In the reaction Zn (s) + NO3 - → (aq) Zn2 + (aq) + NH4 + (aq) which element is reduced?

Answer: N (the Nitrogen)

Explanation:

Reduction refers to a decrease in oxidation number/state due to the gaining of electrons. As such the species that is being reduced will show a decrease in oxidation state.

Based on the redox rules,

Zn (s) has oxidation number of 0 [rule 1: the oxidation number of an element in its free (uncombined) state is zero]

Zn²⁺ has oxidation number of + 2 [rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion]

Now, since Nitrogen is enbedded in a polyatomic ion in both cases, you have to do a bit a calculation to obtain the oxidation state.

For NO₃⁻ : N + (-2 * 3) = - 1

N - 6 = - 1

N = 5

[Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 6: The oxidation state of hydrogen in a compound is usually + 1]

For NH₄⁺:

N + (4 x 1) = 1

N + 4 = 1

N = - 3

[Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 5: The oxidation number of oxygen in a compound is usually - 2]

Therefore, Zn moves from oxidation state of 0 to + 2 (oxidation), while N moves from + 5 to - 3 (reduction).