A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following halfreaction: $$\mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)$$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

Short Answer

Expert verified
The cathode of the cell is the electrode with a higher concentration of \(Cl^-\) ions, which is electrode 2 with a \(2.55 M\) concentration. The standard emf is 0 volts. The cell emf for the given concentrations is approximately \(0.126 V\), and as the cell operates, the concentration of \(Cl^-\) ions will increase at the anode and decrease at the cathode.

Step by step solution

01

Identify the Cathode

In a voltaic cell, the electrode with the higher concentration of reducing agent ions (in this case Cl-) acts as the cathode, as it has a greater tendency to be reduced. Since the concentration of Cl- ions is higher in the electrode 2 with \(2.55M\), this electrode is the cathode of the cell.
02

Calculate the Standard EMF

For the silver-silver chloride electrode, the half reaction is: \[\mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)\] The standard emf (electromotive force) for this cell is 0 volts because both electrodes have the same half-reaction potentials.
03

Calculate Cell EMF for Given Concentrations

Using the Nernst equation, we can calculate the cell EMF for given concentrations: \[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\] Since the standard cell EMF is 0, and Q, the reaction quotient, can be calculated as \(Q = \frac{[Cl^-]_{cathode}}{[Cl^-]_{anode}}\), we get: \[E_{cell} = -\frac{RT}{nF}\ln \frac{[Cl^-]_{cathode}}{[Cl^-]_{anode}}\] We are given the concentrations in the problem, as well as that R is the ideal gas constant, T is the temperature, n is the number of moles of electrons transferred, and F is Faraday's constant. Assuming standard conditions, we have \(R = 8.314 J K^{-1} mol^{-1}\), \(T = 298K\), \(n = 1\), and \(F = 96485 C mol^{-1}\). Plugging these values into the equation, we get: \[E_{cell} = -\frac{8.314 * 298}{1 * 96485}\ln \frac{2.55}{0.0150} \approx 0.126 V\]
04

Predict Changes in Cl- Concentration

As the cell operates, there occurs a transfer of electrons from the anode to the cathode, which can be seen from the half-reaction, and thus a transfer of silver ions from the anode to the cathode. This means that the concentration of Cl- ions at the anode will increase, while at the cathode, it will decrease. In summary: - Anode: \([\mathrm{Cl}^{-}]\) will increase. - Cathode: \([\mathrm{Cl}^{-}]\) will decrease.

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Key Concepts

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

Electrochemical Cell
An electrochemical cell is a fundamental component in the realm of chemistry that enables the conversion of chemical energy into electrical energy through redox reactions. Essentially, this device harnesses the power of electrons moving between two electrodes: an anode and a cathode.

At the anode, oxidation occurs where electrons are lost, and at the cathode, reduction takes place with electrons being gained. These half-reactions are separated into different compartments called half-cells that are often linked by a salt bridge or porous membrane that allows ions to move freely, balancing the electric charge while preventing the two solutions from mixing completely.

In the voltaic cell example from the exercise, silver chloride electrodes are used to induce the half-reactions, and the movement of electrons from the anode to the cathode through an external circuit is what generates electricity.
Nernst Equation
The Nernst equation is an expression in physical chemistry that relates the reduction potential of a half-cell in an electrochemical cell to the standard electrode potential, temperature, activity, and concentrations of the chemical species undergoing reduction and oxidation. It is instrumental in determining the cell's electromotive force (EMF) under non-standard conditions.

The equation is given by: \begin{center} \(E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\), \begin{center}where \(E_{cell}\) is the cell EMF, \(E^0_{cell}\) is the standard EMF, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.

Applying the Nernst equation in our example, you can understand how the cell voltage changes with concentration differences of chloride ions, as is demonstrated when calculating the EMF for the given concentrations of chloride ions.
Electromotive Force (EMF)
Electromotive force, better known as EMF, represents the energy provided by a cell or a battery per coulomb of charge passing through it. It's the driving force that pushes electrons through the circuit, often measured in volts. The EMF is determined by the nature of the electrodes and the electrolyte.

Standard EMF \(E^0_{cell}\) references conditions with each solute at 1 molar concentration and gases at 1 atmosphere pressure. However, real-life scenarios often deviate from these ideal conditions, necessitating calculations like in the Nernst equation to find the actual EMF as substances change concentration during the cell's operation, such as the case in the exercise provided.
Half-Reaction Potential
Half-reaction potentials are the voltages associated with each half-reaction occurring at the electrodes during the operation of an electrochemical cell. These potentials, when measured under standard conditions, determine whether a substance can be readily oxidized or reduced.

Understanding the half-reaction potentials is crucial for predicting the behavior of each electrode, especially in identifying the cathode and the anode in a cell, as seen in the provided exercise. The half-reaction potentials help us understand that the electrode with the higher tendency to gain electrons (higher reduction potential) will serve as the cathode where reduction takes place.
Concentration Cell
A concentration cell is a form of electrochemical cell where both electrodes are made of the same material but are immersed in solutions of differing concentrations. The EMF of the cell is generated due to the concentration gradient between the two half-cells.

The operation and behavior of concentration cells can be explored using Nernst equation since the standard EMF \(E^0_{cell}\) would be zero given identical electrodes. In our exercise example, the voltaic cell created with silver-silver chloride electrodes in different chloride ion concentrations can be considered a concentration cell. As the cell operates, the concentration of chloride ions will equilibrate between the two half-cells, driving the potential to diminish over time.

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Most popular questions from this chapter

A voltaic cell utilizes the following reaction: $$\mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s)$$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode half-cell, increasing the quantity of \(\mathrm{Ag}^{+}\) but not changing its concentration. (d) \(\mathrm{HCl}\) is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\) as \(\mathrm{AgCl}\).

Using the standard reduction potentials listed in Appendix \(\mathrm{E}_{2}\) calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : $$ \begin{array}{l} \text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s) \\ \text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g) \\ \text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l) \end{array} $$

Metallic magnesium can be made by the electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) What mass of \(\mathrm{Mg}\) is formed by passing a current of 4.55 A through molten \(\mathrm{MgCl}_{2}\), for 4.50 days? (b) How many minutes are needed to plate out \(25.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\) using \(3.50 \mathrm{~A}\) of current?

Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(a q),\) (b) \(\mathrm{Ca}(s),\) (c) \(\mathrm{ClO}_{3}^{-}(a q)_{\text {, }}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ?

(a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?

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