मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

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) - Chemistry

Advertisements
Advertisements

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
Advertisements

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.

shaalaa.com
Balancing Redox Reactions in Terms of Loss and Gain of Electrons
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

संबंधित प्रश्‍न

The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?


Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.


How do you count for the following observations?

Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.


Balance the following redox reactions by ion-electron method:

  1. \[\ce{MnO-_4 (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium)}\]
  2. \[\ce{MnO-_4 (aq) + SO2 (g) → Mn^{2+} (aq) + HSO-_4  (aq) (in acidic solution)}\]
  3. \[\ce{H2O2 (aq) + Fe^{2+} (aq) → Fe^{3+} (aq) + H2O (l) (in acidic solution)}\]
  4. \[\ce{Cr_2O^{2-}_7 + SO2(g) → Cr^{3+} (aq) + SO^{2-}_4 (aq) (in acidic solution)}\]

The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.


Choose the correct option.

For the following redox reactions, find the correct statement.

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


Balance the following reaction by oxidation number method.

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


Balance the following reaction by oxidation number method.

\[\ce{Bi(OH)_{3(s)} + Sn(OH)^-_{3(aq)}->Bi_{(s)}  + Sn(OH)^2-_{6(aq)}(basic)}\]


Balance the following redox equation by half-reaction method.

\[\ce{H2C2O_{4(aq)} + MnO^-_{4(aq)}->CO2_{(g)} + Mn^2+_{( aq)}(acidic)}\]


Balance the following redox equation by half-reaction method.

\[\ce{Bi(OH)_{3(s)} + SnO^2-_{2(aq)}->SnO^2-_{3(aq)} + Bi^_{(s)}(basic)}\]


Which of the following is a redox reaction?


Balance the following equations by the oxidation number method.

\[\ce{Fe^{2+} + H^{+} + Cr2O^{2-}7 -> Cr^{3+} + Fe^{3+} + H2O}\]


Balance the following equations by the oxidation number method.

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


Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

\[\ce{HgCl2 (aq) + 2KI (aq) -> HgI2 (s) + 2KCl (aq)}\]


Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

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


Balance the following ionic equations.

\[\ce{Cr2O^{2-}7 + H^{+} + I- -> Cr^{3+} + I2 + H2O}\]


Balance the following ionic equations.

\[\ce{Cr2O^{2-}7 + Fe^{2+} + H+ -> Cr^{3+} + Fe^{3+} + H2O}\]


In acidic medium, reaction, \[\ce{MNO^-_4 → Mn^2+}\] an example of ____________.   


In \[\ce{Cu^{2+} + Ag -> Cu + Ag^+}\], oxidation half-reaction is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×