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The following electrochemical cell has been set up: \(\mathrm{Pt}(\mathrm{s}) \mid \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}(\mathrm{a}=1) \| \mathrm{Ce}^{4+}\) \(\mathrm{Ce}^{3+}(\mathrm{a}=1) \mid \mathrm{Pt}(\mathrm{s}) ; E^{\circ}\left(\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}\right)=0.77 \mathrm{~V}\) \(E^{\circ}\left(\mathrm{Ce}^{4} \mid \mathrm{Ce}^{3+}\right)=1.61 \mathrm{~V} .\) If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase on decrease with time? (a) Ce electrode to \(\mathrm{Fe}\) electrode, decrease (b) \(\mathrm{Ce}\) electrode to \(\mathrm{Fe}\) electrode, increase (c) \(\mathrm{Fe}\) electrode to \(\mathrm{Ce}\) electrode, decrease (d) Fe electrode to \(\mathrm{Ce}\) electrode, increase

Short Answer

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Ce electrode to Fe electrode, decrease

Step by step solution

01

Identify the Anode and Cathode

The cell reaction proceeds spontaneously in the direction where electrons flow from the anode (oxidation) to the cathode (reduction). To determine which half-cell acts as the anode and which as the cathode, compare their standard electrode potentials (E° values). The half-cell with the higher E° value will act as the cathode because it has a greater tendency to gain electrons (reduction), and the one with the lower E° value will act as the anode because it has a greater tendency to lose electrons (oxidation). According to the given standard electrode potentials, 1.61 V for \(\mathrm{Ce}^{4+} | \mathrm{Ce}^{3+}\) is greater than 0.77 V for \(\mathrm{Fe}^{3+} | \mathrm{Fe}^{2+}\). Therefore, cerium (Ce) electrode is the cathode and the iron (Fe) electrode is the anode.
02

Predict Direction of Current Flow

Now that we have established that the cerium electrode is the cathode and the iron electrode is the anode, we can predict the flow of current. Electrons flow from anode to cathode externally, so the current will flow from the cerium electrode to the iron electrode, as conventional current flow is opposite to electron flow.
03

Determine the Change in Current Over Time

Over time, the concentrations of the ions will change due to the cell reactions. At the anode, \(\mathrm{Fe}^{3+}\) ions are produced as \(\mathrm{Fe}^{2+}\) ions are consumed. At the cathode, \(\mathrm{Ce}^{4+}\) ions are consumed as \(\mathrm{Ce}^{3+}\) ions are produced. Since the activities are initially 1 for both ions and the reactions will move the system away from standard conditions, the cell potential will decrease, causing the current to decrease with time as well.

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Key Concepts

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

Standard Electrode Potentials
In the realm of electrochemistry, standard electrode potentials (E° values) are indispensable as they provide a quantitative measure of the tendency of a chemical species to lose or gain electrons. These potentials, expressed in volts, are measured under standard conditions, which include a solute concentration of 1 mole per liter, a pressure of 1 atmosphere for gases, and a temperature of 25°C.

Considering a typical electrochemical cell set-up, the electrode with the higher E° value is deemed as the more favorable site for reduction, hence it becomes the cathode. Consequently, the electrode with the lower E° value is prone to oxidation and becomes the anode. The difference in E° values between the cathode and anode provides the electromotive force, or cell potential, driving the electron flow through an external circuit.

Standard electrode potentials also allow us to predict the direction and feasibility of redox reactions, and these predictions are crucial for constructing electrochemical cells, batteries, and understanding corrosion.
Redox Reactions
The heartbeat of an electrochemical cell is the redox reaction, a chemical process involving a transfer of electrons between two substances. Redox reactions are split into two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained.

Within the electrochemical cell, the substance that loses electrons—typically at the anode—undergoes oxidation, while the substance that gains electrons—typically at the cathode—undergoes reduction. These half-reactions occur at separate electrodes but are connected by the cell's medium, allowing for the movement of ions.

Understanding redox reactions is also central to various applications beyond electrochemical cells, including metallurgy, photography, and energy storage technologies. Grasping how electrons are transferred in these reactions allows students to comprehend numerous biological and geological processes.
Electrochemical Cell Current Flow
Delving into how current flows in an electrochemical cell unveils the practical applications of these devices—from powering small electronics to large-scale energy storage systems. By convention, current is described as the flow of positive charge. Therefore, in an electrochemical cell, the current flows from the positive electrode (cathode) to the negative electrode (anode).

In our specific exercise, we observe this principle in action. Electrons move from the iron (Fe) electrode where oxidation occurs to the cerium (Ce) electrode where reduction happens. However, since conventional current is considered the flow of positive charge, we describe the current as moving in the opposite direction—from Ce to Fe. Understanding this concept is crucial for students as it lays the foundation for circuit design and comprehension of electronic devices.
Anode and Cathode Identification
Identifying the anode and cathode in an electrochemical cell is key to unlocking its functionality. The anode is the electrode where oxidation takes place, meaning it is where electrons are donated to the external circuit. The cathode, on the contrary, is where reduction happens, as it accepts electrons from the external circuit.

In the given exercise, by using their standard electrode potentials, we determined the Ce electrode as the cathode and the Fe electrode as the anode. Over time, the varied consumption and production of ions at these electrodes lead to changes in ion concentration, affecting the cell potential and, consequently, the current flow. Such dynamics are not merely academic; they’re mirrored in real-world technologies like rechargeable batteries, where anode and cathode roles can swap depending on the charging or discharging state.

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Most popular questions from this chapter

A Tl'|Tlcouple was prepared by saturating \(0.10 \mathrm{M}-\mathrm{KBr}\) with TIBr and allowing \(\mathrm{Tl}^{+}\) ions form the insoluble bromide to equilibrate. This couple was observed to have a potential \(-0.444 \mathrm{~V}\) with respect to \(\mathrm{Pb}^{2+} \mid \mathrm{Pb}\) couple in which \(\mathrm{Pb}^{2+}\) was \(0.10 \mathrm{M} .\) What is the \(K_{\mathrm{sp}}\) of TIBr. [Given: \(E_{\mathrm{Pb}^{2+} \mid \mathrm{Pb}}^{\circ}=-0.126 \mathrm{~V}, E_{\mathrm{T}^{+} \mid \mathrm{T}}^{o}=-0.336 \mathrm{~V}\) \(\log 2.5=0.4,2.303 R T / F=0.06]\) (a) \(4.0 \times 10^{-6}\) (b) \(2.5 \times 10^{-4}\) (c) \(4.0 \times 10^{-5}\) (d) \(6.3 \times 10^{-3}\)

The best conductor of electricity is a \(0.1 \mathrm{M}\) solution of (a) Boric acid (b) Sulphuric acid (c) Acetic acid (d) Propanoic acid

The electrode through which electrons enter the electrolytic solution is (a) cathode (b) anode (c) may be anode or cathode (d) both, anode and cathode

The conductivity of saturated solution of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(1.2 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1} .\) The limiting equivalent conductivities of \(\mathrm{BaCl}_{2}, \mathrm{~K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KCl}\) are 160,140 and \(100 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\), respectively. The solubility product of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), is (a) \(10^{-5}\) (b) \(1.08 \times 10^{-23}\) (c) \(1.08 \times 10^{-25}\) (d) \(1.08 \times 10^{-27}\)

Assuming that hydrogen behaves as an ideal gas, what is the EMF of the cell at \(25^{\circ} \mathrm{C}\) if \(P_{1}=600 \mathrm{~mm}\) and \(P_{2}=420 \mathrm{~mm}\) : \(\mathrm{Pt}\left|\mathrm{H}_{2}\left(P_{1}\right)\right| \mathrm{HCl}\left|\mathrm{H}_{2}\left(P_{2}\right)\right| \mathrm{Pt} ?\) [Given: \(2.303 R T / F=0.06, \log 7=0.85]\) (a) \(-0.0045 \mathrm{~V}\) (b) \(-0.0 \mathrm{~V}\) (c) \(+0.0045 \mathrm{~V}\) (d) \(+0.0015 \mathrm{~V}\)

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