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Hydrogen peroxide can function either as an oxidizing agent or as a reducing agent. At standard conditions, is \(\mathrm{H}_{2} \mathrm{O}_{2}\) a better oxidizing agent or reducing agent? Explain.

Short Answer

Expert verified
At standard conditions, hydrogen peroxide (H2O2) is a better oxidizing agent than a reducing agent. This is because its standard reduction potential (Eº) for the reduction half-reaction (+1.77 V) is greater than that of the oxidation half-reaction (+0.68 V), indicating a higher tendency to undergo reduction and thus act as an oxidizing agent.

Step by step solution

01

Write the half-reactions

To analyze H2O2's behavior as an oxidizing or reducing agent, we first need to write the redox half-reactions for its oxidation and reduction. Oxidation half-reaction (H2O2 acts as a reducing agent): \[ \mathrm{H}_{2}\mathrm{O}_{2} \to \mathrm{O}_{2} + 2\mathrm{H}^{+} + 2\mathrm{e}^{-} \] Reduction half-reaction (H2O2 acts as an oxidizing agent): \[ \mathrm{H}_{2}\mathrm{O}_{2} + 2\mathrm{H}^{+} + 2\mathrm{e}^{-} \to 2\mathrm{H}_{2}\mathrm{O} \]
02

Find the standard reduction potentials (Eº) for the half-reactions

Next, we need to find the standard reduction potentials (Eº) for these half-reactions. Eº values can be found in standard reduction potential tables. For the oxidation half-reaction (H2O2 acts as a reducing agent): Eº = +0.68 V For the reduction half-reaction (H2O2 acts as an oxidizing agent): Eº = +1.77 V
03

Compare the Eº values to determine if H2O2 is a better oxidizing or reducing agent

Now, we can compare the Eº values for both half-reactions to identify if H2O2 is a better oxidizing or reducing agent. As a general rule, the more positive the Eº value, the greater the tendency for the substance to undergo reduction (and thus act as an oxidizing agent). Since +1.77 V (reduction half-reaction) is greater than +0.68 V (oxidation half-reaction), H2O2 has a higher tendency to undergo reduction. With the given information, we can now conclude that at standard conditions, hydrogen peroxide (H2O2) is a better oxidizing agent than a reducing agent.

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Key Concepts

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

Oxidizing Agents
Oxidizing agents are substances that get reduced in a chemical reaction, meaning they gain electrons from other species. These agents cause other substances to lose electrons, hence the name "oxidizing".

In any redox reaction, the oxidizing agent is essential because it drives the oxidation process by accepting electrons. They are often identified by their high standard reduction potential, which suggests a strong affinity for electrons. Common oxidizing agents include oxygen, chlorine, and hydrogen peroxide (H₂O₂).

Understanding oxidizing agents is crucial because it helps predict the direction of a redox reaction. The stronger the oxidizing agent, the more readily it will gain electrons. This property is quantitatively expressed through the standard reduction potential (Eº), where a more positive Eº value indicates a stronger oxidizing agent.
Reducing Agents
Reducing agents do the opposite of oxidizing agents; they donate electrons to another substance and get oxidized in the process. This means they lose electrons themselves while facilitating the gain of electrons by their reaction partner.

The ability of a substance to act as a reducing agent correlates with its tendency to have a lower (or more negative) standard reduction potential because these substances more readily lose electrons. Metals typically serve as strong reducing agents as they have a propensity to release electrons.

A good understanding of reducing agents is important in various fields such as industrial chemistry and biochemistry, where redox reactions are common. Knowing which substance acts as a reducing agent can help in designing processes that involve electron transfer, ensuring the desired products are obtained.
Standard Reduction Potential
The standard reduction potential is a critical concept in electrochemistry. Represented as Eº, it measures the tendency of a chemical species to be reduced, i.e., to gain electrons. Measured in volts (V), it is determined under standard conditions (1 M concentration, 1 atm pressure, and 25°C).

Each half-reaction has its own Eº value, which is tabulated in standard reduction potential tables. These values allow us to predict the direction in which a redox reaction will naturally proceed.
  • A higher Eº value indicates a stronger oxidizing agent and a greater eagerness to gain electrons.
  • Conversely, a lower Eº value suggests a stronger reducing agent, indicating a greater tendency to lose electrons.
Understanding standard reduction potentials is essential for balancing redox reactions and predicting reaction spontaneity. This concept is vital in fields such as electrolysis, battery design, and corrosion prevention.
Half-Reactions
In redox chemistry, a half-reaction breaks the overall redox reaction into two distinct parts: one where oxidation occurs and another where reduction takes place. Half-reactions help chemists understand the electron transfer process in more detail.

The oxidation half-reaction involves the loss of electrons, whereas the reduction half-reaction involves the gain of electrons. By writing these half-reactions, chemists can see exactly how electrons are transferred from the reducing agent to the oxidizing agent.

Using half-reactions is especially useful when calculating the overall cell potential in electrochemical cells. In our example with hydrogen peroxide, writing the half-reactions helped determine whether H₂O₂ acts more effectively as an oxidizing or reducing agent under standard conditions. This level of analysis is crucial in research and industry for correctly understanding and predicting the outcomes of complex redox reactions.

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Most popular questions from this chapter

Which of the following statement(s) is(are) true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow\) Al and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}\) , aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 M \mathrm{Cd}^{2+}, 1.0 M \mathrm{Ag}^{+}, 1.0 \mathrm{M}\) \(\mathrm{Au}^{3+},\) and 1.0 \(\mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

Consider the following half-reactions: $$\begin{array}{ll}{\mathrm{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.77 \mathrm{V}} \\\ {\mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.73 \mathrm{V}} \\\ {\mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.62 \mathrm{V}}\end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0\(M\) in chloride ion and 0.020\(M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that 99\(\%\) of a metal must be plated out before another metal begins to plate out.)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is 1.62 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

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