Question

In balancing the reaction
$xZn+NO_{3}^{-}+y{{H}^{+}}\to xZ{{n}^{2+}}+NH_{4}^{+}+z{{H}_{2}}O$
x, y, and z are respectively:
(A) 4, 10, 3
(B) 3, 8, 3
(C) 3, 10, 3
(D) 4, 3, 10

Answer 1

To solve this question, we first need to balance this equation. This is a type of redox reaction. In a redox reaction or an oxidation-reduction, the electrons are transferred between two species.

Complete answer:
A balanced equation is an equation in which the coefficients of atoms for each element of the products are equal to the coefficients of atoms for each element of the reactants.
Let us balance the equation using the half-reaction method.
- Write each atom's oxidation number above the atom in the equation.
$Z{{n}^{0}}+{{({{N}^{+5}}O_{3}^{2-})}^{-}}+{{H}^{1+}}\to Z{{n}^{2+}}+{{({{N}^{-3}}H_{4}^{1+})}^{+}}+H_{2}^{1+}{{O}^{2-}}$
- Then the atoms which are being oxidized and reduced are identified.
Here, we can see that the Zn atom is oxidized from $Z{{n}^{0}}$ to $Z{{n}^{2+}}$ i.e., loss of two electrons. And N atoms are reduced from ${{N}^{+5}}$ to ${{N}^{-3}}$ i.e., the gain of 8 electrons.

& Z{{n}^{0}}-2{{e}^{-}}\to Z{{n}^{2+}}\text{ (oxidation)} \\\ & {{N}^{+5}}+8{{e}^{-}}\to {{N}^{3-}}\text{ (reduction)} \\\ \end{aligned}$$ \- Now in a balanced redox reaction, the number of electrons gained is equal to the number of electrons lost. So, to equalize the change in the oxidation number we need to multiply the oxidation reaction by 4. $$\begin{aligned} & 4Z{{n}^{0}}\xrightarrow{-8{{e}^{-}}}4Z{{n}^{2+}}\text{ (oxidation)} \\\ & {{N}^{+5}}\xrightarrow{+8{{e}^{-}}}{{N}^{3-}}\text{ (reduction)} \\\ \end{aligned}$$ \- Insert these coefficients in the reaction equation. $$4Zn+NO_{3}^{-}+{{H}^{+}}\to 4Z{{n}^{2+}}+NH_{4}^{+}+{{H}_{2}}O$$ \- Place required coefficients for molecules that do not participate in the redox process. Since the oxygen atoms on the reactants side are 3, we multiply the water molecule coefficient by 3. To equate the hydrogen atoms on both sides, we multiply the reactant hydrogen ion by 10. $$4Zn+NO_{3}^{-}+10{{H}^{+}}\to 4Z{{n}^{2+}}+NH_{4}^{+}+3{{H}_{2}}O$$ Since all the atoms and the charges are equalized, the equation is balanced. **So, in balancing the reaction x, y, and z are respectively option (A) 4, 10, 3.** **Note:** It should be noted that there are two methods of balancing a redox reaction. In a reaction in which the substances are in the aqueous solution, the half-reaction method is preferred over the oxidation number method.