NCERT Class XI Chemistry Redox Reactions Solutions
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Question : 19
Total: 30
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)P 4 ( s ) + O H ( a q ) – P H 3 ( g ) + H 2 P O 2 – ( a q )
(b)N 2 H 4 ( l ) + C l O 3 – ( a q ) N O ( g ) + C l ( a q ) –
(c)C l 2 O 7 ( g ) + H 2 O 2 ( a q ) C l O ( a q ) – + O 2 ( g ) + H ( a q ) +
(a)
(b)
(c)
Solution:
(a) P 4 ( s ) + O H ( a q ) – P H 3 ( g ) + H 2 P O 2 – ( a q )
Oxidation number method :
Total increase in O.N. of P fromP 4 H 2 P O 2
– = 1 × 4 = 4
Total decrease in O.N. of P fromP 4 P H 3
Therefore, to balance increase/decrease in O.N. multiplyP H 3 H 2 P O 2 –
P 4 ( s ) P H 3 ( g ) + 3 H 2 P O 2 – ( a q )
To balance O atoms, add6 O H –
P 4 ( s ) + 6 O H ( a q ) – P H 3 ( g ) + 3 H 2 P O 2 – ( a q )
To balance H atoms, add3 H 2 O 3 O H –
P 4 ( s ) + 6 O H ( a q ) – + 3 H 2 O ( l ) P H 3 ( g ) + 3 H 2 P O 2 – ( a q ) + 3 O H ( a q ) –
or,P 4 ( s ) + 3 O H ( a q ) – + 3 H 2 O ( l ) P H 3 ( g ) + 3 H 2 P O 2 – ( a q )
Ion electron method : The two half reactions are :
Oxidation half reaction :P 4 ( s ) + 8 O H ( a q ) – 4 H 2 P O 2 – ( a q ) + 4 e –
Reduction half reaction :
P 4 ( s ) + 12 H 2 O ( l ) + 12 e – 4 P H 3 ( g ) + 12 O H ( a q ) –
Multiply eq. (i) by 3 and add it to eq. (ii), we get
4 P 4 ( s ) + 24 O H ( a q ) – + 12 H 2 O ( l ) 4 P H 3 ( g ) + 12 H 2 P O 2 – ( a q ) + 12 O H ( a q ) –
or,P 4 ( s ) + 3 O H ( a q ) – + 3 H 2 O ( l ) P H 3 ( g ) + 3 H 2 P O 2 – ( a q )
Reductant - phosphorus; oxidant-phosphorus
(b)N 2 H 4 ( l ) + C l O 3 – ( a q ) N O ( g ) + C l ( a q ) –
Oxidation number method :
Net reaction is
6 N 2 H 4 + 8 C l O – 3 12 N O + 8 C l – + 12 H 2 O
3 N 2 H 4 + 4 C l O – 3 6 N O + 4 C l – + 6 H 2 O
Ion-electron method :
Oxidation half-reaction : [N 2 H 4 + 8 O H – 2 N O + 8 e – + 6 H 2 O
Reduction half-reaction : [C l O 3 – + 6 e – + 3 H 2 O C l – + 6 O H –
Net reaction is
6 N 2 H 4 + 8 C l O 3 – 12 N O + 8 C l – + 12 H 2 O
3 N 2 H 4 + 4 C l O 3 – 6 N O + 4 C l – + 6 H 2 O
Reductant :N 2 H 4 C l O 3 –
(c)C l 2 O 7 ( g ) + H 2 O 2 ( a q ) C l O 2 – ( a q ) + O 2 ( g ) + H ( a q ) +
Oxidation number method :
Net reaction isC l 2 O 7 + 4 H 2 O 2 + 2 O H – 2 C l O 2 – + 5 H 2 O + 4 O 2
Ion-electron method :
Oxidation half-reaction : [H 2 O 2 + 2 O H – O 2 + 2 e – + 2 H 2 O
Reduction half-reaction :C l 2 O 7 + 8 e – + 3 H 2 O 2 C l O 2 – + 6 O H –
Net reaction isC l 2 O 7 + 4 H 2 O 2 + 2 O H – 2 C l O 2 – + 5 H 2 O + 4 O 2
Reductant :H 2 O 2 C l 2 O 7
Oxidation number method :
Total increase in O.N. of P from
– = 1 × 4 = 4
Total decrease in O.N. of P from
Therefore, to balance increase/decrease in O.N. multiply
To balance O atoms, add
To balance H atoms, add
or,
Ion electron method : The two half reactions are :
Oxidation half reaction :
Reduction half reaction :
Multiply eq. (i) by 3 and add it to eq. (ii), we get
or,
Reductant - phosphorus; oxidant-phosphorus
(b)
Oxidation number method :
Net reaction is
Ion-electron method :
Oxidation half-reaction : [
Reduction half-reaction : [
Net reaction is
Reductant :
(c)
Oxidation number method :
Net reaction is
Ion-electron method :
Oxidation half-reaction : [
Reduction half-reaction :
Net reaction is
Reductant :
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