Balance the following equations by the oxidation number method

(i) Fe²⁺ + H⁺ + Cr₂^2–O₇ → Cr₃⁺ + Fe₃⁺ + H₂O

(ii) I₂ + NO₃^–  → NO₂ + IO₃ ^–

(iii) I ₂ + S₂O₃^2–  → I^– + S₄O₆^2–

^{(iv) MnO₂ + C₂ O₄2– → Mn₂+ + CO₂}

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

(i) We can balance the given equation by oxidation number method- 

(a)Balance the increase and decrease in O.N.

  (b) Balancing H and O atoms by adding H⁺ and H₂O molecules

 

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

Total decrease in O.N. = 1

To equilize O.N. multiply NO₃^-, by 10

I₂ + 10 NO₃^-? 10NO₂ + IO₃^-

Balancing atoms other than O and H

I₂ + 10 NO₃^-? 10NO₂ + 2 IO₃^-

Balancing O and H

I₂ + 10 NO₃^- + 8H⁺? 10NO₂ + 2 IO₃^- + 4H₂O

 

(iii) We can balance the given equation by oxidation number method

Total increase in O.N. =

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

(i) We can balance the given equation by oxidation number method- 

(a)Balance the increase and decrease in O.N.

  (b) Balancing H and O atoms by adding H⁺ and H₂O molecules

 

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

Total decrease in O.N. = 1

To equilize O.N. multiply NO₃^-, by 10

I₂ + 10 NO₃^-? 10NO₂ + IO₃^-

Balancing atoms other than O and H

I₂ + 10 NO₃^-? 10NO₂ + 2 IO₃^-

Balancing O and H

I₂ + 10 NO₃^- + 8H⁺? 10NO₂ + 2 IO₃^- + 4H₂O

 

(iii) We can balance the given equation by oxidation number method

Total increase in O.N. = 0.5 × 4 = 2

Total increase in O.N. = 1 × 2 = 2

to equilize O.N. multiply S₂O₃^2– and I^- by 2.

 

(iv) We can balance the given equation by oxidation number method-

The balanced chemical reaction is given as-

Total increase in O.N. = 5 × 2= 10

To equalize O.N. multiply CO₂ by 2. 

MnO₂   + C₂O₄^2-   ?  Mn₂⁺   + 2CO₂

Balance H and O by adding 2H₂O   on right side, and 4H+  on left side of equation.

MnO₂   + C₂O₄^2-   + 4H⁺   ?  Mn₂⁺   + 2CO₂   + 2H₂O .

 

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<p>This is a Short answer type question as classified in NCERT Exemplar</p><p><strong> (i)&nbsp;</strong>We can balance the given equation by oxidation number method-&nbsp;</p><div><div><div><div><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743578017phpOK42Pc_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577742phpv3QuOA.jpeg" alt="" width="299" height="46"></picture></div></div><p> (a)Balance the increase and decrease in O.N.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743577777php2xlYy3_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577777php2xlYy3.jpeg" alt="" width="333" height="44"></picture></div></div><p>&nbsp; (b) Balancing H and O atoms by adding H⁺ and H₂O molecules</p><p>&nbsp;</p><p> (ii) We can balance the given equation by oxidation number method-&nbsp;</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743578017phpOK42Pc_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577890php6DoGKq.jpeg" alt="" width="255" height="165"></picture></div><div><p>Total decrease in O.N. = 1</p><p>To equilize O.N. multiply NO₃^-, by 10</p><p>I₂ + 10 NO₃^-? 10NO₂ + IO₃^-</p><p>Balancing atoms other than O and H</p><p>I₂ + 10 NO₃^-? 10NO₂ + 2 IO₃^-</p><p>Balancing O and H</p><p>I₂ + 10 NO₃^- + 8H⁺? 10NO₂ + 2 IO₃^- + 4H₂O</p><p>&nbsp;</p><p><strong> (iii) </strong>We can balance the given equation by oxidation number method</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743577955php8b1T0j_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577955php8b1T0j.jpeg" alt="" width="251" height="55"></picture></div></div><p>Total increase in O.N. = 0.5 × 4 = 2</p><p>Total increase in O.N. = 1 × 2 = 2</p><p>to equilize O.N. multiply <strong>S₂O₃^2–</strong> and I^- by 2.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743577974phpNtoZml_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743577974phpNtoZml.jpeg" alt="" width="244" height="41"></picture></div><div>&nbsp;</div><div><p><strong> (iv) </strong>We can balance the given equation by oxidation number method-</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743578017phpOK42Pc_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743578017phpOK42Pc.jpeg" alt="" width="284" height="45"></picture></div></div><p>The balanced chemical reaction is given as-</p><p>Total increase in O.N. = 5&nbsp;×&nbsp;2= 10</p><p>To equalize O.N. multiply CO₂&nbsp;by 2.&nbsp;</p><p>MnO₂&nbsp;&nbsp;&nbsp;+&nbsp;C₂O₄^2-&nbsp;&nbsp;&nbsp;? &nbsp;Mn₂⁺&nbsp;&nbsp;&nbsp;+ 2CO₂</p><p>Balance H and O by adding 2H₂O&nbsp;&nbsp;&nbsp;on right side, and 4H+&nbsp;&nbsp;on left side of equation.</p><p>MnO₂&nbsp;&nbsp;&nbsp;+&nbsp;C₂O₄^2-&nbsp;&nbsp;&nbsp;+ 4H⁺&nbsp;&nbsp;&nbsp;? &nbsp;Mn₂^+&nbsp;&nbsp;&nbsp;+ 2CO₂&nbsp;&nbsp;&nbsp;+ 2H₂O&nbsp;.</p></div></div></div></div></div></div><div><div>&nbsp;</div></div></div></div>

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