Problem 102 - Chemistry: The Central Science in SI Units, Global Edition

Question

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{l} \mathrm{Ag}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{array} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Ag}^{+}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) With \(\left[\mathrm{Ag}^{+}\right]\) equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

Step-by-Step Solution

Answer

The short version of the answer: a) The cell potential at these concentrations is \(0.357V\). b) The anode of the cell is the iodine electrode. c) Yes, the answer for part (b) is the same as it would be if the cell were operated under standard conditions. d) At [Ag+] = \(0.15M\), the cell would have zero potential when [I⁻] = \(0.218M\).

Step 1: Identifying the half-cell reactions

For each half-cell reaction, we can determine the standard reduction potentials. Using these we can find the overall cell reaction and calculate the standard cell potential, E°. Half-cell reactions: 1. \(Ag^+(aq) + 2e^- -> Ag(s)\) - Silver Half-cell 2. \(I_2(s) + 2e^- -> 2I^-(aq)\) - Iodine Half-cell

Step 2: Calculate the standard cell potential (E°)

Access the standard reduction potentials for each half-cell reaction from a table. We find: 1. \(E°_{Ag} = +0.80V\) 2. \(E°_{I2} = +0.54V\) As the Silver half-cell has higher reduction potential, it is more likely to be reduced. Therefore, the Iodine half-cell will be oxidized and thus, our silver electrode will be the cathode, and our iodine electrode will be the anode. The standard cell potential can be calculated as follows: \(E°_{cell} = E°_{cathode} - E°_{anode} = E°_{Ag} - E°_{I2} = 0.80V - 0.54V = 0.26V\)

Step 3: Apply the Nernst Equation to find the cell potential (E)

The Nernst Equation is given by \(E = E° - \frac{RT}{nF} \ln Q\), where \(E\) is the cell potential, \(E°\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of transferred electrons, \(F\) is the Faraday's constant, and \(Q\) is the reaction quotient. For our cell, we have \(T=298K\), \(n=2\), and \(R=8.314J.mol^{-1}.K^{-1}\), and \(F = 96,485 C.mol^{-1}\). We can also find Q using the concentrations given. Q can be calculated as follows: \(Q = \frac{[Ag^+]^2}{[I^-]^2}\). Now, substitute the known values into the Nernst Equation and solve for E: \(E = 0.26V - \frac{8.314*298}{2*96485} \ln{\frac{0.15^2}{0.035^2}}\)

Step 4: Calculate cell potential (E)

On solving the Nernst Equation, we get the cell potential E: \(E \approx 0.357V\) Our final answers are: a) The cell potential at these concentrations is 0.357V. b) The anode of the cell is the iodine electrode. c) Yes, the answer for part (b) is the same as it would be if the cell were operated under standard conditions. d) To find the concentration of I⁻ when the cell has zero potential, set E = 0 in the Nernst Equation and solve for [I⁻]: \(0 = 0.26V - \frac{8.314*298}{2*96485} \ln{\frac{0.15^2}{[I^-]^2}}\) On solving this equation for [I⁻], we find: \([I^-] \approx 0.218M\) Therefore, with [Ag+] = 0.15M, the cell would have zero potential when [I⁻] = 0.218M.

Key Concepts

Standard Reduction Potentials

In the world of electrochemistry, standard reduction potentials play a pivotal role. Think of these as a measure of the tendency of a chemical species to gain electrons and be reduced. Each half-reaction has its own standard reduction potential, usually found in tables, and it tells us how easily a species can be reduced compared to the standard hydrogen electrode. Standard reduction potentials are measured under specific conditions: 1 M concentration, 1 atm pressure, and 298 K temperature. In our voltaic cell exercise, the silver and iodine half-reactions have standard reduction potentials of +0.80 V and +0.54 V, respectively. The silver reaction has a higher potential, indicating it is more likely to gain electrons and be reduced. This higher reduction potential of silver makes the silver electrode our cathode, where the reduction occurs.To calculate the \[E°_{cell}\], the standard cell potential, we use the formula:\[E°_{cell} = E°_{cathode} - E°_{anode}\]In this case, it results in 0.80 V - 0.54 V = 0.26 V. This potential is crucial for understanding the overall ability of the voltaic cell to perform work under standard conditions.

Nernst Equation

The Nernst Equation is a powerful tool to determine the potential of an electrochemical cell under non-standard conditions. It adjusts the standard potential to account for the actual concentrations of the reactants and products present in the cell. This is immensely helpful in practical applications, where conditions rarely match the "standard" setup.Here's the equation:\[E = E° - \frac{RT}{nF} \ln{Q}\]Where:- \(E\) is the cell potential under non-standard conditions.- \(E°\) is the standard cell potential.- \(R\) is the gas constant (8.314 J/mol·K).- \(T\) is the temperature in Kelvin.- \(n\) is the number of moles of electrons transferred.- \(F\) is Farday's constant (96,485 C/mol).- \(Q\) is the reaction quotient, describing the ratio of concentrations of products over reactants.For our voltaic cell, we used the concentrations given: \([Ag^+] = 0.15 M\) and \([I^-] = 0.035 M\). The reaction quotient, \(Q\), is found using:\[Q = \frac{[Ag^+]^2}{[I^-]^2}\]Substituting all this into the Nernst Equation gives a cell potential \(E\) of approximately 0.357 V. This showcases how concentration changes impact cell potential, emphasizing the dynamic nature of real-world conditions.

Electrode Potentials

Electrode potentials define the voltage or electromotive force (emf) developed by a cell. They depict the cell's ability to drive the flow of electric current through an external circuit. When we consider a voltaic cell, there are two electrodes involved: the cathode and the anode, each with differing potentials. The cathode has a higher electrode potential, meaning it attracts electrons. As we've seen in this particular scenario, the silver electrode serves as the cathode due to its higher standard reduction potential when compared to iodine. On the other hand, the anode hosts oxidation, where electrons are released. In our case, the iodine electrode plays this role. Electrode potential essentially describes these tendencies and the ensuing electron journey between these terminals. Using these basic principles, we can understand why the cell operates the way it does. The potential difference between the cathode and anode is what makes the flow of electrons possible, and hence, the cell's ability to do work. Understanding electrode potentials helps in predicting the direction of electron flow, assessing the feasibility of reactions, and in practical applications like designing batteries and electroplating processes. Together, they form the backbone of various electrochemical technologies.