Question

For the following half-reactions, answer the questions below: $$ \begin{array}{cc} \mathrm{Ce}^{4+}(a q)+e^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Ni}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Ni}(s) & E^{\circ}=-0.24 \mathrm{~V} \\ \mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) & E^{o}=-1.68 \mathrm{~V} \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Sn}(s)\) reduce \(\mathrm{Ag}^{+}(\mathrm{aq})\) to \(\mathrm{Ag}(s) ?\) (f) Will \(\mathrm{Hg}(l)\) reduce \(\mathrm{Sn}^{2+}(a q)\) to \(\mathrm{Sn}(s) ?\) (g) Which ion(s) can be reduced by \(\operatorname{Sn}(s)\) ? (h) Which metal(s) can be oxidized by \(\mathrm{Ag}^{+}(a q)\) ?

Short answer

Answer: Al³⁺

Step-by-step solution

Find the weakest oxidizing agent

An oxidizing agent is a species that gains electrons and causes the oxidation of another species. The weakest oxidizing agents are the least likely to gain electrons or have the lowest reduction potential. In our case, the species with the lowest reduction potential is Al³⁺ with a value of -1.68 V. Therefore, Al³⁺ is the weakest oxidizing agent. The answer to (a) is Al³⁺.

Find the strongest oxidizing agent

The strongest oxidizing agents are the most likely to gain electrons or have the highest reduction potential. In our case, the species with the highest reduction potential is Ce⁴⁺ with a value of +1.61 V. Therefore, Ce⁴⁺ is the strongest oxidizing agent. The answer to (b) is Ce⁴⁺.

Find the strongest reducing agent

A reducing agent is a species that loses electrons and causes the reduction of another species. The most powerful reducing agents have the lowest reduction potential in their reverse redox reactions. In our case, the reverse reaction of Al³⁺, which is the oxidation of Al(s), has the lowest reduction potential. Thus, Al(s) is the strongest reducing agent. The answer to (c) is Al(s).

Find the weakest reducing agent

The weakest reducing agents have the highest reduction potential in their reverse redox reactions. In our case, the reverse reaction of Ce⁴⁺, which is the oxidation of Ce³⁺, has the highest reduction potential. Thus, Ce³⁺ is the weakest reducing agent. The answer to (d) is Ce³⁺.

Test if Sn(s) will reduce Ag⁺(aq) to Ag(s)

To determine if Sn(s) will reduce Ag⁺(aq) to Ag(s), we need to consider their reduction potentials. If the reduction potential of Sn²⁺/Sn is lower than Ag⁺/Ag, Sn(s) will act as a reducing agent and reduce Ag⁺(aq) to Ag(s). Sn²⁺/Sn has a reduction potential of -0.14 V, while Ag⁺/Ag has one of 0.80 V. Since the reduction potential of Sn²⁺/Sn is lower, Sn(s) will reduce Ag⁺(aq) to Ag(s). The answer to (e) is yes.

Test if Hg(l) will reduce Sn²⁺(aq) to Sn(s)

The same logic applies to determine if Hg(l) will reduce Sn²⁺(aq) to Sn(s). If the reduction potential of Hg²²⁺/Hg is lower than Sn²⁺/Sn, Hg(l) will act as a reducing agent and reduce Sn²⁺(aq) to Sn(s). Hg²²⁺/Hg has a reduction potential of 0.80 V, while Sn²⁺/Sn has one of -0.14 V. Since the reduction potential of Hg²²⁺/Hg is not lower, Hg(l) will not reduce Sn²⁺(aq) to Sn(s). The answer to (f) is no.

Find which ions can be reduced by Sn(s)

Sn(s) can reduce ions with higher reduction potentials than Sn²⁺/Sn. Comparing Sn²⁺/Sn (-0.14 V) with the given species, we find that Sn(s) can reduce Ce⁴⁺(aq) (1.61 V), Ag⁺(aq) (0.80 V), and Hg²²⁺(aq) (0.80 V). The answer to (g) is Ce⁴⁺, Ag⁺, and Hg²²⁺.

Find which metals can be oxidized by Ag⁺(aq)

Ag⁺(aq) can oxidize metals with lower reduction potentials than Ag⁺/Ag. Comparing Ag⁺/Ag (0.80 V) with the given species, we find that Ag⁺(aq) can oxidize Sn(s) (-0.14 V), Ni(s) (-0.24 V), and Al(s) (-1.68 V). The answer to (h) is Sn, Ni, and Al.

Key concepts

Oxidizing Agent

In the realm of redox reactions, an oxidizing agent is the substance that breathes life into the process by gaining electrons. Imagine it as a character in our chemical story that takes electrons away from another, thereby causing that other substance to lose electrons, a process we call oxidation. The strength of an oxidizing agent is best understood through its appetite for electrons, measured by something called the standard reduction potential. A high positive value indicates a strong desire to gain electrons and hence a stronger oxidizing power. For example, in our textbook problem, Ce⁴⁺ with its generous +1.61 V of standard reduction potential is the strongest oxidizing agent, very eager to snag electrons and get reduced to Ce³⁺.

On the other hand, Al³⁺, bearing a less enthusiastic -1.68 V, is seen as the least desirous of electrons, making it our weakest oxidizing agent. This measure helps us predict which substances will be more likely to drive the redox reaction forward by acting as the electron-hungry entity.

Reducing Agent

Reducing agents, the electron donors in the chemistry dance, do the opposite of their oxidizing partners: they give away electrons to others. This generosity allows them to reduce others while they themselves become oxidized. Their strength is inversely related to that of the corresponding oxidized form in the electrochemical series, with a lower or more negative standard reduction potential indicating a mightier reducing prowess. In our exercise, Al(s) is heralded as the strongest reducing agent — it has the lowest potential when its reaction is flipped (-1.68 V), showing a distinct readiness to let go of its electrons.

Ce³⁺, with the highest (though still negative) potential in its reverse reaction, is considered the weakest reducing agent. This helps us to compare different substances and their abilities to push a redox reaction by surrendering electrons.

Standard Reduction Potential

The standard reduction potential (E°) is a numerical value that indicates just how willing a substance is to be reduced. In other words, it tells us how much a substance would like to gain electrons. Measured in volts (V), these values serve as a sort of 'chemical currency' for the electron exchange rate in redox reactions. The higher the positive value, the greater the substance's desire to adopt electrons and be reduced, functioning as an oxidizing agent. Conversely, a lower or negative value indicates a stronger tendency to donate electrons, characteristic of reducing agents.

This concept is crucial for predicting the direction of redox reactions and for solving exercises like ours. For example, the question about whether Sn(s) can reduce Ag⁺ is answered by comparing their potential: Sn(s), at -0.14 V, can reduce Ag⁺, which has a higher potential of +0.80 V.

Electrochemical Series

The electrochemical series is essentially a leaderboard of substances ranked by their standard reduction potential. It's a powerful tool for students to predict the outcome of redox reactions. Substances at the top with higher E° values are strong oxidizing agents, eager to gain electrons. Those at the bottom, with lower or negative E° values, are strong reducing agents, ready to donate electrons.

Let's apply this to our textbook problem. With the electrochemical series, we can easily deduce that Al(s), sitting at lower ranks due to its -1.68 V potential, can be oxidized by just about any substance higher up in the series, such as Ag⁺ with its +0.80 V. Conversely, Ce⁴⁺ (+1.61 V) reigns over most others with its high rank, asserting it can reduce many substances below it in the series. This ranking serves as an invaluable cheat sheet for determining the likelihood of various redox reactions occurring.