Question

At one time on Earth, iron was present mostly as iron(II). Later, once plants had produced a significant quantity of oxygen in the atmosphere, the iron became oxidized to iron(III). Show that \(\mathrm{Fe}^{2+}(a q)\) can be spontaneously oxidized to \(\mathrm{Fe}^{3+}(a q)\) by \(\mathrm{O}_{2}(g)\) at \(25^{\circ}\) Cassuming the following reasonable environmental conditions: \(\left[\mathrm{Fe}^{2+}\right]=\left[\mathrm{Fe}^{3+}\right]=1 \times 10^{-7} \mathrm{M} ; \mathrm{pH}=7.0 ; P_{\mathrm{O}_{2}}=160 \mathrm{~mm} \mathrm{Hg}\).

Short answer

The reaction is spontaneous, as the cell potential is positive (+0.443 V).

Step-by-step solution

Write the Half-Reactions

For the oxidation of iron, the half-reaction is: \[ ext{Fe}^{2+} ightarrow ext{Fe}^{3+} + e^-\]For the reduction of oxygen, the balanced half-reaction at pH 7 is: \[ ext{O}_2 + 4H^+ + 4e^- ightarrow 2H_2O\]

Determine Standard Electrode Potentials

Using a standard table, you find: - The standard reduction potential \(E^ heta\) for \(Fe^{3+}/Fe^{2+}\) is \(+0.77 \text{ V}\).- The standard reduction potential \(E^ heta\) for \(O_2/H_2O\) half-reaction is \(+1.23 \text{ V}\).

Write the Nernst Equation for Each Half-Reaction

For the oxidation half-reaction, \[E_{Fe} = E^ heta_{Fe} - \frac{0.0591}{n} \log \frac{[ ext{Fe}^{3+}]}{[ ext{Fe}^{2+}]} \]where \(E^ heta_{Fe} = +0.77 \text{ V}\) and \(n=1\).Substitute values:\[E_{Fe} = +0.77 - \frac{0.0591}{1} \log \frac{1 imes 10^{-7}}{1 imes 10^{-7}} = +0.77 \text{ V}\]

Calculate the Effective Potential for the Reduction Reaction

For the oxygen reduction reaction, the Nernst equation is:\[E_{O_2} = E^ heta_{O_2} - \frac{0.0591}{4} \log \left(\frac{1}{P_{O_2}}\right)\]Given \(P_{O_2} = 160 \text{ mm Hg}\), convert to atm:\[P_{O_2} = \frac{160}{760} \approx 0.211 \text{ atm}\]Substitute values and solve:\[E_{O_2} = +1.23 - \frac{0.0591}{4} \log \left(\frac{1}{0.211}\right) = +1.23 - \frac{0.0591}{4} (0.676) \]\[E_{O_2} \approx +1.23 - 0.017 = +1.213 \text{ V}\]

Calculate the Cell Potential

The overall cell potential \(E_{cell}\) becomes:\[E_{cell} = E_{cathode} - E_{anode} = E_{O_2} - E_{Fe}\]Substitute:\[E_{cell} = +1.213 - 0.77 = +0.443 \text{ V}\]Since \(E_{cell} > 0\), the reaction is spontaneous.

Key concepts

Standard Electrode Potential

The standard electrode potential is a crucial aspect of electrochemistry. It helps to determine how likely a particular half-reaction is to occur. Simply put, it's a measure of the "voltage" associated with a reduction reaction when all reactants and products are in their standard states. This concept is essential because it allows us to predict if a reaction will be spontaneous under standard conditions.

• For \( ext{Fe}^{3+}/ ext{Fe}^{2+}\), the standard reduction potential, \(E^\theta\), is \(+0.77\) V.
• For \( ext{O}_2/ ext{H}_2O\), the standard reduction potential is \(+1.23\) V.

Having a more positive standard electrode potential indicates a greater tendency to gain electrons. In the context of the iron oxidation problem, these potentials help determine if the reaction of iron being oxidized by oxygen is favorable. Comparing these values is a key step in solving the problem, as it helps establish which half-reaction will act as the anode and which will be the cathode in any electrochemical cell.

Nernst Equation

To delve deeper into electrochemistry and reaction spontaneity, the Nernst Equation comes into play. This equation allows us to calculate the potential of a half-reaction under non-standard conditions. It's especially useful because real-world conditions rarely match the idealized standard conditions.
The Nernst Equation is given by:

• \(E = E^\theta - \frac{0.0591}{n} \log Q\)

Here:

• \(E\) represents the potential of the half-reaction under non-standard conditions.
• \(n\) is the number of moles of electrons exchanged.
• \(Q\) is the reaction quotient.

This formula helps us adjust the standard potentials we've discussed earlier, to reflect actual environmental conditions like concentration and pressure.
In our exercise, the Nernst Equation is used to account for the non-standard concentrations of iron ions and the partial pressure of oxygen, showcasing how dynamic and adaptable the equation is to different scenarios.

Oxidation-Reduction Reaction

Also known as redox reactions, these are chemical processes where atoms have their oxidation state changed. This occurs through the transfer of electrons between elements or compounds. Redox reactions are integral to many natural and industrial processes, including metabolism, combustion, and corrosion.
In any redox reaction, there are two complementary processes:

• Oxidation: A substance loses electrons, increasing its oxidation state.
• Reduction: A substance gains electrons, decreasing its oxidation state.

In the exercise we're examining, iron(II) is oxidized (loses electrons) to become iron(III), while the oxygen molecule is reduced (gains electrons) to form water. Understanding these two simultaneous processes is crucial, as they determine the flow of electrons and how energy is transferred in the system.
Recognizing which components undergo oxidation and which undergo reduction is pivotal in solving electrochemical problems.

Iron Oxidation

Iron oxidation is a specific type of redox reaction where iron undergoes a change in its oxidation state. It involves the transformation of iron from a lower oxidation state (Fe\(^{2+}\)) to a higher oxidation state (Fe\(^{3+}\)), which is a common process both in natural environments and industrial applications.
Key factors about this process:

• Fe\(^{2+}\) is oxidized to Fe\(^{3+}\), losing an electron in the process.
• This type of reaction is critical in the formation of rust, which is essentially iron(III) oxide.
• Environmental conditions such as pH and the presence of oxygen greatly affect the rate and extent of iron oxidation.

In the context of the exercise, the oxidation of Fe\(^{2+}\) by oxygen in the presence of a neutral pH is considered spontaneous. This is due to the positive cell potential, calculated through the standard electrode potentials and adjusted with the Nernst Equation. This process not only highlights the importance of redox reactions in nature but also demonstrates the interconnectedness of various chemical principles.