Question

You have the following cell process: \(\mathrm{Fe}(\mathrm{s})+\mathrm{Co}^{2+}(.5 \mathrm{M}) \rightarrow \mathrm{Fe}^{2+}(1.0 \mathrm{M})+\mathrm{Co}(\mathrm{s})\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \leftrightarrows \mathrm{Fe}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.44 \mathrm{e}\) and \(\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \leftrightharpoons \mathrm{Co}(\mathrm{s})\) with \(\mathrm{E}^{\circ}=-.28\), find the standard cell potential \(\Delta \mathrm{E}\), the cell potential \(\Delta \mathrm{E}\) and the concentration ratio at which the potential generated by the cell is exactly zero. which the potential generated by the cell is exactly zero.

Short answer

For the given redox reaction, we find the standard cell potential (\( \Delta E^{\circ} \)) to be \(0.16 \, V\), the cell potential (\( \Delta E \)) to be approximately \(0.152 \, V\), and the concentration ratio at which the potential generated by the cell is exactly zero to be approximately \(6.14\).

Step-by-step solution

Calculating the standard cell potential

To calculate the standard cell potential (\( \Delta E^{\circ} \)), we need to identify the reduction half-reactions and their respective standard reduction potentials: Iron half-reaction: \( \mathrm{Fe}^{2+} + 2 \mathrm{e}^{-} \leftrightarrows \mathrm{Fe}(s) \\\ E^{\circ}_{Fe} = - 0.44 \, V \) Cobalt half-reaction: \( \mathrm{Co}^{2+} + 2 \mathrm{e}^{-} \leftrightharpoons \mathrm{Co}(s) \\\ E^{\circ}_{Co} = - 0.28 \, V \) Now, we need to determine which metal will undergo oxidation and which metal will undergo reduction. To do this, we compare their standard reduction potentials: Since \( E^{\circ}_{Fe} < E^{\circ}_{Co} \), iron will oxidize while cobalt will reduce. Finally, we can calculate the standard cell potential as follows: \( \Delta E^{\circ} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Co} - E^{\circ}_{Fe} = (-0.28 \, V) - (-0.44 \, V) = 0.16 \, V \)

Calculating the cell potential using the Nernst equation

Now, we will find the cell potential (\( \Delta E \)) using the Nernst equation: \( \Delta E = \Delta E^{\circ} - \frac{RT}{nF} \ln Q \) In this case, we have: \( Q = \frac{[Fe^{2+}]}{[Co^{2+}]} = \frac{1.0 \, M}{0.5 \, M} = 2 \) \( T = 298 \, K \) (standard room temperature) \( n = 2 \) (moles of electrons) \( R = 8.314 \, J \cdot mol^{-1} \cdot K^{-1} \) (gas constant) \( F = 96,485 \, C \cdot mol^{-1} \) (Faraday's constant) Plug these values into the Nernst equation to get: \( \Delta E = 0.16 \, V - \frac{(8.314 \, J \cdot mol^{-1} \cdot K^{-1})(298 \, K)}{(2)(96,485 \, C \cdot mol^{-1})} \ln 2 \) \( \Delta E \approx 0.152 \, V \)

Determining the concentration ratio for zero cell potential

We'll use the Nernst equation to find the concentration ratio when the cell potential is zero: \( \Delta E = \Delta E^{\circ} - \frac{RT}{nF} \ln Q = 0 \) Therefore: \( Q = \frac{[Fe^{2+}]}{[Co^{2+}]} \) Rearrange the Nernst equation to solve for Q: \( \ln Q = \frac{nF\Delta E^{\circ}}{RT} \) \( Q = e^{\frac{nF\Delta E^{\circ}}{RT}} \) Plugging in the values, we get: \( Q = e^{\frac{(2)(96,485 \, C \cdot mol^{-1})(0.16 \, V)}{(8.314 \, J \cdot mol^{-1} \cdot K^{-1})(298 \, K)}} \) \( Q \approx 6.14 \) Thus, the concentration ratio is: \( \frac{[Fe^{2+}]}{[Co^{2+}]} \approx 6.14 \)

Key concepts

Standard Cell Potential

The standard cell potential, often denoted as \( \Delta E^{\circ} \) or \( E^{\circ}_{cell} \), is a critical concept in electrochemistry. It indicates the voltage difference between two half-cells in an electrochemical cell when they are operating under standard conditions (concentrations of 1 mol/L, 1 atm gas pressure, and a temperature of 25°C or 298K). This potential is derived from the intrinsic tendency of a chemical species to lose or gain electrons, described through half-reactions.

Each half-reaction has its standard reduction potential, which is measured under standard conditions against a standard hydrogen electrode. By comparing the standard reduction potentials of the two half-reactions in a cell, the direction of electron flow can be determined. The species with the higher reduction potential will undergo reduction, and the species with the lower reduction potential will undergo oxidation.

To calculate the overall standard cell potential, we take the difference between the standard reduction potential of the cathode (reduction) and the anode (oxidation):\[\Delta E^{\circ} = E^{\circ}_{cathode} - E^{\circ}_{anode}\].

In the given exercise, we determined that cobalt would be reduced and iron would be oxidized, leading to a positive standard cell potential of \( 0.16 \, V \) for the reaction.

Nernst Equation

The Nernst equation is pivotal in understanding how the cell potential changes with varying conditions like concentration, pressure, and temperature. It refines the standard cell potential to predict the actual working potential of a cell under non-standard conditions.

The general form of the Nernst equation is:\[\Delta E = \Delta E^{\circ} - \frac{RT}{nF} \ln Q\],where \( \Delta E \) is the cell potential at any moment, \( \Delta E^{\circ} \) is the standard cell potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvins, \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient representing the ratio of product activities to reactant activities.

Using this equation, we can calculate how changes in reactant and product concentrations affect the cell's voltage. For the exercise, the Nernst equation helped to find the actual cell potential with given concentrations, determining it to be approximately \( 0.152 \, V \) using room temperature and the provided concentration values.

Concentration Ratio

In electrochemical cells, the concentration ratio plays a significant role, as it influences the electrical potential generated by the cell. When we refer to the concentration ratio in the context of the Nernst equation, we are talking about the ratio of the concentrations of the products to the reactants, denoted in the reaction quotient \( Q \).

For a reaction where the products and reactants are in different solute concentrations, \( Q \) is calculated as:\[Q = \frac{\text{{Concentration of products}}}{\text{{Concentration of reactants}}}\].

In our exercise, we aimed to figure out at what concentration ratio the cell potential would become zero, effectively rendering the reaction in equilibrium. We did this by rearranging the Nernst equation to solve for \( Q \) when \( \Delta E = 0 \) and found that the specific concentration ratio where the cell produces no voltage was approximately 6.14.

Redox Reactions

Redox reactions are chemical reactions involving the transfer of electrons between two species. These reactions are comprised of two half-reactions: oxidation, where a species loses electrons, and reduction, where a species gains electrons. The species that gets oxidized is termed the reducing agent, and the one that gets reduced is called the oxidizing agent.

In electrochemical cells, redox reactions drive the flow of electrons from the anode to the cathode through an external circuit. This electron flow generates electrical energy, which can be harnessed for various applications. The standard cell potential of the redox reaction gives us an idea of the electromotive force (EMF) or voltage that the reaction can produce under standard conditions.

In the provided exercise, we had iron getting oxidized to \(Fe^{2+} \) while cobalt was reduced from \(Co^{2+} \) to cobalt metal. This exchange of electrons between iron and cobalt set up a potential difference, calculated using standard potentials and modified for actual conditions using the Nernst equation.