Question

for the reaction, $N{{H}_{3}}+OC{{l}^{-}}\to {{N}_{2}}{{H}_{4}}+C{{l}^{-}}$ occurring in basic medium, the coefficient of ${{N}_{2}}{{H}_{4}}$ in the balanced equation will be:
A. 1
B.2
C. 3
D. 4

Answer 1

Hint: Basic medium in any reaction involves the reaction with a base, means that the hydroxide ions are taking part in the chemical equation. A balanced chemical equation tells the stoichiometric coefficients of each species from the reactant as well as the product. so the equation will be balanced in basic medium.

Complete answer:
A balanced chemical equation consists of the same number of the atoms of the reacting species in the product side. The coefficients of the balanced equations tell us the exact amount of that substance obtained in the product, or used in the form of reactant. If a reaction takes place in the basic medium, this means that hydroxide, $O{{H}^{-}}$, ions are added on the side where there is deficiency of negative charges or ions. We can balance the chemical equation in the basic medium, through half reaction method. This method involves an equation to be separately balanced in the reduction and the oxidation half. Then we can find the coefficient of ${{N}_{2}}{{H}_{4}}$.
The reaction given is $N{{H}_{3}}+OC{{l}^{-}}\to {{N}_{2}}{{H}_{4}}+C{{l}^{-}}$, in this reaction, ammonia is being oxidized while, chlorite ion is being reduced.
Writing the oxidation half of the reaction with same amount of reactant and product we have, $2N{{H}_{3}}\to {{N}_{2}}{{H}_{4}}+2{{e}^{-}}$
Now writing the reduction half, $OC{{l}^{-}}+2{{e}^{-}}\to C{{l}^{-}}$
Since, the medium is basic, we will add hydroxide ions on the sides where there is deficiency in negative charge, so,
For oxidation, $2N{{H}_{3}}+2O{{H}^{-}}\to {{N}_{2}}{{H}_{4}}+2{{e}^{-}}$,
For reduction, $OC{{l}^{-}}+2{{e}^{-}}\to C{{l}^{-}}+2O{{H}^{-}}$.
Now, the oxygen atoms in both equations are balanced by adding water molecule, so the equations will become,
$2N{{H}_{3}}+2O{{H}^{-}}\to {{N}_{2}}{{H}_{4}}+2{{e}^{-}}+2{{H}_{2}}O$ , and $OC{{l}^{-}}+2{{e}^{-}}+{{H}_{2}}O\to C{{l}^{-}}+2O{{H}^{-}}$.
Now, cancelling out the similar species, which are 2 electrons, same number of hydroxide ions and 1 water molecule from both the equations, and combining both the half equations, we will get the final equation, as
$2N{{H}_{3}}+OC{{l}^{-}}\to {{N}_{2}}{{H}_{4}}+C{{l}^{-}}+{{H}_{2}}O$, from this balanced equation it can be seen that coefficient of ${{N}_{2}}{{H}_{4}}$is 1.

Hence the coefficient of ${{N}_{2}}{{H}_{4}}$is found to be 1 in a balanced equation, so option A is correct.

Note:
Half reaction method of balancing is used to balance redox reactions, in which self oxidation and reduction occurs. The basic medium provides$O{{H}^{-}}$ions which are negative and fills the deficiency of negative charges, while in the acidic medium hydrogen ions, ${{H}^{+}}$ are added to compensate the deficiency of positive ions.