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Chapter 20: Problem 35

A voltaic cell that uses the reaction $$ \mathrm{T1}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tr}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) $$ has a measured standard cell potential of \(+1.19 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(\mathrm{E}\), determine \(E_{\mathrm{ed}}^{0}\) for the reduction of \(\mathrm{Ti}^{3+}(a q)\) to \(\mathrm{Ti}^{+}(a q)\). (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

Short Answer

Expert verified
The half-cell reactions are: 1. Reduction reaction: \(\mathrm{Ti}^{3+}(aq) + 2e^{-} \rightarrow \mathrm{Ti}^{+}(aq)\) 2. Oxidation reaction: \(2 \mathrm{Cr}^{2+}(aq) \rightarrow 2 \mathrm{Cr}^{3+}(aq) + 2e^{-}\) The standard electrode potential for the reduction of \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\) is \(1.60~V\). The sketch of the voltaic cell includes labeled anode (Cr) and cathode (Ti), electrolytic solutions, and a salt bridge. The direction of electron flow is from the anode to the cathode, represented by an arrow pointing from the Cr half-cell to the Ti half-cell.

Step by step solution

01

Identify the half-cell reactions

The given overall redox reaction is: $$ \mathrm{Ti}^{3+} (aq) + 2 \mathrm{Cr}^{2+} (aq) \longrightarrow \mathrm{Ti}^{+}(aq) + 2 \mathrm{Cr}^{3+}(aq) $$ We can break this down into two half-cell reactions: 1. Reduction reaction (for the species gaining electrons): \(\mathrm{Ti}^{3+}(aq) + 2e^{-} \rightarrow \mathrm{Ti}^{+}(aq)\) 2. Oxidation reaction (for the species losing electrons): \(2 \mathrm{Cr}^{2+}(aq) \rightarrow 2 \mathrm{Cr}^{3+}(aq) + 2e^{-}\)
02

Determine the standard electrode potential for \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\) reduction reaction

From the problem, we know that the standard cell potential (\(E^0_{\text{cell}}\)) for the voltaic cell is \(+1.19~V\). According to the cell potential formula, we have: $$ E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}} $$ We can find \(E^0_{\text{Cr}^{2+}/\text{Cr}^{3+}}\) from Appendix E and use it to find the required standard electrode potential. According to Appendix E, the standard potential for \(\text{Cr}^{3+}(aq) + e^- \rightarrow \text{Cr}^{2+}(aq)\) reduction is \(E^0_{\text{Cr}^{2+}/\text{Cr}^{3+}} = -0.41~V\). Since the reduction potential is given, we can find the oxidation potential by changing the sign: $$ E^0_{\text{Cr}^{2+}/\text{Cr}^{3+}} (\text{oxidation}) = +0.41~V $$ Now, using the formula for cell potential, we can find the required standard electrode potential for the reduction of \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\). Since the reduction of \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\) occurs at the cathode, and the oxidation of \(\mathrm{Cr}^{2+}(aq)\) to \(\mathrm{Cr}^{3+}(aq)\) occurs at the anode, we substitute the values in the formula: $$ 1.19~V = E^0_{\text{Ti}^{3+}/\text{Ti}^{+}} - 0.41~V $$ Solving for \(E^0_{\text{Ti}^{3+}/\text{Ti}^{+}}\): $$ E^0_{\text{Ti}^{3+}/\text{Ti}^{+}} = 1.19~V + 0.41~V = 1.60~V $$ So the standard electrode potential for the reduction of \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\) is \(1.60~V\).
03

Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow

To sketch the voltaic cell, we need to represent the electrodes (anode and cathode), the electrolytic solutions, and the salt bridge. 1. Anode: The oxidation of \(\mathrm{Cr}^{2+}(aq)\) to \(\mathrm{Cr}^{3+}(aq)\) takes place at the anode. In the sketch, label this electrode as "Anode (Cr)" and note that its half-cell reaction is \(2 \mathrm{Cr}^{2+} (aq) \rightarrow 2 \mathrm{Cr}^{3+} (aq) + 2 e^{-}\). 2. Cathode: The reduction of \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\) takes place at the cathode. In the sketch, label this electrode as "Cathode (Ti)" and note that its half-cell reaction is \(\mathrm{Ti}^{3+} (aq) + 2 e^{-} \rightarrow \mathrm{Ti}^{+}(aq)\). 3. Electrolytic solutions: Represent the solutions in beakers where the anode and cathode are immersed. Indicate the concentration of ions in the solution, i.e., \(\text{Cr}^{3+}(aq), \text{Cr}^{2+}(aq)\) near the anode and \(\text{Ti}^{3+}(aq), \text{Ti}^{+}(aq)\) near the cathode. 4. Salt bridge: Connect the two beakers with a U-shaped tube filled with an inert electrolyte solution, such as KCl or KNO3. Direction of electron flow: Electrons flow from the anode to the cathode (from the Cr half-cell to the Ti half-cell), which can be represented by an arrow pointing from the anode to the cathode in the sketch.

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Key Concepts

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

Understanding Standard Cell Potential
The standard cell potential, often represented by the notation \(E^0_{cell}\), plays a crucial role in the field of electrochemistry, particularly when it comes to understanding the workings of a voltaic cell, also known as a galvanic cell. This potential is a measure of the electromotive force (EMF) of a voltaic cell when all reactants and products are in their standard states, typically at a concentration of 1M and a pressure of 1 atm.

More simply put, the standard cell potential tells us the maximum voltage the cell can produce under ideal conditions. It is calculated by taking the difference between the standard electrode potentials of the cathode (reduction) and anode (oxidation) half-cells. According to the formula \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\), a positive cell potential indicates a spontaneous reaction, while a negative value suggests a non-spontaneous reaction.

This concept is pivotal because it helps determine the feasibility and direction of the chemical reaction within the cell without actually performing the experiment. It also provides insight into the energy efficiency of the cell in generating electrical power from a chemical reaction.
Deciphering Half-Cell Reactions
Half-cell reactions are the individual processes that occur at each electrode in a voltaic cell. These reactions are essential for understanding the inner workings of electrochemical cells as each half-cell hosts one of the two key reactions 鈥� oxidation or reduction.

In oxidation, a species loses electrons and its oxidation state increases. Conversely, reduction involves a species gaining electrons, resulting in a decrease in oxidation state. Together, these complementary processes constitute the overall redox reaction driving the cell's operation. Writing out the half-cell reactions allows us to see exactly how electrons are transferred during the electrochemical process.

For instance, in the reduction reaction \(\mathrm{Ti}^{3+}(aq) + 2e^{-} \rightarrow \mathrm{Ti}^{+}(aq)\), titanium ions receive electrons and are reduced. Meanwhile, in the oxidation reaction \(2 \mathrm{Cr}^{2+}(aq) \rightarrow 2 \mathrm{Cr}^{3+}(aq) + 2e^{-}\), chromium ions release electrons and are oxidized. These paired reactions exemplify the interconnected nature of oxidation and reduction in maintaining the flow of electrons in the cell.
Standard Electrode Potential Essentials
Standard electrode potential, denoted as \(E^0\), is a fundamental principle that quantifies the tendency of a chemical species to gain or lose electrons. It is measured under standard conditions of pressure and concentration and reflects the intrinsic property of the electrode material.

It's important to realize that \(E^0\) values by themselves are not absolute; they're measured relative to the standard hydrogen electrode, which has an assigned potential of zero volts. In practice, these values are crucial to predict the direction of electron flow in the cell and to calculate the cell's overall potential. For example, a higher standard electrode potential signifies a greater likelihood for reduction to occur, making it the cathode in a voltaic cell.

In the exercise, we calculated the standard electrode potential for the reduction of \(\mathrm{Ti}^{3+}(aq)\) to \(\mathrm{Ti}^{+}(aq)\) as \(1.60~V\). This value suggests that titanium in this oxidation state has a significant tendency to gain electrons and undergo reduction.
Determining Electron Flow Direction in Voltaic Cells
The direction of electron flow in a voltaic cell is an integral aspect of its functionality. Electrons flow from the anode, where oxidation occurs, to the cathode, the site of reduction. This flow is driven by the potential difference between the two half-cells, and it's the movement of these electrons through an external circuit that can be harnessed as electrical energy.

In our example, the electron flow direction is from chromium to titanium, meaning electrons exit the anode where chromium ions are oxidized, and travel towards the cathode where titanium ions are reduced. The understanding of electron flow is not only academic; it's key to the design and implementation of electrochemical cells in practical applications, such as batteries and corrosion prevention systems.

Remember, the electrons move through the external circuit, while positively charged ions move within the cell, from the anode to the cathode through the salt bridge to maintain charge balance. This comprehensive picture enables us to construct a correct and functional representation of the voltaic cell.

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Most popular questions from this chapter

An iron object is plated with a coating of cobalt to protect against corrosion. Does the cobalt protect iron by cathodic protection? Explain.

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), have an atypical oxidation state.) (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) Exercises 901 (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{HCO}_{2} \mathrm{H}(a q)+\) \(\mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{BrO}_{3}{ }^{-}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{N}_{2}(g)\) (acidic solution) (e) \(\mathrm{NO}_{2}{ }^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{AlO}_{2}{ }^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution) Voltaic Cells (Section 20.3)

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\mathrm{e}}=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\text {red }}^{e}=+0.49 \mathrm{~V} \end{array} $$ (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.

A common shorthand way to represent a voltaic cell is anode | anode solution || cathode solution |cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such AS from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fe \(\left|\mathrm{Fe}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag}_{\mathrm{g}}\) sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{H}^{+}\right| \mathrm{H}_{3}\) s sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\)and Cl. Sketch the cell.

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(\mathrm{g})\) (acidic solution) (d) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (acidic solution) (e) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(\) aq \() \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}\) (aq) (basic solution)
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