To complete and balance the given chemical equation:

$$Zn + NO_3^- \to Zn^{2+} + NH_4^+$$

We start by looking at the oxidation and reduction processes separately. In this reaction:

• Zinc (Zn) is oxidized to zinc ions ($$Zn^{2+}$$).
• The nitrate ion ($$NO_3^-$$) is reduced to ammonium ions ($$NH_4^+$$).

To balance this equation, we should keep in mind that both mass and charge must be conserved. Therefore, we have to balance not only the atoms but also the charges throughout the equation.

First, let's write the half-reactions:

• Oxidation (loss of electrons): $$Zn \to Zn^{2+} + 2e^-$$
• Reduction (gain of electrons): Since nitrate ($$NO_3^-$$) is reducing to ammonia ($$NH_4^+$$), let's assume a general reduction half-reaction that will be balanced later: $$NO_3^- + xH^+ + ye^- \to NH_4^+ + zH_2O$$

Now, balancing the reduction half-reaction involves getting the nitrogen from a +5 oxidation state in $$NO_3^-$$ to a -3 state in $$NH_4^+$$. This process consumes 8 electrons (e-) to reduce each nitrogen atom, also hydrogen ions ($$H^+$$) are required to form $$NH_4^+$$, and water ($$H_2O$$) is a product due to the oxygen from the nitrate.

The correct balanced reduction half-reaction, considering water as the source of $$H^+$$ and the formation of $$H_2O$$, is:

$$NO_3^- + 10H^+ + 8e^- \to NH_4^+ + 3H_2O$$

Thus, combining our half-reactions and ensuring the electrons are balanced:

• For the zinc oxidation: $$Zn \to Zn^{2+} + 2e^-$$
• For the reduction of nitrate: $$NO_3^- + 10H^+ + 8e^- \to NH_4^+ + 3H_2O$$

Since the reduction half-reaction consumes 8 electrons and the oxidation produces only 2, we need to multiply the oxidation half-reaction by 4 to balance the electrons:

$$4Zn \to 4Zn^{2+} + 8e^-$$

The full reaction, adding the two half-reactions together and balancing mass and charge, now looks like this:

$$4Zn + 10H^+ + NO_3^- \to 4Zn^{2+} + NH_4^+ + 3H_2O$$

It is important to note that this equation assumes acidic conditions due to the presence of $$H^+$$ ions. However, the original equation did not specify the reaction environment, so this balanced equation is based on typical conditions that allow for $$NO_3^-$$ reduction to $$NH_4^+$$.