Question

Given the reaction $N{{O}_{3}}^{-}+Zn\to Z{{n}^{2+}}+N{{H}_{4}}^{+}$
The balanced chemical reaction is $N{{O}_{3}}^{-}+7{{H}_{2}}O+4Zn\to N{{H}_{4}}^{+}+10O{{H}^{-}}+4Z{{n}^{2+}}$
A. True
B. False

Answer 1

Hint: Think about how ionic reactions are balanced and how the charges and excess atoms of hydrogen and oxygen are accounted for. If, after balancing the given reaction, the answer turns out to be the second reaction given, then the answer is true.

Complete answer:
Here, note which atom is getting oxidized and which is getting reduced. The oxidation state of $Zn$ is changing from 0 to 2+. The oxidation state of $N$ is changing from 5 to -3. Now, the oxidation state of $N$ is changing by 8 units, but the oxidation state of $Zn$ is changing only by 2 units.
Thus, we have to multiply the zinc atoms on both sides by 4 to equate the oxidation states.
$N{{O}_{3}}^{-}+4Zn\to 4Z{{n}^{2+}}+N{{H}_{4}}^{+}$
Now, we have to equate the $O$ atoms. To do this, add 3 molecules of water to the right-hand side.
$N{{O}_{3}}^{-}+4Zn\to 4Z{{n}^{2+}}+N{{H}_{4}}^{+}+3{{H}_{2}}O$
Now, to balance the $H$ atoms, add 10 protons to the left-hand side.
$N{{O}_{3}}^{-}+4Zn+10{{H}^{+}}\to 4Z{{n}^{2+}}+N{{H}_{4}}^{+}+3{{H}_{2}}O$
Since, the reaction is occurring in a basic medium, add 10 hydroxide ions to both sides and check if the charge is balanced.
$N{{O}_{3}}^{-}+4Zn+10{{H}^{+}}+10O{{H}^{-}}\to 4Z{{n}^{2+}}+N{{H}_{4}}^{+}+3{{H}_{2}}O+10O{{H}^{-}}$
The protons and the hydroxide ions on the left-hand side will combine to form water. Three of them will be cancelled with the three water molecules on the right-hand side. Thus, the net balanced reaction will be:
$N{{O}_{3}}^{-}+4Zn+7{{H}_{2}}O\to 4Z{{n}^{2+}}+N{{H}_{4}}^{+}+10O{{H}^{-}}$
If we compare this reaction to the reaction given in the problem, we will find that it is the same.
Hence the answer is ‘A. True’

Note: Be careful while balancing the oxidation states. The net exchange of electrons should remain equal at both the oxidation and the reduction end. Here, nitrogen was gaining 8 electrons, but zinc was losing only 2 electrons, so we multiplied the zinc atoms on both sides by 4 so that they would lose 8 electrons altogether.