Question

The stoichiometric coefficient n in the following reaction is:
$n{{H}_{2}}C{{O}_{2}}+2KMn{{O}_{4}}\to nC{{O}_{2}}+{{K}_{2}}O+2MnO+{{H}_{2}}O$

Answer 1

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

Complete answer:
A balanced equation is an equation in which the coefficients of atoms for each element of the products is 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.
$H_{2}^{+1}{{C}^{+2}}O_{2}^{-2}+{{K}^{+1}}M{{n}^{+7}}O_{4}^{-2}\to {{C}^{+4}}O_{2}^{-2}+K_{2}^{+1}{{O}^{-2}}+M{{n}^{+2}}{{O}^{-2}}+H_{2}^{+1}{{O}^{-2}}$
- Identify the atoms which are being reduced and oxidized.
Here, we can see that the Mn atom is reduced from $M{{n}^{+7}}$ to $M{{n}^{+2}}$ (gain of 5 electrons) and the carbon atom is oxidized from ${{C}^{+2}}$ to ${{C}^{+4}}$ (loss of 2 electrons).

& M{{n}^{+7}}+5{{e}^{-}}\to M{{n}^{+2}}\text{ (reduction)} \\\ & {{C}^{+2}}-2{{e}^{-}}\to {{C}^{+4}}\text{ (oxidation)} \\\ \end{aligned}$$ \- Now in a balanced redox reaction, the number of electrons gained is equal to the number of electrons lost. So, to equalize the change in the oxidation number we need to multiply the oxidation reaction with 5 and the reduction reaction with 2. $$\begin{aligned} & 2M{{n}^{+7}}\xrightarrow{+10{{e}^{-}}}2M{{n}^{+2}}\text{ (reduction)} \\\ & 5{{C}^{+2}}\xrightarrow{-10{{e}^{-}}}5{{C}^{+4}}\text{ (oxidation)} \\\ \end{aligned}$$ \- Insert these coefficients in the reaction equation. $$5{{H}_{2}}C{{O}_{2}}+2KMn{{O}_{4}}\to 5C{{O}_{2}}+{{K}_{2}}O+2MnO+{{H}_{2}}O$$ \- Place required coefficients for molecules that do not participate in the redox process. Since the hydrogen atoms on the reactants side are 10, we multiply the water molecule coefficient by 5. $$5{{H}_{2}}C{{O}_{2}}+2KMn{{O}_{4}}\to 5C{{O}_{2}}+{{K}_{2}}O+2MnO+5{{H}_{2}}O$$ Since all the atoms and the charges are equalized, the equation is balanced. So, we can see that the stoichiometric coefficient n in the reaction is 5. **Note:** It should be noted that there are two methods of balancing a redox reaction. The half-reaction method is preferred over the oxidation number method in a reaction in which the substances are in the aqueous solution.