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Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: $$ \begin{array}{lr} \hline \text { Reduction Half-Reaction } & \multicolumn{1}{c} {E^{\circ}(\mathbf{V})} \\ \hline \mathrm{A}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{A}(s) & 1.33 \\\ \mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s) & 0.87 \\\ \mathrm{C}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{C}^{2+}(a q) & -0.12 \\ \mathrm{D}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{D}(s) & -1.59 \\\ \hline \end{array} $$ (a) Which substance is the strongest oxidizing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize \(\mathrm{C}^{2+}\) ? [Sections 20.4 and 20.5\(]\)

Short Answer

Expert verified
(a) The strongest oxidizing agent is A鈦(aq), and the weakest oxidizing agent is D鲁鈦(aq). (b) The strongest reducing agent is D(s), and the weakest reducing agent is A(s). (c) The substances that can oxidize C虏鈦 are A鈦(aq) and B虏鈦(aq).

Step by step solution

01

(a) Identifying the strongest and weakest oxidizing agents

To find the strongest and weakest oxidizing agents, we need to look at the reduction half-reactions and their corresponding standard electrode potentials, \(E^\circ\). The strongest oxidizing agent will have the highest reduction potential, while the weakest oxidizing agent will have the lowest reduction potential. So, we can compare \(E^\circ\) values and find out the strongest and weakest oxidizing agents: - A鈦(aq) + e鈦 鈫 A(s), \(E^\circ = 1.33 \ V\) - B虏鈦(aq) + 2e鈦 鈫 B(s), \(E^\circ = 0.87 \ V\) - C鲁鈦(aq) + e鈦 鈫 C虏鈦(aq), \(E^\circ = -0.12 \ V\) - D鲁鈦(aq) + 3e鈦 鈫 D(s), \(E^\circ = -1.59 \ V\) The strongest oxidizing agent is A鈦(aq), and the weakest oxidizing agent is D鲁鈦(aq).
02

(b) Identifying the strongest and weakest reducing agents

To find the strongest and weakest reducing agents, we need to consider the reverse of the reduction half-reactions, which are the oxidation half-reactions. The strongest reducing agent will have the lowest reduction potential (or the highest negative oxidation potential), while the weakest reducing agent will have the highest reduction potential (or the lowest negative oxidation potential) when written as an oxidation process. So, we can compare \(E^\circ\) values (but with opposite signs) and find out the strongest and weakest reducing agents: - A(s) 鈫 A鈦(aq) + e鈦 , \(E^\circ = -1.33 \ V\) (opposite sign of A鈦 reduction) - B(s) 鈫 B虏鈦(aq) + 2e鈦 , \(E^\circ = -0.87 \ V\) (opposite sign of B虏鈦 reduction) - C虏鈦(aq) 鈫 C鲁鈦(aq) + e鈦 , \(E^\circ = 0.12 \ V\) (opposite sign of C鲁鈦 reduction) - D(s) 鈫 D鲁鈦(aq) + 3e鈦 , \(E^\circ = 1.59 \ V\) (opposite sign of D鲁鈦 reduction) The strongest reducing agent is D(s), and the weakest reducing agent is A(s).
03

(c) Identifying substances that can oxidize C虏鈦

To identify the substance(s) that can oxidize \(\mathrm{C}^{2+}\), we must find a substance that has a higher reduction potential than the reduction potential of the reverse reaction of \(\mathrm{C}^{2+}\) to \(\mathrm{C}^{3+}\). This means we are looking for a reaction where \(\Delta E^\circ = E_r^\circ - E_ox^\circ > 0\) Comparing \(\mathrm{C}^{2+}\) oxidation potential \(0.12 \ V\) with other reduction potentials: - A鈦(aq): \(1.33 \ V > 0.12 \ V\) - B虏鈦(aq): \(0.87 \ V > 0.12 \ V\) - D鲁鈦(aq): \(-1.59 \ V < 0.12 \ V\) Both A鈦(aq) and B虏鈦(aq) have higher reduction potentials than the oxidation potential of \(\mathrm{C}^{2+}\). Therefore, the substances that can oxidize \(\mathrm{C}^{2+}\) are A鈦(aq) and B虏鈦(aq).

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Key Concepts

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

Oxidizing and Reducing Agents
Understanding the role of oxidizing and reducing agents in electrochemistry is pivotal for analyzing reactions and determining their feasibility. An oxidizing agent, often called an oxidant, gains electrons in a reaction. As it accepts electrons, it causes another species to be oxidized and is therefore reduced itself. Conversely, a reducing agent, or reductant, donates electrons and thereby reduces the other substance and is itself oxidized.

In the given exercise where we have a series of standard electrode potentials for hypothetical reactions, the substances with the higher positive electrode potentials, such as A鈦(aq) with an E鈦 value of 1.33 V, are strong oxidizers because they have a greater tendency to gain electrons. In contrast, substances with negative standards potentials, like D鲁鈦(aq) with -1.59 V, are strong reducers because they are more inclined to donate electrons. This tendency is crucial for predicting the direction of electron flow and the spontaneity of redox reactions.
Half-Reactions in Electrochemistry
At the heart of electrochemical processes are half-reactions. These are the two parts of a whole redox reaction, where one substance loses electrons (oxidation) and another gains them (reduction). Electrochemistry splits these into reduction half-reactions, showing the gain of electrons, and oxidation half-reactions, showing the loss.

In the table provided in the exercise, the reduction half-reactions are listed with their associated standard electrode potentials. To analyze an oxidation half-reaction, one must reverse the reduction half-reaction and change the sign of the standard potential, as seen in the solution step discussing reducing agents. For example, the reduction half-reaction for A is A鈦(aq) + e鈦 鈫 A(s), and the corresponding oxidation half-reaction is A(s) 鈫 A鈦(aq) + e鈦 with an E鈦 that is the negative of the original (from +1.33 V to -1.33 V). Half-reactions are integral to understanding the electron transfer process and are essential when balancing redox reactions.
Electrochemical Cell Potentials
The electrochemical cell potential, represented by E鈦, measures the ability of a reaction to move electrons鈥攖hat is, its propensity to occur spontaneously as an electrochemical reaction. These potentials are given under standard conditions, typically designated at 1 M concentration, 1 atm pressure, and 298 K (25 掳C). A positive E鈦 indicates a reaction that tends to occur spontaneously, while a negative E鈦 suggests non-spontaneity under standard conditions.

To determine whether a specific reaction, like the oxidation of \(\mathrm{C}^{2+}\), can occur, we look for other reaction partners with higher reduction potentials. In our exercise, we compare the reduction potential of \(\mathrm{C}^{2+}\) with the reduction potentials of other species. A鈦(aq) and B虏鈦(aq) both have higher standard potentials than the reverse oxidation of \(\mathrm{C}^{2+}\); thus, they can oxidize it, resulting in an overall cell potential which is positive and signifies a spontaneous reaction.

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Most popular questions from this chapter

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {red }}^{\circ}\).

(a) What happens to the emf of a battery as it is used? Why does this happen? (b) The AA-size and D-size alkaline batteries are both \(1.5-\mathrm{V}\) batteries that are based on the same electrode reactions. What is the major difference between the two batteries? What performance feature is most affected by this difference?

Using the standard reduction potentials listed in Appendix \(\mathrm{E}\), calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : $$ \text { (a) } \mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ 3 \mathrm{Ce}^{3+}(a q)+\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q) $$ (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}{ }^{3-}(a q) \longrightarrow\) $$ \mathrm{N}_{2}(g)+5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q) $$

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

(a) Write the reactions for the discharge and charge of a nickelcadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+\begin{array}{c} 2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{~V} \end{array} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

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