Identify the oxidizing and reducing agents in the reaction \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}\)

Before identifying the oxidizing and reducing agents, we need to assign oxidation numbers to each element in the reaction. Use the rules for determining oxidation numbers to do this: \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}\) Oxidation numbers: \(\mathrm{N} = -3\), \(\mathrm{H} = +1\), \(\mathrm{Cr} = +6\) in \(\mathrm{Cr}_{2}\mathrm{O}_{7}\), \(\mathrm{Cr} = +3\) in \(\mathrm{Cr}_{2}\mathrm{O}_{3}\), \(\mathrm{O} = -2\)

Now that we have assigned oxidation numbers, we can determine which elements undergo a change in oxidation state during the reaction: N goes from -3 to 0 (increases by 3), Cr goes from +6 to +3 (decreases by 3).

The element that loses electrons (increase in oxidation state) is the reducing agent, while the element that gains electrons (decrease in oxidation state) is the oxidizing agent. In our reaction, N loses electrons (oxidation state increases by 3), making it the reducing agent. Cr gains electrons (oxidation state decreases by 3), making it the oxidizing agent. It is important to note that the entire compound containing the element should be considered as the oxidizing or reducing agent. Thus, the reducing agent is \(\mathrm{NH}_{4}\), and the oxidizing agent is \(\mathrm{Cr}_{2}\mathrm{O}_{7}\). Finally, we can write our conclusion: The reducing agent in the given reaction is \(\mathrm{NH}_{4}\), and the oxidizing agent is \(\mathrm{Cr}_{2}\mathrm{O}_{7}\).