Problem 80 - Chemistry An Atoms First Approach

Question

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Step-by-Step Solution

Answer

The potential of this cell at \(25°C\), with given concentrations of \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\), is approximately \(1.47\,\text{V}\) (part a). The concentration of \(\mathrm{Al^{3+}}\) in the unknown solution when the measured cell potential at \(25°C\) is \(1.62\,\text{V}\) is approximately \(6.5 \times 10 ^{-4}\,\text{M}\) (part b).

Step 1: Identify the half-reactions and standard electrode potentials

To find the potential of this electrochemical cell, first, we need to identify the half-reactions occurring at the nickel and aluminum electrodes and their standard electrode potentials (E°). From a table of standard electrode potentials, we have: For Nickel: \[Ni^{2+} + 2e^- \rightleftharpoons Ni(s); E°_{Ni} = -0.26\,\text{V}\] For Aluminum: \[ Al^{3+} + 3e^- \rightleftharpoons Al(s); E°_{Al} = -1.66\,\text{V}\]

Step 2: Calculate the standard cell potential

In order to calculate the potential of this cell under standard conditions, we need to calculate the cell potential using the standard electrode potentials we found in step 1. The cell potential under standard conditions, \(E°_{cell}\) is the difference between the standard electrode potentials of the reduction and oxidation half-reactions: \[E°_{cell} = E°_{cathode} - E°_{anode}\] As aluminum has the more negative standard reduction potential, it will be oxidized, and nickel will be reduced. \[E°_{cell} = (-0.26\,\text{V}) - (-1.66\,\text{V})\]

Step 3: Calculate the actual cell potential at given concentrations

To find the actual cell potential at given concentrations of \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\), we use the Nernst equation: \[E_{cell} = E°_{cell} - \frac{RT}{nF} \ln \frac{[\text{Reductant}]}{[\text{Oxidant}]}\] At \(25°C\), \(R = 8.314\,\text{J mol}^{-1}\,\text{K}^{-1}\), \(T = 298.15\,\text{K}\), and \(F = 96,500\,\text{C mol}^{-1}\). Using the Nernst Equation for part (a), \[E_{cell} = E°_{cell} - \frac{(8.314)(298.15)}{6(96500)} \ln \frac{ \left(1.0\,\text{M} \right)}{\left(7.2 \times 10^{-3}\,\text{M} \right)}\]

Step 4: Evaluate and display the results for part (a)

Now, we will calculate the cell potential, \[E_{cell} = (1.4\,\text{V}) - \frac{(8.314)(298.15)}{6(96500)} \ln \frac{(1.0\,\text{M})}{(7.2 \times 10^{-3}\,\text{M})} = 1.47\,\text{V}\] The potential of this cell at \(25°C\), with given concentrations of \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\), is approximately \(1.47\,\text{V}\).

Step 5: Calculate the concentration of \(\mathrm{Al^{3+}}\) for part (b)

Given a cell potential of \(1.62\,\text{V}\), we need to find the concentration of \(\mathrm{Al^{3+}}\). We can rearrange the Nernst equation from step 3 to solve for \(\mathrm{[Al^{3+}]}\): \[[\mathrm{Al^{3+}}] = \frac{[\mathrm{Ni^{2+}}]}{\text{exp} \left[ \frac{nF (E_{cell} - E°_{cell})}{RT} \right]}\] Plugging in the given values, \[[\mathrm{Al^{3+}}] = \frac{1\,\text{M}}{\text{exp} \left[ \frac{6(96500)(1.62\,\text{V} - 1.4\,\text{V})}{(8.314)(298.15)} \right]} = 6.5 \times 10^{-4}\,\text{M}\] The concentration of \(\mathrm{Al^{3+}}\) in the unknown solution when the measured cell potential at \(25°C\) is \(1.62\,\text{V}\) is approximately \(6.5 \times 10 ^{-4}\,\text{M}\).

Key Concepts

Nernst equation

The Nernst equation is a fundamental tool in electrochemistry used to determine the actual cell potential of an electrochemical cell under non-standard conditions. It accounts for the changes in concentration of reactants and products which can affect the cell potential. This equation becomes crucial when the concentrations of ions are not at the standard state, which is 1 M for solutes.

The Nernst equation is written as follows:

• \[ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln \frac{[\text{Reductant}]}{[\text{Oxidant}]} \]

Here, \( E^{\circ}_{cell} \) is the standard cell potential, \( R \) is the universal gas constant equal to \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \), \( T \) is the absolute temperature in Kelvin, \( n \) is the number of moles of electrons exchanged in the redox reaction, and \( F \) is the Faraday constant, approximately \( 96500 \text{ C mol}^{-1} \).

The term \( \ln \frac{[\text{Reductant}]}{[\text{Oxidant}]} \) represents the natural logarithm of the relative concentration of reactants over products. It is essential to adjust the concentrations in the expression to ensure accurate cell potential calculations under specific conditions.

standard electrode potentials

Standard electrode potentials are a measure of the intrinsic voltage or energy change associated with a redox reaction at standard conditions, which is generally 298 K, 1 atm, and 1 M concentration for aqueous ions. They are usually denoted by \( E^{\circ} \) and are crucial in understanding the tendency of a species to gain or lose electrons. Every half-reaction has its own standard electrode potential and is referenced to the standard hydrogen electrode (SHE), which has an assigned potential of 0 V.

In this exercise, we found the standard electrode potentials for both nickel and aluminum half-reactions:

• For Nickel: \( Ni^{2+} + 2e^- \rightarrow Ni(s); \quad E^{\circ}_{Ni} = -0.26 \text{ V} \)
• For Aluminum: \( Al^{3+} + 3e^- \rightarrow Al(s); \quad E^{\circ}_{Al} = -1.66 \text{ V} \)

These values highlight that aluminum has a more negative potential than nickel, indicating aluminum more readily loses electrons than nickel under standard conditions. Understanding and comparing these potentials helps determine which species will undergo reduction or oxidation in an electrochemical cell.

cell potential calculation

Calculating the cell potential involves both the use of standard electrode potentials and the Nernst equation, especially when ion concentrations deviate from the standard conditions.

Firstly, you find the \( E^{\circ}_{cell} \), which is the standard potential for the whole cell, by subtracting the anode's potential from the cathode's. In our case from the step-by-step solution:

• \[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.26 \text{ V} - (-1.66 \text{ V}) = 1.4 \text{ V} \]

Once \( E^{\circ}_{cell} \) is determined, the Nernst equation helps adjust for the given concentrations:

• \[ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln \frac{[\text{Reductant}]}{[\text{Oxidant}]} \]

For part (a) of this exercise, the actual cell potential with given ion concentrations is calculated. The measured cell potential, when known, can inversely infer unknown concentrations, as shown in part (b). Understanding how to derive the actual potentials and what impacts them ensures accurate predictions in electrochemical processes.