl in ClO₃⁻: +5; in Cl⁻: −1 (change −6).

2. Balance electron transfer.

   LCM(8, 6) = 24. Need 3 N₂H₄ (3×8 = 24 e⁻ lost) and 4 ClO₃⁻ (4×6 = 24 e⁻ gained):

3. Balance O and H. Right has 6 O (in NO), left has 12 O (in ClO₃⁻). Excess 6 O on left → add 6 H₂O on right. Left has 12 H (in 3 N₂H₄), right has 12 H (in 6 H₂O). Balanced.

Oxidising agent: ClO₃⁻ (reduced to Cl⁻)

Reducing agent: N₂H₄ (oxidised to NO)

(c) $\ce{Cl2O7(g) + H2O2(aq) -> ClO2^-(aq) + O2(g) + H^+}$ in acidic medium (then convert to basic)

Chlorine in Cl₂O₇ (+7) is reduced to +3 in ClO₂⁻; oxygen in H₂O₂ (−1) is oxidised to 0 in O₂.

Ion-electron method

1. Half-reactions.

   • Reduction: $\ce{Cl2O7 -> ClO2^-}$
   • Oxidation: $\ce{H2O2 -> O2}$

2. Balance atoms other than O and H.

   • Reduction: $\ce{Cl2O7 -> 2 ClO2^-}$
   • Oxidation: already balanced for O. …