Question

Balance equations by oxidation number method or ion electron method.

S+HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+N{{O}_{2}}+{{H}_{2}}O \\\ F{{e}^{2+}}+MnO_{4}^{-}+{{H}^{+}}\to F{{e}^{3+}}+M{{n}^{2+}}+{{H}_{2}}O \\\ \end{array}$$

Answer 1

Oxidation number method is the method used to balance the chemical reaction based on their oxidation state. The idea behind the method is electrons transferred between the charged atoms or molecules.

Complete answer:
- In the question it is given that there are two chemical redox reactions.
- We have to balance the given redox reactions by using the oxidation number method.
- Coming to the first chemical reaction
$S+HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+N{{O}_{2}}+{{H}_{2}}O$
- In the first step we have to assign the oxidation number of all the atoms in the given chemical reaction.
$\overset{0}{\mathop{S}}\,+\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+\overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}+{{\overset{+1}{\mathop{H}}\,}_{2}}\overset{-2}{\mathop{O}}\,$
- Divide the given chemical reaction into two half-reactions (one half-reaction for oxidation and other for reduction reaction).
Oxidation reaction: $\overset{0}{\mathop{S}}\,\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+6{{e}^{-}}$
Reduction reaction: $\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+{{e}^{-}}\to \overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}$
- In the next step balance the charge on both sides.
Oxidation reaction: $\overset{0}{\mathop{S}}\,\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+6{{e}^{-}}+6{{H}^{+}}$
Reduction reaction: $\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+{{e}^{-}}+{{H}^{+}}\to \overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}$
- Now balance the oxygen atoms on both the sides.
Oxidation reaction: $\overset{0}{\mathop{S}}\,+4{{H}_{2}}O\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+6{{e}^{-}}+6{{H}^{+}}$
Reduction reaction: $\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+{{e}^{-}}+{{H}^{+}}\to \overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}+{{H}_{2}}O$
- In the next step we have to make the electrons gained or lost in oxidation and reduction to be equal.
Oxidation reaction: $\overset{0}{\mathop{S}}\,+4{{H}_{2}}O\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+6{{e}^{-}}+6{{H}^{+}}$
Reduction reaction: $6\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+6{{e}^{-}}+{{H}^{+}}\to 6\overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}+6{{H}_{2}}O$
- Now add the two half-reactions and make a single reaction.
$\overset{0}{\mathop{S}}\,+4{{H}_{2}}O+6\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+6{{e}^{-}}+{{H}^{+}}\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+6{{e}^{-}}+6{{H}^{+}}+6\overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}+6{{H}_{2}}O$
- Simplify the above equation by removing the common things on both sides. $$$$
$\overset{0}{\mathop{S}}\,+\overset{+1}{\mathop{6H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}\to {{\overset{+1}{\mathop{H}}\,}_{2}}\overset{+6}{\mathop{S}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}+6\overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}+2{{\overset{+1}{\mathop{H}}\,}_{2}}\overset{-2}{\mathop{O}}\,$
- Therefore the balanced chemical equation by using the oxidation number method is as follows.
$S+6HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+6N{{O}_{2}}+2{{H}_{2}}O$
- Now coming to the second given equation
$F{{e}^{2+}}+MnO_{4}^{-}+{{H}^{+}}\to F{{e}^{3+}}+M{{n}^{2+}}+{{H}_{2}}O$
- In the first step we have to assign the oxidation number of all the atoms in the given chemical reaction.
${{\overset{+2}{\mathop{Fe}}\,}^{2+}}+\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+{{H}^{+}}\to {{\overset{+3}{\mathop{Fe}}\,}^{3+}}+{{\overset{2+}{\mathop{Mn}}\,}^{2+}}+{{\overset{+1}{\mathop{H}}\,}_{2}}\overset{-2}{\mathop{O}}\,$
- Divide the given chemical reaction into two half-reactions (one half-reaction for oxidation and other for reduction reaction).
Oxidation reaction: ${{\overset{+2}{\mathop{Fe}}\,}^{2+}}\to {{\overset{+3}{\mathop{Fe}}\,}^{3+}}+{{e}^{-}}$
Reduction reaction: $\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+5{{e}^{-}}\to {{\overset{2+}{\mathop{Mn}}\,}^{2+}}$
- In the next step balance the charge on both sides.
Oxidation reaction: ${{\overset{+2}{\mathop{Fe}}\,}^{2+}}\to {{\overset{+3}{\mathop{Fe}}\,}^{3+}}+{{e}^{-}}$
Reduction reaction: $\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+5{{e}^{-}}+8{{H}^{+}}\to {{\overset{2+}{\mathop{Mn}}\,}^{2+}}$
- Now balance the oxygen atoms on both the sides.
Oxidation reaction: ${{\overset{+2}{\mathop{Fe}}\,}^{2+}}\to {{\overset{+3}{\mathop{Fe}}\,}^{3+}}+{{e}^{-}}$
Reduction reaction: $\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+5{{e}^{-}}+8{{H}^{+}}\to {{\overset{2+}{\mathop{Mn}}\,}^{2+}}+4{{H}_{2}}O$
- In the next step we have to make the electrons gained or lost in oxidation and reduction to be equal.
Oxidation reaction: $5{{\overset{+2}{\mathop{Fe}}\,}^{2+}}\to 5{{\overset{+3}{\mathop{Fe}}\,}^{3+}}+5{{e}^{-}}$
Reduction reaction: $\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+5{{e}^{-}}+8{{H}^{+}}\to {{\overset{2+}{\mathop{Mn}}\,}^{2+}}+4{{H}_{2}}O$
- Now add the two half-reactions and make a single reaction.
$5{{\overset{+2}{\mathop{Fe}}\,}^{2+}}+\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+5{{e}^{-}}+8{{H}^{+}}\to 5{{\overset{+3}{\mathop{Fe}}\,}^{3+}}+5{{e}^{-}}+{{\overset{2+}{\mathop{Mn}}\,}^{2+}}+4{{H}_{2}}O$
- Simplify the above equation by removing the common things on both sides.
$5{{\overset{+2}{\mathop{Fe}}\,}^{2+}}+\overset{+7}{\mathop{Mn}}\,\overset{-2}{\mathop{O}}\,_{4}^{-}+8{{H}^{+}}\to 5{{\overset{+3}{\mathop{Fe}}\,}^{3+}}+{{\overset{2+}{\mathop{Mn}}\,}^{2+}}+4{{H}_{2}}O$
- Therefore the balanced chemical equation by using the oxidation number method is as follows.
$F{{e}^{2+}}+MnO_{4}^{-}+8{{H}^{+}}\to 5F{{e}^{3+}}+M{{n}^{2+}}+4{{H}_{2}}O$

Note:
In the oxidation number method first we have to divide the chemical reaction into two halves (oxidation and reduction) then we have to follow each and every step carefully at the time of balancing the charge, and balancing the oxygen atoms.