Question

Balance the following reaction by oxidation number method:
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{H}_{2}}O$

Answer 1

Hint: Oxidation number or oxidation state, of an atom in a molecule or compound is defined as the actual charge on the atom if it exists in a monoatomic ion.
In an oxidation number method in a balance redox reaction, the total increase in oxidation number must be equal to the total decrease in oxidation number and that makes the basis for balancing a redox reaction.

Complete step by step answer:
 To balance a redox reaction by oxidation method we follow the following five steps –
(i) Write the oxidation number of all the atoms.
(ii) Select the oxidizing and reducing agent.
(iii) Multiply oxidizing agent by the number of loss of electrons and reducing agents by the number of gain of electrons.
(iv) Balance the number of atoms on both sides whose oxidation number changes in the reaction.
(v) To balance oxygen atoms, add ${{H}_{2}}O$ to the deficient side. Then balance the number of $H$ atoms by adding ${{H}^{+}}$ ions to the hydrogen deficient side.
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{H}_{2}}O$
We will balance the given redox reaction by this method -
Firstly let us see the oxidation number of all the central atoms.
On the reactant side, chromium in ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ has +6 oxidation state and similarly, iron in ferrous sulphate has +2 state, sulphur in sulphuric acid has +6 state. On the product side, sulphur in ${{K}_{2}}S{{O}_{4}}$ has +6 state, chromium in $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$ has +3 state and iron of$F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ has +3 state.
Now we will convert this reaction into ionic form by eliminating the elements which are not going either oxidation or reduction.
$\begin{align}
  & \\
 & \,\,{{\,}_{+2}}\,\,\,\,\,\,\,\,\,\,{{\,}_{+6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3}}\, \\
 & Fe\,\,\,\,+\,\,\,C{{r}_{2}}{{O}_{7}}\to \,\,F{{e}_{2}}\,\,+\,\,C{{r}_{2}}\, \\
\end{align}$
In this reaction $Fe$ is getting oxidised so it will act as a reductant and $Cr$ is getting reduced so it will act as an oxidant.
$\begin{align}
 & \\
 & \,\,{{\,}_{+6}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\,}_{+3}}\,\,\,\,\,\,\,\,\,\,\, \\
 & C{{r}_{2}}{{O}_{7}}\to \,\,\,C{{r}_{2}}\, \\
\end{align}$ ………………………..(i) (Oxidation number of $Cr$ decreases by 3)
$\begin{align}
 & _{+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3\,\,} \\
 & Fe\,\,\to \,\,F{{e}_{2}} \\
\end{align}$ …………………………….(ii) (Oxidation number of $Fe$ increases by 1)
Equalise the increase or decrease by multiplying equation (i) by 1 and equation (ii) by 3, so we get
$\begin{align}
  & \\
 & \,\,{{\,}_{+2}}\,\,\,\,\,\,\,\,\,\,{{\,}_{+6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3}}\, \\
 & Fe\,\,\,\,+\,\,\,C{{r}_{2}}{{O}_{7}}\to \,\,3F{{e}_{2}}\,\,+\,\,C{{r}_{2}}\, \\
\end{align}$
Balance oxygen atom by adding ${{H}_{2}}O$ molecules and $H$ ions by adding ${{H}^{+}}$ ions.
$\begin{align}
  & \\
 & \,\,{{\,}_{+2}}\,\,\,\,\,\,\,\,\,\,{{\,}_{+6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3}}\, \\
 & 6Fe\,\,\,\,+\,\,\,C{{r}_{2}}{{O}_{7}}\,+\,14{{H}^{+}}\to \,\,3F{{e}_{2}}\,\,+\,\,C{{r}_{2}}\,+\,7{{H}_{2}}O \\
\end{align}$
So overall balance reaction is -
\[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+6FeS{{O}_{4}}+7{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+7{{H}_{2}}O\]

Note: -In acidic medium chromium undergo reduction and its oxidation state change from +6 to +3.
-When metal participates in a chemical reaction, it is always oxidized, thus metal always behaves like reducing agents.
-The oxidation state of $Fe$ in Ferro compounds is +2 while in ferric compounds, it is in +3 oxidation state.