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# Balance the equation:\begin{align} & A{{s}_{2}}{{S}_{3}}+N{{O}_{3}}^{-}\to As{{O}_{4}}^{3-}N{{O}_{2}}+S\text{ (Acidic medium)} \\ & \text{Al+N}{{\text{O}}_{3}}^{-}\to Al{{\left( OH \right)}_{4}}^{-}+N{{H}_{3}}\text{ }(Basic\text{ medium)} \\ \end{align}

Last updated date: 13th Jun 2024
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HINT: These reactions are redox reactions. Balance the reaction by balancing the oxidation and the reduction half reaction separately. While balancing the reaction in an acidic medium, remember that in this medium, ${{H}^{+}}$ ions are added. In the basic medium, remember that $O{{H}^{-}}$ ions are added and thus the ${{H}^{+}}$ ions are converted to ${{H}_{2}}O$ .

Complete step by step solution:
In the first equation, we have $A{{s}_{2}}{{S}_{3}}+N{{O}_{3}}^{-}\to As{{O}_{4}}^{3-}NO+S$ in acidic medium. So firstly, we can write down the basic equation as-
$A{{s}_{2}}{{S}_{3}}+N{{O}_{3}}^{-}+{{H}^{+}}\to As{{O}_{4}}^{3-}NO+S+{{H}_{2}}O$
Now, we will write down the oxidation number of each of the elements as-
${{\overset{+3}{\mathop{As}}\,}_{2}}{{\overset{-2}{\mathop{S}}\,}_{3}}+\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}^{-}\to \overset{+5}{\mathop{As}}\,{{\overset{-2}{\mathop{O}}\,}_{4}}^{3-}+\overset{+4}{\mathop{N}}\,{{\overset{-4}{\mathop{O}}\,}_{2}}+S$
We can see that for nitrogen, the oxidation number changes from +5 to +4 and thus we can say it is reduced and arsenic is oxidised as the oxidation number changes from +3 to +5.
So, we can write the reduction half reaction for nitrogen as-$N{{O}_{3}}^{-}\to N{{O}_{2}}$
Nitrogen is already balanced so we have to balance oxygen here. We know that in an acidic medium, we will have ${{H}^{+}}$ ions. When we balance an equation in acidic medium, we add ${{H}^{+}}$ ions to the reactant side and ${{H}_{2}}O$ to the product side. So, we can write the balanced equation as-
$N{{O}_{3}}^{-}+2{{H}^{+}}+{{e}^{-}}\to N{{O}_{2}}+{{H}_{2}}O$
Now, for the oxidation half reaction, both arsenic and sulphur are oxidised. The final reaction equation is-
$A{{s}_{2}}{{S}_{3}}\to 2As{{O}_{4}}^{3-}+3S$
Now we have to balance the oxygen atoms too. For that, we will add water to the reactant side and balance the hydrogen atoms by adding ${{H}^{+}}$ ions double the number of water molecules on the product side-
$A{{s}_{2}}{{S}_{3}}+8{{H}_{2}}O\to 2As{{O}_{4}}^{3-}+3S+16{{H}^{+}}$
And then we have to balance the charge too which will give us the final equation as-
$A{{s}_{2}}{{S}_{3}}+8{{H}_{2}}O\to 2As{{O}_{4}}^{3-}+3S+16{{H}^{+}}+10{{e}^{-}}$
Now, we can see that the oxidation reaction has 10 electrons and the reduction reaction has 1 electron. So, to balance the charge, we need to multiply the reduction reaction by 10. Thus, the half reactions will be -
\begin{align} & 10N{{O}_{3}}^{-}+20{{H}^{+}}+10{{e}^{-}}\to 10N{{O}_{2}}+10{{H}_{2}}O \\ & A{{s}_{2}}{{S}_{3}}+8{{H}_{2}}O\to 2As{{O}_{4}}^{3-}+3S+16{{H}^{+}}+10{{e}^{-}} \\ \end{align}
Now, combining the two reactions will give us the final balanced equation as-
$A{{s}_{2}}{{S}_{3}}+10N{{O}_{3}}^{-}+4{{H}^{+}}\to 2As{{O}_{4}}^{3-}+3S+2{{H}_{2}}O+10N{{O}_{2}}$
Now, the second reaction is: $\text{Al+N}{{\text{O}}_{3}}^{-}\to Al{{\left( OH \right)}_{4}}^{-}+N{{H}_{3}}$ in basic medium
Firstly, we will write the oxidation and reduction half reactions for the above reaction. Here, Al is oxidised to $Al{{(OH)}_{4}}^{-}$ as its oxidation number changes from 0 to +3 and nitrogen is reduced as $N{{O}_{3}}^{-}$ is converted to $N{{H}_{3}}$ and oxidation number changes from +5 to +3.
So, we can write down the half reactions as-
\begin{align} & Al\to Al{{\left( OH \right)}_{4}}^{-}+{{e}^{-}} \\ & N{{O}_{3}}^{-}+{{e}^{-}}\to N{{H}_{3}} \\ \end{align}
Now, we will balance all the atoms on both the sides and add water and ${{H}^{+}}$ ions to balance H and O. So, we can write the reactions as-
\begin{align} & Al+4{{H}_{2}}O\to Al{{\left( OH \right)}_{4}}^{-}+4{{H}^{+}}+{{e}^{-}} \\ & N{{O}_{3}}^{-}+{{e}^{-}}+9{{H}^{+}}\to N{{H}_{3}}+3{{H}_{2}}O \\ \end{align}
Now, we have to balance charge on both sides.
For the oxidation reaction, the charge on the reactant side is 0 and that on the product side is +2. If we increase the coefficient of the electron by 3, the charge on the reactant side will become 0. So, the reaction becomes-
$Al+4{{H}_{2}}O\to Al{{\left( OH \right)}_{4}}^{-}+4{{H}^{+}}+3{{e}^{-}}$
Now, similarly for the reduction reaction, increasing the coefficient of electron by 8 will equate the charge. So, the reaction becomes-
$N{{O}_{3}}^{-}+8{{e}^{-}}+9{{H}^{+}}\to N{{H}_{3}}+3{{H}_{2}}O$
The reaction is taking place in basic medium so we will add $O{{H}^{-}}$ ions to both the sides and ${{H}^{+}}$ will be converted to ${{H}_{2}}O$ . So, the reactions will become-
\begin{align} & Al+4{{H}_{2}}O+4O{{H}^{-}}\to Al{{\left( OH \right)}_{4}}^{-}+4{{H}_{2}}O+3{{e}^{-}} \\ & Or,Al+4O{{H}^{-}}\to Al{{\left( OH \right)}_{4}}^{-}+3{{e}^{-}} \\ \end{align}
\begin{align} & N{{O}_{3}}^{-}+8{{e}^{-}}+9{{H}_{2}}O\to N{{H}_{3}}+3{{H}_{2}}O+9O{{H}^{-}} \\ & Or,N{{O}_{3}}^{-}+8{{e}^{-}}+6{{H}_{2}}O\to N{{H}_{3}}+9O{{H}^{-}} \\ \end{align}
We can see that the number of electrons in the oxidation reaction is 3 and that in reduction reaction is 8. So, to balance them, we multiply the oxidation reaction by 8 and the reduction reaction by 3. Thus, the equation becomes-
\begin{align} & 8Al+32O{{H}^{-}}\to 8Al{{\left( OH \right)}_{4}}^{-}+24{{e}^{-}} \\ & 3N{{O}_{3}}^{-}+24{{e}^{-}}+18{{H}_{2}}O\to 3N{{H}_{3}}+27O{{H}^{-}} \\ \end{align}
So, combining the two reaction, we will get the final balanced equation as-
$8Al+5O{{H}^{-}}+3N{{O}_{3}}^{-}+18{{H}_{2}}O\to 8Al{{\left( OH \right)}_{4}}^{-}+3N{{H}_{3}}$
The above reaction is the required balanced equation.

NOTE: The above given reactions are redox reactions. To balance the redox reactions we have to balance charge as well as mass as we have done above. There are several methods that we can use to balance redox reactions like half-reaction method and oxidation number method. In the above question, we have used a half-reaction method.