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Nitric acid is a strong oxidizing agent. State which of the following species is least likely to be produced when nitric acid reacts with a strong reducing agent such as zinc metal, and explain why: \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}, \mathrm{NO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4},\) \(\mathrm{N}_{2} \mathrm{O}_{5}, \mathrm{NH}_{4}^{+}\).

Short Answer

Expert verified
\(NH_4^+\) is least likely because it requires extensive reduction.

Step by step solution

01

Understand the Oxidation States

To solve this problem, first consider the oxidation states of nitrogen in the different nitrogen oxides and ions provided: \(N_2O\) (-1), \(NO\) (+2), \(NO_2\) (+4), \(N_2O_4\) (+4), \(N_2O_5\) (+5), and \(NH_4^+\) (-3).
02

Assess Likely Redox Pathways

Next, understand that in a reaction with a strong reducing agent such as zinc, nitric acid will undergo reduction. Generally, more oxidized forms of nitrogen (higher oxidation states) are reduced to less oxidized states.
03

Analyze Each Product

Evaluate each potential product. Lower oxidation states indicate a greater likelihood of being a reduction product with a strong reducing agent. \(NH_4^+\) has an oxidation state of -3, which is far lower than others and indicates complete reduction.
04

Identify Unlikely Product Formation

Considering the significant reduction required to form \(NH_4^+\) from +5 (in nitric acid), \(NH_4^+\) is less likely to be generated due to extensive reduction. Compounds where nitrogen is in a higher oxidation state are more probable products.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nitric Acid in Redox Reactions
Nitric acid is well-known for its strong oxidizing properties. When we refer to it as an oxidizing agent, it means that nitric acid tends to accept electrons from other substances. In redox reactions, nitric acid gets reduced by gaining electrons, while the other substance, often a metal or reducing agent, gets oxidized by losing electrons.

In a typical reaction involving nitric acid and a reducing agent like zinc, the nitric acid itself changes, often leading to different products such as nitrogen oxides. The oxidation state of nitrogen in nitric acid is +5, which indicates it's a highly oxidized form of nitrogen. This means when it interacts with a reducing agent, it is prone to being reduced to a lower oxidation state form, thus leading to products with lower oxidation states than +5. Understanding these changes helps predict the likelihood of forming specific nitrogen-containing products.
Reducing Agent Properties
A reducing agent, such as zinc, plays a crucial role in redox reactions by donating electrons to another substance. In this process, the reducing agent itself gets oxidized. When zinc acts as a reducing agent in reactions with nitric acid, it can significantly alter the oxidation state of nitrogen.

It’s important to recognize that the strength of the reducing agent determines the extent of the reduction in the other substance. Zinc, being relatively reactive, is capable of transferring electrons effectively to nitric acid, causing the nitric acid to convert into various forms of lower oxidation states of nitrogen.
  • Oxidation state changes: Indicates how reduced the nitrogen becomes.
  • Strength of reducing agent: Determines the extent of electron transfer.
When assessing potential outcomes of such reactions, consider both the oxidation states and the ability of zinc to maintain transformation into less oxidized forms of nitrogen.
Basics of Redox Reactions
Redox, short for reduction-oxidation reactions, involves the transfer of electrons between two substances. One substance gets oxidized (loses electrons) while the other gets reduced (gains electrons).

Redox reactions are characterized by changes in oxidation states of the reactants involved. In the context of nitric acid reacting with zinc, the nitric acid is reduced by receiving electrons from the zinc. As each electron transfer occurs, the oxidation state of nitrogen in nitric acid transitions from +5 toward lower values depending on the potential final product.
  • Oxidation involves loss of electrons.
  • Reduction involves gain of electrons.
  • The relative positions of substances in a redox reaction impact the outcome.
By assessing these exchanges, one can determine the likely products formed, such as nitrogen oxides with lower oxidation states, and predict the unlikely production of more extensively reduced forms like ammonium ions in the presence of very strong reducing conditions.

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Most popular questions from this chapter

Give the oxidation numbers for the underlined atoms in the following molecules and ions: (a) \(\mathrm{ClF},(\mathrm{b}) \mathrm{IF}_{7}\) (c) \(\underline{\mathrm{C}} \mathrm{H}_{4}\) (d) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{2}\) (e) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{4}\) (f) \(\mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4},(\mathrm{~g}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (h) \(\mathrm{KMnO}_{4}\), (i) \(\mathrm{NaHCO}_{3},(\mathrm{j}) \mathrm{Li}_{2},(\mathrm{k}) \mathrm{NaIO}_{3},\) (I) \(\mathrm{KO}_{2}\), \((\mathrm{m}) \mathrm{PF}_{6}^{-},(\mathrm{n}) \mathrm{K} \mathrm{AuCl}_{4}\)

How many moles of \(\mathrm{MgCl}_{2}\) are present in \(60.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{MgCl}_{2}\) solution?

A sample of \(0.6760 \mathrm{~g}\) of an unknown compound containing barium ions \(\left(\mathrm{Ba}^{2+}\right)\) is dissolved in water and treated with an excess of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). If the mass of the \(\mathrm{BaSO}_{4}\) precipitate formed is \(0.4105 \mathrm{~g}\), what is the percent by mass of Ba in the original unknown compound?

Give the oxidation numbers for the underlined atoms in the following molecules and ions: (a) \(\mathrm{Mg}_{3} \mathrm{~N}_{2},\) (b) \(\mathrm{Cs} \underline{\mathrm{O}}_{2},\) (c) \(\mathrm{Ca} \underline{\mathrm{C}}_{2}\) (d) \(\mathrm{CO}_{3}^{2-}\), (e) \(\underline{\mathrm{C}}_{2} \mathrm{O}_{4}^{2-}\) (f) \(\mathrm{ZnO}_{2}^{2-},(\mathrm{g}) \mathrm{Na} \underline{\mathrm{B}} \mathrm{H}_{4}\) (h) \(\underline{\mathrm{W}} \mathrm{O}_{4}^{2-}\)

A volume of \(35.2 \mathrm{~mL}\) of a \(1.66 \mathrm{M} \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of a \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

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