Question

Balance the following equation by oxidation number method.
\[HN{{O}_{3}}+C{{u}_{2}}O\to Cu{{(N{{O}_{3}})}_{2}}+NO+{{H}_{2}}O\]

Answer 1

Hint: To verify whether a given equation is balanced or not, check if it contains the same type and number of atoms on both sides of the equation and also calculate that the sum of the charges on one side of the equation is equal to the sum of the charges on the other side.

Complete answer:
Step 1. Let us start with writing the unbalanced equation first and assign the respective oxidation numbers for each atom.
\[HN{{O}_{3}}+C{{u}_{2}}O\to Cu{{(N{{O}_{3}})}_{2}}+NO+{{H}_{2}}O\]
\[\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,\to \overset{+2}{\mathop{Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+\overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,+{{\overset{+1}{\mathop{H}}\,}_{2}}\overset{-2}{\mathop{O}}\,\]

Step 2. Now, identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number decreases) and separate the reaction process into half-reaction for oxidation ad reduction, respectively. Also, write down the transfer of electrons. Add coefficients wherever necessary in order to make the numbers of oxidized and reduced atoms equal on both sides.
\[\begin{align}
  & O\;\;\;\;\;\;\;\;{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,\to \overset{+2}{\mathop{Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+2{{e}^{-}}Cu \\
 & R\;\;\;\;\;\;\;\;\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+3{{e}^{-}}\to \overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,N \\
\end{align}\]

Step 3. After that, we are going to balance the atoms in each of the half-reaction, as a chemical equation must have the same number of atoms of each element on both sides of the equation. Here also, add coefficients wherever necessary (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms.
a)Just balance the other atoms, let us leave hydrogen and oxygen as it is for now. Always be sure to add the reactants on the left side ant the products on the right side.
\[\begin{align}
  & O\;\;\;\;\;\;\;\;{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+4HN{{O}_{3}}\to \overset{+2}{\mathop{Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+2{{e}^{-}} \\
 & R\;\;\;\;\;\;\;\;\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+3{{e}^{-}}\to \overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\, \\
\end{align}\]
b)Now we will balance the charges. We have assumed that the reaction is occurring in an acidic solution, so we have to balance the charge such that both sides have the same amount of total charge. We will do that by adding an H+ ion to the side, which needs a positive charge.
\[\begin{align}
  & O\;\;\;\;\;\;\;\;{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+4HN{{O}_{3}}\to \overset{+2}{\mathop{Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+2{{e}^{-}}+2{{H}^{+}} \\
 & R\;\;\;\;\;\;\;\;\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+3{{e}^{-}}+3{{H}^{+}}\to \overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\, \\
\end{align}\]
c)Now we will balance the number of oxygen atoms. Check for oxygen atoms on both sides, if they aren't equal, make them equal by adding water molecules.
\[\begin{align}
  & O\;\;\;\;\;\;\;\;{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+4HN{{O}_{3}}\to \overset{+2}{\mathop{2Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+2{{e}^{-}}+2{{H}^{+}}+{{H}_{2}}O \\
 & R\;\;\;\;\;\;\;\;\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+3{{e}^{-}}+3{{H}^{+}}\to \overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,+2{{H}_{2}}O \\
\end{align}\]
In a chemical reaction, the number of electrons gained s always equal to the number of electrons lost. In other words, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction. We will make both of them equal by multiplying the coefficients of all species by the LCM between the half-reactions.
\[\begin{align}
  & O\;\;\;\;\;\;\;\;{{\overset{+1}{\mathop{[Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+4HN{{O}_{3}}\to \overset{+2}{\mathop{2Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+2{{e}^{-}}+2{{H}^{+}}+{{H}_{2}}O]\,\times \,3 \\
 & R\;\;\;\;\;\;\;\; [\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+3{{e}^{-}}+3{{H}^{+}}\to \overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,+2{{H}_{2}}O]\,\times \,2 \\
\end{align}\]
\[\begin{align}
  & O\;\;\;\;\;\;\;\;3{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+12HN{{O}_{3}}\to \overset{+2}{\mathop{6Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+6{{e}^{-}}+6{{H}^{+}}+3{{H}_{2}}O \\
 & R\;\;\;\;\;\;\;\;\overset{+1}{\mathop{2H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+6{{e}^{-}}+6{{H}^{+}}\to \overset{+2}{\mathop{2N}}\,\overset{-2}{\mathop{O}}\,+4{{H}_{2}}O \\
\end{align}\]

Step 5. After that, we will put both of the half-reactions together by adding all the reactants together on the left and all of the products on the right side.
\[3{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+14\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}+6{{e}^{-}}+6{{H}^{+}}\to \overset{+2}{\mathop{6Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+\overset{+2}{\mathop{2N}}\,\overset{-2}{\mathop{O}}\,+6{{e}^{-}}+7{{H}_{2}}O+6{{H}^{+}}\]

Step 6. At last, we have to just simplify the obtained the final equation. If there are same species on either side of the equation, they will be cancelled out. And write the whole balanced equation.
\[3{{\overset{+1}{\mathop{Cu}}\,}_{2}}\overset{-2}{\mathop{O}}\,+14\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}\to \overset{+2}{\mathop{6Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+\overset{+2}{\mathop{2N}}\,\overset{-2}{\mathop{O}}\,+7{{H}_{2}}O\]

Note: We have done this balancing in acidic medium, for reactions in a basic solution, balance the charge so that both sides have the same total charge by adding an OH- ion to the side, which needs a negative charge.