SELECT * FROM question_mgmt as q WHERE id=3239 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=49 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

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

(a) P4(s) + OH – (aq)  →  PH3(g) + HPO2 – (aq)

(b) N2H4(l) + ClO3 – (aq)  →  NO(g)  +  Cl–(g)

(c) Cl2O7 (g)  +  H2O2(aq)  →  ClO – 2(aq)   +  O2(g)  +  H + (aq)

Answer:

(a) The O.N. (oxidation number) of P decreases from 0 in P4 to -3 in PH3 and increases from 0 in P4 to + 2 in HPO-2. Hence, P4 acts both as an oxidizing agent as well as a reducing agent in this reaction.

Ion-electron method:

The oxidation half equation is:

P4(s) →  H2PO-(aq)

The P atom is balanced as:

P0 4(s)  →  4H2P+1O-(aq)

The O.N. is balanced by adding 4 electrons as:

P4(s)    →   4H2PO-(aq)    +   4e-

The charge is balanced by adding 8OH- as:

P4(s)    +  8OH - (aq)  →   4H2PO-2(aq)

The O and H atoms are already balanced. The reduction half equation is:

P4(s)    →    PH3(g)

The P atom is balanced as

P04(s)    →   4 P-3H3(g)

The O.N. is balanced by adding 12 electrons as:

P4(s)  +  12e-   →    4 PH3(g)

The charge is balanced by adding 12OH- as:

P4(s)  +  12e-   →    4 PH3(g)   +  12OH-(aq)  .....(i)

The O and H atoms are balanced by adding 12H2O as:

P4(s) + 12H2O(l)  +  12e-   →    4 PH3(g)   +  12OH-(aq)  -- (ii)

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

P4(s) + 3OH-(aq) + 3H2O  →  PH3   +   3H2PO-2(aq)

 

(b)

The oxidation number of N increases from -2 in N2H4 to +2 in NO and the oxidation number of Cl decreases from + 5 in CIO-3 to -1 in Cl-. Hence, in this reaction, N2H4 is the reducing agent and CIO-3 is the oxidizing agent. Ion-electron method:

The oxidation half equation is:

N-22 H4(l)  →  N+2 O(g)

The N atoms are balanced as:

N2H4(l)   →   2NO(g)

The oxidation number is balanced by adding 8 electrons as:

N2H4(l)   →    2NO(g)  +  8e-

The charge is balanced by adding 8 OH-ions as:

N2H4(l)  + 8OH-(aq)  →  2NO(g)  +  8e-

The O atoms are balanced by adding 6H2O as:

N2H4(l)  + 8OH-(aq)  → 2NO(g)  + 6H2O(l) +  8e- .... (i)

The reduction half equation is:

C+5IO-3(aq)  →  C-1l-(aq)

The oxidation number is balanced by adding 6 electrons as:

CIO-3(aq)  +  6e-  → Cl-(aq)

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

CIO-3(aq)  +  6e-  →    Cl-(aq) + 6OH-(aq) 

The O atoms are balanced by adding 3H2O as:

CIO-3(aq)  + 3H2O(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:

3N2H4(l)  +  4CIO-3(aq)   →  6NO(g)  +  4Cl-(aq) + 6H2O(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 CIO-3 with 4 to balance the increase and decrease in O.N., we get:

3N2H4(l)  + 4CIO-3(aq)  → NO(g)  + Cl-(aq)

The N and Cl atoms are balanced as:

3N2H4(l)  + 4CIO-3(aq)  → 6NO(g)  + 4Cl-(aq)

The O atoms are balanced by adding 6H2O as:

3N2H4(l)  + 4CIO-3(aq)  → 6NO(g)  + 4Cl-(aq) + 6H2O(l)

This is the required balanced equation.

 

(c)

The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in CIO-2and the oxidation number of O increases from -1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent.

Ion-electron method:

The oxidation half equation is:

H2O-12(aq)  →  O02(g)

The oxidation number is balanced by adding 2 electrons as:

H2O2(aq)  →  O2(g)  + 2e-

The charge is balanced by adding 2OH-ions as:

H2O2(aq)  +  2OH-(aq)  →   O2(g)  + 2e-

The oxygen atoms are balanced by adding 2H2O as:

H2O2(aq)  +  2OH-(aq)  →   O2(g)  + 2H2O(l) + 2e- ... (i)

The reduction half equation is:

C+7l2O7(g)   →  C+3lO-2(g)

The Cl atoms are balanced as:

Cl2O7(g)   →   2ClO-2(g)

The oxidation number is balanced by adding 8 electrons as:

Cl2O7(g)    +   8e- →   2ClO-2(g)

The charge is balanced by adding 6OH- as:

Cl2O7(g)    +   8e- →   2ClO-2(g)  +  6OH- (aq)

The oxygen atoms are balanced by adding 3H2O as:

Cl2O7(g)   +    3H2O(l) +  8e- →   2ClO-2(g)  +  6OH- (aq)  .... (ii)

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:

Cl2O7(g)   +    4H2O2(aq) + 2OH- (aq)    →  2ClO-2(aq)  +  4O2(g)  + 5H2O(l)

 

Oxidation number method:

Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8

Total increase in oxidation number of H2O2 = 2 × 1 = 2

By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get:

Cl2O7(g)   +    4H2O2(aq)  →   CIO-2(aq)  +  4O2(g)

The Cl atoms are balanced as:

Cl2O7(g)   +    4H2O2(aq)  →   2CIO-2(aq)  +  4O2(g)

The O atoms are balanced by adding 3H2O as:

Cl2O7(g)   +    4H2O2(aq)  →   2CIO-2(aq)  +  4O2(g)  +  3H2O(l)

The H atoms are balanced by adding 2OH- and 2H2O as:

Cl2O7(g)   +   4H2O2(aq)   +   2OH-(aq) →   2CIO-2(aq)  +  4O2(g)  +  5H2O(l)

This is the required balanced equation.


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