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Chapter 1: Problem 11

For the decomposition of hydrogen peroxide, which element (if any) is being reduced, and which is being oxidized? \(\begin{array}{|c|c|c|}\hline & {\text { Oxidized }} & {\text { Reduced }} \\\ \hline \text { (A) Hydrogen } & {\text { Oxygen }} \\ \hline \text { (B) Oxygen } & {\text { None }} \\ \hline \text { (C) None }\\\ \hline \text { (D) Oxygen} & {\text { Oxygen}} \\ \hline\end{array}\)

Short Answer

Expert verified
The element oxygen (O) is being oxidized. None of the elements in this reaction are being reduced. So, the answer is (B) Oxygen is oxidized and None is reduced.

Step by step solution

01

Identify Initial and Final Oxidation States

In the reactant hydrogen peroxide (H2O2), hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -1. In the products water (H2O) and oxygen (O2), hydrogen still has an oxidation state of +1, but oxygen has an oxidation state of 0 in O2.
02

Compare Oxidation States

Comparing the initial and final oxidation states of both hydrogen and oxygen, we can see that the oxidation state of hydrogen does not change (+1 to +1), but the oxidation state of oxygen changes from -1 in H2O2 to 0 in O2.
03

Identify Oxidation and Reduction

As the oxidation state of oxygen increased (from -1 to 0), this means it has lost electrons and is therefore oxidized. Since there is no reduction in oxidation state, the reduction process does not occur, meaning that 'None' is being reduced.

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Key Concepts

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

Oxidation States
Understanding oxidation states is crucial for deciphering redox reactions. An oxidation state, also known as an oxidation number, reflects the total number of electrons that an atom gains or loses to form a chemical bond. These states help track how electrons move within a compound during reactions.

When determining oxidation states, it helps to follow these key guidelines:
  • The oxidation state of a free element (like O鈧� or H鈧�) is always 0.
  • For monoatomic ions, the oxidation state is equal to the ion's charge.
  • In compounds, hydrogen typically has an oxidation state of +1, and oxygen usually has a state of -2. Exceptions exist, such as in peroxides like H鈧侽鈧�, where oxygen has a -1 oxidation state.
  • The sum of oxidation states for all atoms in a neutral molecule must equal 0, while in ions, it should equal the ion's charge.
These principles allow us to analyze how atoms undergo oxidation (increase in oxidation state) or reduction (decrease in oxidation state) during reactions.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are fundamental in chemistry where electron transfer occurs between substances. In any redox reaction, one substance loses electrons, and another gains electrons. This process involves:
  • Oxidation: The loss of electrons by a molecule, atom, or ion, resulting in an increase in oxidation state.
  • Reduction: The gain of electrons by a molecule, atom, or ion, leading to a decrease in oxidation state.
These reactions are crucial as they underpin many biological and industrial processes. Identifying the substance being oxidized and the one being reduced is essential to solve redox problems.

In the decomposition of hydrogen peroxide, for example, oxygen exhibits a change in oxidation state from -1 in H鈧侽鈧� to 0 in O鈧�, showing it undergoes oxidation by losing electrons. Here, no reduction occurs, indicating there is no simultaneous electron gain by another element.
Decomposition of Hydrogen Peroxide
The decomposition of hydrogen peroxide ( H鈧侽鈧�) is an excellent example of a redox reaction in which only one element undergoes a change in oxidation state. Hydrogen peroxide breaks down into water ( H鈧侽) and oxygen gas ( O鈧�), showcasing a simple yet crucial chemical transformation.

Starting with H鈧侽鈧�:
  • Hydrogen maintains an oxidation state of +1 before and after the reaction.
  • Oxygen changes from an oxidation state of -1 to 0.
  • The molecule undergoes a redox reaction with oxygen being oxidized, while hydrogen sees no change in its oxidation state.
These changes highlight the oxidizing nature of hydrogen peroxide, making it useful for cleaning and antiseptic purposes.

Understanding decomposition reactions like that of hydrogen peroxide aids in grasping broader chemical concepts. It demonstrates how molecules can spontaneously transform, releasing products that we utilize daily.

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Most popular questions from this chapter

Which gas has the strongest IMFs? (A) He (B) Ne (C) NO (D) All gases have identical IMFs.

The following reaction is found to be at equilibrium at 25掳C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause a reduction in the value for the equilibrium constant? (A) Increasing the amount of \(\mathrm{SO}_{3}\) (B) Reducing the amount of \(\mathrm{O}_{2}\) (C) Raising the temperature (D) Lowering the temperature

Which substance would have the highest boiling point? (A) Ethanol, because it is the most asymmetrical (B) Acetone, because of the double bond (C) Ethylene glycol, because it has the most hydrogen bonding (D) All three substances would have very similar boiling points because their molar masses are similar.

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

A solution of \(\mathrm{Co}^{2+}\) ions appears red when viewed under white light. Which of the following statements is true about the solution? (A) A spectrophotometer set to the wavelength of red light would read a high absorbance. (B) If the solution is diluted, the amount of light reflected by the solution will decrease. (C) All light with a frequency that is lower than that of red light will be absorbed by it. (D) Electronic transmissions within the solution match the wavelength of red light.
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