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Balance the Following Equations in the Basic Medium by Ion-electron Method and Oxidation Number Methods and Identify the Oxidising Agent and the Reducing Agent. N2h4(L) + Clo-3(Aq) → No(G) + Cl–(G)

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प्रश्न

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

\[\ce{N2H4(l) + ClO^-_3 (aq) → NO(g) + Cl–(g)}\]

सविस्तर उत्तर
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उत्तर

The oxidation number of N increases from – 2 in N2H4 to + 2 in NO and the oxidation number of Cl decreases from + 5 in `"ClO"_3^-` to - 1 in Cl. Hence, in this reaction, N2H4 is the reducing agent and `"ClO"_3^-` is the oxidizing agent.

Ion–electron method:

The oxidation half equation is:

\[\ce{^{-2}N2H_{4(l)} -> ^{+2}NO_{(g)}}\]

The N atoms are balanced as:

\[\ce{N_2H_{4(l)} -> 2NO_{(g)}}\]

The oxidation number is balanced by adding 8 electrons as:

\[\ce{N_2H_{4(l)} -> 2NO_{(g)} + 8e-}\]

The charge is balanced by adding 8 OHions as:

\[\ce{N_2H_{4(l)} -> 2NO_{(g)} + 8e-}\]

The O atoms are balanced by adding 6H2O as:

\[\ce{N_2H_{4(l)} + 8OH-_{(aq)} -> 2NO_{(g)} + 6H_2O_{(l)} + 8e-}\]  .....(i)

The reduction half equation is:

\[\ce{^{+5}ClO^-_{3(aq)} -> ^{-1}Cl^-_{(aq)}}\]

The oxidation number is balanced by adding 6 electrons as:

\[\ce{ClO^-_3_{(aq)} + 6e- -> Cl-_{(aq)}}\]

The charge is balanced by adding 6OH– ions as:

\[\ce{ClO^-_3_{(aq)} + 6e-  -> Cl-_{(aq)} + 6OH-_{(aq)}}\]

The O atoms are balanced by adding 3H2O as:

\[\ce{ClO^-_3_{(aq)} + 3H_2O_{(l)} + 6e- -> Cl-_{(aq)} + 6OH-_{(aq)}}\] .....(ii)

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> 6NO_{(g)} + 4Cl-_{(aq)} + 6H_2O_{(l)}}\]

Oxidation number method:

Total decrease in oxidation number of N = 2 × 4 = 8

Total increase in oxidation number of Cl = 1 × 6 = 6

On multiplying N2H4 with 3 and `"ClO"_3^-` with 4 to balance the increase and decrease in O.N., we get:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> NO_{(g)} + Cl-_{(aq)}}\]

The N and Cl atoms are balanced as:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> 6NO_{(g)} + 4l-_{(aq)}}\]

The O atoms are balanced by adding 6H2O as:

\[\ce{3N_2H_{4(l)} + 4ClO^-_3_{(aq)} -> 6NO_{(g)} + 4Cl-_{(aq)} + 6H_2O_{(l)}}\]

This is the required balanced equation.

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पाठ 7: Redox Reactions - EXERCISES [पृष्ठ २८२]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 11
पाठ 7 Redox Reactions
EXERCISES | Q 8.19 - (b) | पृष्ठ २८२

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  1. \[\ce{MnO-_4 (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium)}\]
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  3. \[\ce{H2O2 (aq) + Fe^{2+} (aq) → Fe^{3+} (aq) + H2O (l) (in acidic solution)}\]
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\[\ce{Cl_2O_{7(g)} + H_2O_{2(aq)} -> ClO-_{2(aq)} + O_{2(g)} + H+_{(aq)}}\]


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\[\ce{Sn^{2⊕} + 2Fe^{3⊕}->Sn^{4⊕} + 2Fe^{2⊕}}\]


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\[\ce{Cr2O^2-_{7(aq)} + SO^2-_{3(aq)}->Cr^3+_{ (aq)} + SO^2-_{4(aq)}(acidic)}\]


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\[\ce{2Zn_{(s)} + O2_{(g)} -> 2ZnO_{(s)}}\]


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\[\ce{I2 + NO^{-}3 -> NO2 + IO^{-}3}\]


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\[\ce{MnO2 + C2O^{2-}4 -> Mn^{2+} + CO2}\]


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\[\ce{3HCl (aq) + HNO3 (aq) -> Cl2 (g) + NOCl (g) + 2H2O (l)}\]


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\[\ce{PCl3 (l) + 3H2O (l) -> 3HCl (aq) + H3PO3 (aq)}\]


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\[\ce{4NH3 (g) + 3O2 (g) -> 2N2 (g) + 6H2O (g)}\]


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