8.18. Balance the following redox reactions by ion-electron method. (a) MnO4–(aq) +I–(aq) → MnO2 (s) + I2 (s) (in basic medium) (b) MnO4–(aq) + SO2(g) → Mn2+(aq) +H2SO4–(in acidic solution) (c) H2O2 (aq) + Fe2+(aq) → Fe3+(aq) + H2O(l) (in acidic solution) (d) Cr2O72- (aq) + SO2 (g) → Cr3+ (aq) + SO42-(aq) (in acidic solution)

(a) The balanced half reaction equations are:Oxidation half equation:I− (aq) → I2 (s) - (i)Reduction half reaction equation:MnO4− (aq) → MnO2 (aq) - (ii)Balance I atoms and charges in the oxidation half reaction.2I− (aq) → I2 (s) + 2e−In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.MnO4− (aq) + 3e−→ MnO2 (aq)Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.MnO4− (aq) + 3e− → MnO2 (aq)+4OH−To balance O atoms, add 2 water molecules to reactant side.MnO4− (aq)+ 3e− +2H2O→MnO2 (aq) + 4OH−To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.6I− (aq) → 3I2 (s)+6e−2MnO4− (aq) + 6e−+4H2O → 2MnO2 (aq)+8OH−Add two half-cell reactions to obtain the balanced equation.2MnO4−? (aq) + 6I− (aq) + 4H2? O2? (l) → 2MnO2? (s) + 3I2? (s) + 8OH− (b) The balanced half reaction equations are:Oxidation half equation:SO2 (g) + 2H2O (l) → HS04– (aq) + 3H+ (aq) +2e– … (i)Reduction half equation:MnO4– (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (l) ………. (ii)Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have, 2MnO4– (aq) + 5S02 (g) + 2H20 (l) + H+ (aq) → 2Mn2+ (aq) + 5HSO4– (aq) (c) Oxidation half equation: Fe2+ (aq) → Fe3+ (aq) + e– … (i)Reduction half equation: H2O2 (aq) + 2H+ (aq) + 2e– → 2H2O (l) … (ii)Multiply Eq. (i) by 2 and add it to Eq. (ii), we have, H2O2 (aq) +2Fe2+ (aq) +2H+ (aq) → 2Fe3+ (aq) + 2H2O (l) (d)Oxidation half equation:SO2 (g) + 2H2O (l) → SO42- (aq) + 4H+ (aq) + 2 e– … (i)Reduction half equation:Cr2O72– (aq) + 14H+ (aq) + 6e– → 2Cr3+ (aq) + 7H2O (l) … (ii)Multiply Eq. (i) by 3 and add it to Eq. (ii), we have, Cr2O72– (aq) + 3SO2 (q) + 2H+ (aq) → 2Cr3+ (aq) + 3SO42- (aq) + H2O (l)

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Redox Reactions Ncert Solutions Chemistry Class 11th Chemistry Class 11th Redox Reactions

8.18. Balance the following redox reactions by ion-electron method.
(a) MnO₄^–(aq) +I^–(aq) → MnO₂(s) + I₂ (s) (in basic medium)
(b) MnO₄^–(aq) + SO₂(g) → Mn²⁺(aq) +H₂SO₄^–(in acidic solution)
(c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution)
(d) Cr₂O₇^2-(aq) + SO₂ (g) → Cr³⁺(aq) + SO₄^2-(aq) (in acidic solution)

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Vishal Baghel | Contributor-Level 10

4 months ago

(a) The balanced half reaction equations are:
Oxidation half equation:

I⁻ (aq) → I₂ (s) - (i)

Reduction half reaction equation:

MnO₄⁻ (aq) → MnO₂ (aq) - (ii)

Balance I atoms and charges in the oxidation half reaction.

2I⁻ (aq) → I₂ (s) + 2e⁻

In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.

MnO₄⁻ (aq) + 3e⁻→ MnO₂ (aq)

Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.

MnO4⁻ (aq) + 3e⁻→ MnO₂ (aq)+4OH⁻

To b

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(a) The balanced half reaction equations are: Oxidation half equation:I− (aq) → I2 (s)       - (i)Reduction half reaction equation:MnO4− (aq) → MnO2 (aq) - (ii)Balance I atoms and charges in the oxidation half reaction.2I− (aq) → I2 (s) + 2e−In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.MnO4− (aq) + 3e−→ MnO2 (aq)Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.MnO4− (aq) + 3e− → MnO2 (aq)+4OH−To balance O atoms, add 2 water molecules to reactant side.MnO4− (aq)+ 3e− +2H2O→MnO2 (aq) + 4OH−To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.6I− (aq) → 3I2 (s)+6e−2MnO4− (aq) + 6e−+4H2O → 2MnO2 (aq)+8OH−Add two half-cell reactions to obtain the balanced equation.2MnO4−? (aq) + 6I− (aq) + 4H2? O2? (l) → 2MnO2? (s) + 3I2? (s) + 8OH−  (b) The balanced half reaction equations are: Oxidation half equation:SO2 (g) + 2H2O (l) → HS04–  (aq) + 3H+ (aq) +2e–                        … (i) Reduction half equation:MnO4– (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (l) ………. (ii) Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have, 2MnO4– (aq) + 5S02 (g) + 2H20 (l) + H+ (aq) → 2Mn2+ (aq) + 5HSO4– (aq) (c) Oxidation half equation: Fe2+ (aq) → Fe3+ (aq) + e– … (i) Reduction half equation: H2O2 (aq) + 2H+ (aq) + 2e– → 2H2O (l) … (ii) Multiply Eq. (i) by 2 and add it to Eq. (ii), we have, H2O2 (aq) +2Fe2+ (aq) +2H+ (aq) → 2Fe3+ (aq) + 2H2O (l) (d)Oxidation half equation:SO2 (g) + 2H2O (l) → SO42- (aq) + 4H+ (aq) + 2 e–  … (i) Reduction half equation:Cr2O72– (aq) + 14H+ (aq) + 6e– → 2Cr3+ (aq) + 7H2O (l) … (ii) Multiply Eq. (i) by 3 and add it to Eq. (ii), we have, Cr2O72– (aq) + 3SO2 (q) + 2H+ (aq) → 2Cr3+ (aq) + 3SO42- (aq) + H2O (l)

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