Question

When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation:

$x{\rm{Cu}} + y{\rm{HN}}{{\rm{O}}_{\rm{3}}} \to {\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_2} + {\rm{NO}} + {\rm{N}}{{\rm{O}}_2} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$

The sum of coefficients i.e. x and y is:

Answer 1

Hint: A reaction is termed as redox reaction if there is change in oxidation number (increase as well as decrease) of some reacting species which are involved in the reaction. The increase of oxidation number is termed as oxidation whereas decrease of oxidation number is reduction.

Complete step by step answer:
Here, we have to balance the given reaction to know the coefficient x and y. The reaction is,

 $x{\rm{Cu}} + y{\rm{HN}}{{\rm{O}}_{\rm{3}}} \to {\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_2} + {\rm{NO}} + {\rm{N}}{{\rm{O}}_2} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$


In the reaction, copper has zero oxidation state in the reactant side.

Now, we calculate the oxidation state of copper in${\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_2}$. If we take x as the oxidation state of copper then,

$x - 2 = 0$

$ \Rightarrow x = 2$


So, the oxidation number increases. That means oxidation reaction takes place.

Now, we compare the oxidation state of nitrogen in reactants and products.

For ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$, oxidation state of nitrogen is,

$x + 1 - 6 = 0$

$ \Rightarrow x = 5$


And in ${\rm{N}}{{\rm{O}}_{\rm{2}}}$, oxidation state of nitrogen is +4 and in NO, oxidation state is +2. So, a reduction reaction takes place.

Now, we have to write half reactions for oxidation and reduction reactions and balance them.

Oxidation reaction:

${\rm{Cu}} \to {\rm{C}}{{\rm{u}}^{2 + }} + 2{e^ - }$ ……. (1)

Reduction reaction:

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - \to {\rm{NO}}$

Oxidation state of N is LHS is +5 and in RHS is +2. So, to balance we have to add 3 electrons to LHS.

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - + 3{e^ - } \to {\rm{NO}}$

To balance oxygen atoms, we have to add 2 water molecules in RHS.

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - + 3{e^ - } \to {\rm{NO}} + {\rm{2}}{{\rm{H}}_2}{\rm{O}}$

Now, to balance hydrogen atoms we have to add four protons to LHS.

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - + 3{e^ - } + 4{{\rm{H}}^ + } \to {\rm{NO}} + {\rm{2}}{{\rm{H}}_2}{\rm{O}}$ …… (2)

The other reduction reaction is,

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - \to {\rm{N}}{{\rm{O}}_{\rm{2}}}$

Here, the LHS oxidation state of N is +5 and in RHS oxidation state is +4. So, to balance charge we have to add one electron to LHS.

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - + {e^ - } \to {\rm{N}}{{\rm{O}}_{\rm{2}}}$

Now, to balance oxygen atoms, we have to add one water molecule in RHS.

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - + {e^ - } \to {\rm{N}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$


Now, to balance hydrogen atoms, we have to add two protons to LHS.

${\rm{N}}{{\rm{O}}_{\rm{3}}}^ - + {e^ - } + 2{{\rm{H}}^ + } \to {\rm{N}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$ …… (3)

Now, we have to multiply equation (1) by 2 to equate the total number of electrons of oxidation and reduction reaction. The equation obtained is,

${\rm{2Cu}} \to 2{\rm{C}}{{\rm{u}}^{2 + }} + 4{e^ - }$ …… (4)

Now, we have to add equation (2), (3) and (4).

${\rm{2Cu}} + {\rm{4}}{{\rm{e}}^ - } + 4{{\rm{H}}^ + } + 2{{\rm{H}}^ + } + 2{\rm{N}}{{\rm{O}}_{\rm{3}}}^ - \to 2{\rm{C}}{{\rm{u}}^{2 + }} + 4{e^ - } + {\rm{N}}{{\rm{O}}_{\rm{2}}} + {\rm{NO}} + {\rm{3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

${\rm{2Cu}} + 4{{\rm{H}}^ + } + 2{\rm{HN}}{{\rm{O}}_{\rm{3}}} \to 2{\rm{C}}{{\rm{u}}^{2 + }} + {\rm{N}}{{\rm{O}}_{\rm{2}}} + {\rm{NO}} + {\rm{3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Or

${\rm{2Cu}} + 6{\rm{HN}}{{\rm{O}}_{\rm{3}}} \to 2{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_2} + {\rm{N}}{{\rm{O}}_{\rm{2}}} + {\rm{NO}} + {\rm{3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Therefore, the coefficient of Cu is 2 (x) and ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$
is 6(y).


So, the summation of x and y is 8.

Note: Balanced chemical equation is the chemical equation in which the number of atoms of all elements and charges of both reactant and product side is equal. The principle behind the balanced chemical equation is the law of conservation of mass.