Write balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO₄ ^– ) reacts with sulphur dioxide gas in acidic medium to produce Mn₂⁺ and hydrogensulphate ion. (Balance by ion electron method).

(ii) Reaction of liquid hydrazine (N₂H₄ ) with chlorate ion (ClO₃^– ) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)

(iii) Dichlorine heptaoxide (Cl₂O₇ ) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO₂^– ) and oxygen gas. (Balance by ion electron method)

This is a Short answer type question as classified in NCERT Exemplar

(i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺?5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

(ii) We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given

3N₂H₄ + 4ClO₃^- → 6NO + 4Cl^- + 6H₂O

The balanced chemical is given as-

(iii) We can balance the given reaction by ion electron method as

Cl₂O₇(g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

Balancing bu ion electron method

2 × { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  × { H₂O₂ → O₂ + 2H⁺ + 2e

...more

This is a Short answer type question as classified in NCERT Exemplar

(i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺?5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

(ii) We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given

3N₂H₄ + 4ClO₃^- → 6NO + 4Cl^- + 6H₂O

The balanced chemical is given as-

(iii) We can balance the given reaction by ion electron method as

Cl₂O₇(g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

Balancing bu ion electron method

2 × { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  × { H₂O₂ → O₂ + 2H⁺ + 2e^- 

The balanced chemical is given as-

2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O

                         8H₂O₂ → 8O₂ + 16H⁺ + 16e^-

          2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺

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<p>This is a Short answer type question as classified in NCERT Exemplar</p><p><strong>(i) </strong>2Mn0^–₄&nbsp;+ 5S0₂&nbsp;+ 2H₂0 + H⁺?5HS0^–₄&nbsp;+ 2Mn^2+<br>Balancing by ion-electron method:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743577009phpRxjJFK_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577009phpRxjJFK.jpeg" alt="" width="387" height="333"></picture></div><div><p><strong>(ii) </strong>We can balance the given reaction by oxidation number method-</p><p>Balancing by oxidation number method as to make the electron gain and loss equal as given</p><p>3N₂H₄ + 4ClO₃^-→ 6NO + 4Cl^-+ 6H₂O</p><p>The balanced chemical is given as-</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743577085php8E1qxR_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577085php8E1qxR.jpeg" alt="" width="275" height="162"></picture></div><div><p><strong>(iii) </strong>We can balance the given reaction by ion electron method as</p><p>Cl₂O₇(g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)</p><p>Balancing bu ion electron method</p><p>2 × { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O</p><p>8^&nbsp;× { H₂O₂ → O₂ + 2H⁺ + 2e^-&nbsp;</p><p>The balanced chemical is given as-</p><p>2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;8H₂O₂ → 8O₂ + 16H⁺ + 16e^-</p><p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺</p></div></div></div></div>

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