Question

Given a reaction: $N{O}_3^{\ominus }+Zn⟶{Zn}^{2+}+N{H}_4^{\oplus }$
The balanced reaction is $N{O}_3^{\ominus }+7H_{2}O+4Zn⟶N{H}_4^{\oplus }+10\stackrel{\ominus }{O}H+4{Zn}^{2+}$.
If true enter 1, else enter 0.

• A. 1

Answer 1

The correct option is A 1
The unbalanced chemical equation is as follows:
$N{O}_3^{\ominus }+Zn⟶{Zn}^{2+}+N{H}_4^{\oplus }$
All atoms other than H and O are balanced.
The oxidation number of Zn changes from 0 to +2. Thus, it is oxidized.
The oxidation number of N changes from +5 to -3. Thus, it is reduced.
To balance the change in the oxidation number, multiply the Zn and $Z{n}^{2+}$ ions with coefficient 4.
$N{O}_3^{\ominus }+4Zn⟶4{Zn}^{2+}+N{H}_4^{\oplus }$
Add 3 water molecules on RHS to balance O atoms.
$N{O}_3^{\ominus }+4Zn⟶4{Zn}^{2+}+N{H}_4^{\oplus }+3H_{2}O$
Add ten protons on LHS to balance H atoms.
$N{O}_3^{\ominus }+4Zn+10H^{+}⟶4{Zn}^{2+}+N{H}_4^{\oplus }+3H_{2}O$
Since, the reaction occurs in basic medium, add 10 hydroxide ions on both sides.
$N{O}_3^{\ominus }+4Zn+10H^{+}+10O{H}^{-}⟶4{Zn}^{2+}+N{H}_4^{\oplus }+3H_{2}O+10O{H}^{-}$
On LHS 10 protons and 10 hydroxide ions combine to form 10 water molecules.
Three water molecules cancel on LHS and RHS.
The balanced reaction is as follows:
$N{O}_3^{\ominus }+7H_{2}O+4Zn⟶N{H}_4^{\oplus }+10\stackrel{\ominus }{O}H+4{Zn}^{2+}$