Balance these equations by using ion electron /oxidation number method --(underline the oxidising agent and reducing agents ).
1.Mg +NO3- + H+ ----> Mg 2+ +N2O +H2O.
2.Cl2 + OH- ------> Cl- + ClO3- + H2O.
Answers
Answer:
The oxidation number of N increases from −2 to +2. The oxidation number of Cl decreases from +5 to −1. Hence, hydrazine is the reducing agent and chlorate ion is the oxidizing agent.
Ion-electron method:
The oxidation half reaction is
N
2
H
4
(l)→NO(g)
Balance the N atoms.
N
2
H
4
(l)→2NO(g)
To balance oxidation number, add 8 electrons.
N
2
H
4
(l)→2NO(g)+8e
−
Add 8 hydroxide ions are to balance the charge.
N
2
H
4
(l)+8OH
−
(aq)→2NO(g)+8e
−
The reduction half reaction is
ClO
3
−
(aq)→Cl
−
(aq)
Add 6 electrons to balance the oxidation number.
ClO
3
−
(aq)+6e
−
→Cl
−
(aq)
Add 6 hydroxide ions to balance the charge.
ClO
3
−
(aq)+6e
−
→Cl
−
(aq)+6OH
−
(aq)
Multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2. Add two half reactions obtain the
balanced chemical equation.
3N
2
H
4
(aq)+4ClO
3
−
(aq)→6NO(s)+4Cl
−
(aq)+6H
2
O(aq)
Oxidation number method:
Total decrease in oxidation number of N is 8. Total increase in the oxidation number of Cl is 6.
Multiply N
2
H
4
with 3 and multiply ClO
3
−
with 4.
3N
2
H
4
(l)+4ClO
3
−
(aq)→NO(g)+Cl
−
(aq)
Balance N and Cl atoms.
3N
2
H
4
(l)+4ClO
3
−
(aq)→6NO(g)+4Cl
−
(aq)
Balance the O atoms are by adding 6 water molecules.
3N
2
H
4
(l)+4ClO
3
−
(aq)→6NO(g)+4Cl
−
(aq)+6H
2
O(l)
This is the balanced chemical equation.