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Write Brief Answer · Q45

Q.Balance the following equations by the ion-electron method.

i) KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
ii) C₂O₄²⁻ + Cr₂O₇²⁻ → Cr³⁺ + CO₂ (in acid medium)
iii) Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI
iv) Zn + NO₃⁻ → Zn²⁺ + NO (in acid medium)
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Step 1 (i). Reduction: MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O. Oxidation: Sn²⁺ → Sn⁴⁺ + 2e⁻. LCM(5,2)=10: 2×(reduction) + 5×(oxidation) gives 2MnO4⁻ + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 8H2O + 5Sn⁴⁺; restoring K⁺ and Cl⁻ spectators: 2KMnO4 + 5SnCl2 + 16HCl → 2MnCl2 + 5SnCl4 + 8H2O + 2KCl.

Step 2 (ii). Oxidation: C2O4²⁻ → 2CO2 + 2e⁻. Reduction: Cr2O7²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O. LCM(2,6)=6: 3×(oxidation) + 1×(reduction) gives 3C2O4²⁻ + Cr2O7²⁻ + 14H⁺ → 6CO2 + 2Cr³⁺ + 7H2O (charge check: left = 3(−2)+(−2)+14(+1) = +6; right = 6(0)+2(+3) = +6, balanced).

Step 3 (iii). Oxidation: 2S2O3²⁻ → S4O6²⁻ + 2e⁻ (average S oxidation state rises from +2 to +2.5). Reduction: I2 + 2e⁻ → 2I⁻. Electrons already equal (2=2), so simply add: 2S2O3²⁻ + I2 → S4O6²⁻ + 2I⁻; restoring Na⁺: 2Na2S2O3 + I2 → Na2S4O6 + 2NaI. …

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