NCERT Class XI Chemistry Redox Reactions Solutions
© examsnet.com
Question : 18
Total: 30
Balance the following redox reactions by ion - electron method :
(a)M n O 4 – ( a q ) + I ( a q ) – M n O 2 ( s ) + I 2 ( s )
(b)M n O 4 – ( a q ) + S O 2 ( g ) M n ( a q ) 2 + + H S O 4 – ( a q )
(c)H 2 O 2 ( a q ) + F e ( a q ) 2 + F e ( a q ) 3 + + H 2 O ( l )
(d)C r 2 O 7 2 – + S O 2 ( g ) C r ( a q ) 3 + + S O 4 2 – ( a q )
(a)
(b)
(c)
(d)
Solution:
(a) The skeleton equation is : M n O 4 – ( a q ) + I ( a q ) – M n O 2 ( s ) + I 2 ( s )
(i) The O.N. of the atoms involved in the equation is:
(
− ) +
+
(ii) The species involved in the oxidation and reduction half reaction :
(iii) Oxidation half reaction :I – I 2
Reduction half reaction :M n O 4 – M n O 2
(iv) Balancing the oxidation half reaction2 I – I 2 + 2 e –
(v) Balancing the reduction half reaction
(1) As the decrease in O.N. is 3, therefore, adding 3e– on the reactant side
M n O 4 – + 3 e – M n O 2
(2) To balance the oxygen atoms, add two H2O molecules on the product side
M n O 4 – + 3 e – M n O 2 + 2 H 2 O
(3) To balance the charges, add 4O H – H 2 O
M n O 4 – + 3 e – + 4 H 2 O M n O 2 + 4 O H – + 2 H 2 O
M n O 4 – + 3 e – + 2 H 2 O M n O 2 + 4 O H –
Thus, the reduction half reaction is balanced.
(vi) Adding the two half reactions
In order to equate the electrons, multiply eqn. (i) by 3 and eqn. (ii) by 2.
Add the two equations.
[ 2 I − → I 2 + 2 e − ] × 3
[M n O 4 − + 3 e − + 2 H 2 O M n O 2 + 4 O H −
–––––––––––––––––––––––––
2 M n O 4 – + 6 I – + 4 H 2 O 3 I 2 + 2 M n O 2 + 8 O H –
or2 M n O 4 – ( a q ) + 6 I ( a q ) – + 4 H 2 O ( l ) 3 I 2 ( s ) + 2 M n O 2 ( s ) + 8 O H ( a q ) –
(b) The skeleton equation is :M n O 4 – ( a q ) + S O 2 ( g ) M n ( a q ) 2 + + H S O 4 – ( a q )
(i) The O.N. of atoms involved in the equation is :
− +
+
−
(ii) The species involved in the oxidation and reduction half reactions :
(iii) Oxidation half reaction :S O 2 H S O 4 –
Reduction half reaction :M n O 4 – M n 2 +
(iv) Balancing the oxidation half reaction
(1) As the increase in O.N. is 2, therefore, add two electrons on the product side to balance change in O.N.S O 2 H S O 4 – + 2 e –
(2) In order to balance the number of oxygen atoms, add two H2O molecules on the reactant side and then to balance H atoms add 3H+ on the product side.
S O 2 + 2 H 2 O H S O 4 – + 3 H + + 2 e –
(v) Balancing the reduction half reaction
The reduction half reaction is :M n O 4 – M n 2 +
(1) As the decrease in O.N. is 5, therefore add5 e –
M n O 4 – + 5 e – M n 2 +
(2) In order to balance the no. of oxygen atoms, add fourH 2 O H +
M n O 4 – + 8 H + + 5 e – M n 2 + + 4 H 2 O
(vi) Adding the two half, reactions
In order to equate the electrons, multiply eqn. (i) by 5 and eqn. (ii) by 2.
Add the two eqns.
[S O 2 + 2 H 2 O H S O 4 − + 3 H + + 2 e −
[M n O 4 − + 8 H + + 5 e − M b 2 + + 4 H 2 O
––––––––––––––––––––––––
2 M n O 4 − ( a q ) + 5 S O 2 ( g ) + 2 H 2 O ( l ) + H ( a q ) + 5 H S O 4 − ( a q ) + 2 M n ( a q ) 2 +
(c) Oxidation half equation :F e ( a q ) 2 + F e ( a q ) 3 + + e –
Reduction half equation :H 2 O 2 ( a q ) + 2 H ( a q ) + + 2 e – 2 H 2 O ( l )
Multiplying eqn. (i) by 2 and adding it to eqn. (ii), we get
H 2 O 2 ( a q ) + 2 F e ( a q ) 2 + + 2 H ( a q ) + 2 F e ( a q ) 3 + + 2 H 2 O ( l )
(d) Oxidation half equation :S O 2 ( g ) + 2 H 2 O ( l ) S O 4 2 – ( a q ) + 4 H ( a q ) + + 2 e –
Reduction half equation :
C r 2 O 7 2 – ( a q ) + 14 H ( a q ) + + 6 e – 2 C r ( a q ) 3 + + 7 H 2 O ( l )
Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get,
C r 2 O 7 2 – ( a q ) + 3 S O 2 ( g ) + 2 H ( a q ) + 2 C r ( a q ) 3 + + 3 S O 4 2 – ( a q ) + H 2 O ( l )
(i) The O.N. of the atoms involved in the equation is:
(ii) The species involved in the oxidation and reduction half reaction :
(iii) Oxidation half reaction :
Reduction half reaction :
(iv) Balancing the oxidation half reaction
(v) Balancing the reduction half reaction
(1) As the decrease in O.N. is 3, therefore, adding 3e– on the reactant side
(2) To balance the oxygen atoms, add two H2O molecules on the product side
(3) To balance the charges, add 4
Thus, the reduction half reaction is balanced.
(vi) Adding the two half reactions
In order to equate the electrons, multiply eqn. (i) by 3 and eqn. (ii) by 2.
Add the two equations.
[
–––––––––––––––––––––––––
or
(b) The skeleton equation is :
(i) The O.N. of atoms involved in the equation is :
(ii) The species involved in the oxidation and reduction half reactions :
(iii) Oxidation half reaction :
Reduction half reaction :
(iv) Balancing the oxidation half reaction
(1) As the increase in O.N. is 2, therefore, add two electrons on the product side to balance change in O.N.
(2) In order to balance the number of oxygen atoms, add two H2O molecules on the reactant side and then to balance H atoms add 3H+ on the product side.
(v) Balancing the reduction half reaction
The reduction half reaction is :
(1) As the decrease in O.N. is 5, therefore add
(2) In order to balance the no. of oxygen atoms, add four
(vi) Adding the two half, reactions
In order to equate the electrons, multiply eqn. (i) by 5 and eqn. (ii) by 2.
Add the two eqns.
[
[
––––––––––––––––––––––––
(c) Oxidation half equation :
Reduction half equation :
Multiplying eqn. (i) by 2 and adding it to eqn. (ii), we get
(d) Oxidation half equation :
Reduction half equation :
Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get,
© examsnet.com
Go to Question: