# How to balance Zn + HNO_3 -> Zn(NO_3)_2 + NH_4NO_3 + H_2O?

Jun 23, 2015

$4 Z n + 10 H N {O}_{3} \to 4 Z n {\left(N {O}_{3}\right)}_{2} + N {H}_{4} N {O}_{3} + 3 {H}_{2} O$

#### Explanation:

Start by assigning oxidation numbers to the atoms that take part in the reaction.

$\stackrel{\textcolor{b l u e}{0}}{Z n} + \stackrel{\textcolor{b l u e}{+ 1}}{H} \stackrel{\textcolor{b l u e}{+ 5}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z n} {\left(\stackrel{\textcolor{b l u e}{+ 5}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}}\right)}_{2} + \stackrel{\textcolor{b l u e}{- 3}}{N} \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{4}} \stackrel{\textcolor{b l u e}{- 5}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}} + \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{- 2}}{O}$

Notice that the oxidation state of zinc changes from zero on the reactants' side to +2 on the products' side, which means that it is being oxidized.

Now take a look at nitrogen's oxidation state. Notice that some nitrogen atom keep their oxidation state constant, i.e. +5 on both sides of the equation, but that other nitrogen atoms change their oxidation state from +5 to -3, which implies that they are being reduced.

The oxidation and reduction half reactions look like this

• Oxidation half-reaction

$\stackrel{\textcolor{b l u e}{0}}{Z n} + H N {O}_{3} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z n} {\left(N {O}_{3}\right)}_{2} + 2 {e}^{-}$

Balance the nitrogen atoms first by multiplying the nitric acid by 2.

$\stackrel{\textcolor{b l u e}{0}}{Z n} + 2 H N {O}_{3} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z n} {\left(N {O}_{3}\right)}_{2} + 2 {e}^{-}$

Since you're in acidic solution, you can balance the hydrogen atoms by adding protons, ${H}^{+}$, to the side that lacks hydrogen, and oxygen atoms by adding water molecules to the side that lacks oxygen.

SInce you only need to balance the hydrogen atoms present on the reactants' side, you have

$\stackrel{\textcolor{b l u e}{0}}{Z n} + 2 H N {O}_{3} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z n} {\left(N {O}_{3}\right)}_{2} + 2 {e}^{-} + 2 {H}^{+}$

• Reduction half-reaction

$H \stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3} + 8 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 3}}{N} {H}_{4}^{+}$

Balance the oxygen atoms by adding three water molecules to the products' side.

$H \stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3} + 8 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 3}}{N} {H}_{4}^{+} + 3 {H}_{2} O$

Balance the hydrogen atoms by adding protons to the reactants' side.

$9 {H}^{+} + H \stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3} + 8 {e}^{-} \to \stackrel{\textcolor{b l u e}{- 3}}{N} {H}_{4}^{+} + 3 {H}_{2} O$

In any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

Since you have 4 electrons lost and 8 gained, multiply the oxidation half-reaction by 4 to get

$\left\{\begin{matrix}4 Z n + 8 H N {O}_{3} \to 4 Z n {\left(N {O}_{3}\right)}_{2} + 8 {e}^{-} + 8 {H}^{+} \\ 9 {H}^{+} + H N {O}_{3} + 8 {e}^{-} \to N {H}_{4}^{+} + 3 {H}_{2} O\end{matrix}\right.$

Add these two half-reactions together to get

$4 Z n + 8 H N {O}_{3} + 9 {H}^{+} + H N {O}_{3} + \cancel{8 {e}^{-}} \to 4 Z n {\left(N {O}_{3}\right)}_{2} + \cancel{8 {e}^{-}} + N {H}_{4} N {O}_{3} + 8 {H}^{+} + 3 {H}_{2} O$

If you reduce like terms that are present on both sides of the equation, you'll get

$4 Z n + 9 H N {O}_{3} + {H}^{+} \to 4 Z n {\left(N {O}_{3}\right)}_{2} + N {H}_{4} N {O}_{3} + 3 {H}_{2} O$

SInce you're in acidic solution, the proton present on the reactants' side can only come from the nitric acid, which implies that you actually have 10 molecules of acid, instead of 9.

Thus, the balanced chemical equation is

$4 Z n + 10 H N {O}_{3} \to 4 Z n {\left(N {O}_{3}\right)}_{2} + N {H}_{4} N {O}_{3} + 3 {H}_{2} O$