Question

Complete and balance the following reactions:
$Zn+N{{O}_{3}}^{-}\to Z{{n}^{2+}}+N{{H}_{4}}^{+}$

Answer 1

Hint: Balancing and completion of an ionic equation requires identification of reduction and oxidation half of the equation. Reduction half is the part which gains electrons, and oxidation part is the one which loses it.
-We follow a set of rules to balance a chemical equation, which includes balancing of the charges and oxygen as well as hydrogen atoms separately in a step by step manner. At the end of this process we add up both the oxidation as well as reduction half to get the complete equation.

Complete answer:
There are certain rules which we follow, while balancing an ionic chemical equation.
At first we write the net ionic equation for the unbalanced reaction. If we are given a word equation for the work of balancing, we'll need to be able to identify insoluble compounds, weak electrolytes, and strong electrolytes. Strong electrolytes are those which dissociate completely into their ions in water. For instance, strong bases, strong acids, as well as soluble salts. Weak electrolytes dissociate very few ions in aqueous solution, so they are denoted by their molecular formula.
In the next step, we separate the net ionic equation into the two half-reactions. This means the identification and separation of the reaction into a reduction and oxidation half-reaction.
Now we focus on one of the half-reactions, and balance the atoms except for $O$ and $H$. We want the same number of atoms of every element on both of the sides of the equation.
Then we will repeat this with the other half-reaction.
Add \[{{H}_{2}}O\] in order to balance the number of $O$ atoms. Add \[{{H}^{+}}\] in order to balance the $H$ atoms. The atomic mass of the atoms should balance out now.
Now we balance the charge present in both sides. Add \[{{e}^{-}}\](electrons) to one side of each half-reaction in order to balance charges. You may need to multiply the electrons by the two half-reactions to get the charge to balance out. It's fine to change coefficients as long as you change them on both sides of the equation.
Then we must add the two half-reactions together. And then inspect the final equation in order to make sure that it is well balanced. Electrons or the charges present on both sides of the ionic equation should cancel out.
The value of oxidation state of each element is shown below,
\[\overset{0}{\mathop{Zn}}\,~+~\overset{+5}{\mathop{N}}\,{{\overset{^{-2}}{\mathop{{{O}^{-}}}}\,}_{3}}~\to~\overset{-3}{\mathop{N}}\,\overset{^{+1}}{\mathop{{{H}_{4}}^{+}}}\,~+~\overset{+2}{\mathop{Z{{n}^{2+}}}}\,\]
Now we will write the individual redox couples involved in the reaction
\[Zn\to ~Z{{n}^{2+}}~+{ }2{{e}^{-}}\], which is the oxidation half and
\[N{{O}_{3}}^{-}~+{ }8{{e}^{-}}~\to ~N{{H}_{4}}^{+}\], is the reduction half of the reaction.
In the next step we will balance the charges in each of the half reactions,
\[Zn\to ~Z{{n}^{2+}}~+{ }2{{e}^{-}}\] as we already know this is the oxidation half and it is already balanced.
\[N{{O}_{3}}^{-}~+{ }8{{e}^{-}}~+{ }10{{H}^{+}}~\to ~N{{H}_{4}}^{+}\], we balanced the charge by adding \[{{H}^{+}}\] ion so that both sides have the equal value of total charge to the side deficient in positive charge.
And then we will Balance the oxygen atoms by adding water molecules.
\[N{{O}_{3}}^{-}~+{ }8{{e}^{-}}~+{ }10{{H}^{+}}~\to ~N{{H}_{4}}^{+}~+{ }3{{H}_{2}}O\]
We wrote the reduction half only, because the oxidation half will remain the same.
Now we Make gain of electron equivalent to loss of electron,
\[4Zn~\to ~4Z{{n}^{2+}}~+{ }8{{e}^{-}}\], we multiplied the oxidation half with four, in order to make the number of electrons eight.
\[N{{O}_{3}}^{-}~+{ }8{{e}^{-}}~+{ }10{{H}^{+}}~\to ~N{{H}_{4}}^{+}~+{ }3{{H}_{2}}O\]
Now finally we will add both the half equations,
\[4Zn0+~N{{O}_{3}}^{-}~+{ }8{{e}^{-}}~+{ }10{{H}^{+}}~\to ~4Z{{n}^{2+}}~+~N{{H}_{4}}^{+}~+{ }8{{e}^{-}}~+{ }3{{H}_{2}}O\]
Now we cancel out the eight electrons on both the sides to get,
$4Zn+N{{O}_{3}}^{-}+10{{H}^{+}}\to 4Z{{n}^{2+}}+N{{H}_{4}}^{+}+3{{H}_{2}}O$.
Which is the final complete and balanced equation.

Note:To complete and balance a given chemical equation, we divide the equation into oxidation and reduction half. Then balance all the elements except oxygen and hydrogen on both the sides.
-Then we write their oxidation states and balance the number of oxygen by adding water molecules and the number of hydrogen by adding ${{H}^{+}}$ ion. Finally we balance the number of electrons, and add both half reactions and cancel out the electrons on both the sides.