Problem 88 - Chemistry Principles and Practice

Question

The electrochemical processes that occur in the corrosion of iron are represented by the half-reactions $$ \begin{array}{l} \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \\ \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) \end{array} $$ (a) Write the overall reaction. (b) What is the standard potential for the overall chemical reaction? (c) Natural water has a pH of about \(5.9,\) and air is 0.21 mol fraction oxygen. If the concentration of iron(II) in the water is \(5 \times 10^{-5} M\), what is the potential of the corrosion reaction in the presence of air and natural water at 1 atm pressure and \(298 \mathrm{~K} ?\)

Step-by-Step Solution

Answer

The overall reaction is \( 2 \mathrm{Fe} + \mathrm{O}_2 + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Fe}^{2+} + 2 \mathrm{H}_2 \mathrm{O} \), the standard potential is 1.67 V, and the potential under given conditions is calculated using the Nernst equation.

Step 1: Write the Overall Reaction

First, we need to combine the given half-reactions. The oxidation half-reaction is \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2 \mathrm{e}^{-} \), and the reduction half-reaction is \( \mathrm{O}_2 + 4 \mathrm{H}^{+} + 4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \). To combine them, we must balance the number of electrons. The oxidation reaction is multiplied by 2: \( 2 \mathrm{Fe} \rightarrow 2 \mathrm{Fe}^{2+} + 4 \mathrm{e}^{-} \). Now, the overall reaction by adding both balanced processes gives us \( 2 \mathrm{Fe} + \mathrm{O}_2 + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Fe}^{2+} + 2 \mathrm{H}_2 \mathrm{O} \).

Step 2: Determine Standard Cell Potential

The standard cell potential can be calculated using standard reduction potentials. For this process, remember that the standard reduction potential for \( \mathrm{Fe}^{2+}/\mathrm{Fe} \) is \(-0.44 \text{ V} \) and for \( \mathrm{O}_2 + 4\mathrm{H}^{+} + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O} \) is \( +1.23 \text{ V} \). Calculate the standard potential by adding the potentials of the reduction half-reaction (since oxidation potential is the negative of the reduction): \( E^\circ_{\text{cell}} = 1.23 \text{ V} - (-0.44 \text{ V}) = 1.67 \text{ V} \).

Step 3: Use Nernst Equation for Potential

Using the Nernst Equation for the potential of the cell under non-standard conditions, we can write the equation as: \[ E = E^\circ - \frac{RT}{nF} \ln Q \]. Here \( E^\circ = 1.67 \text{ V} \), number of electrons \( n = 4 \), \( R = 8.314 \text{ J/mol·K} \), \( F = 96485 \text{ C/mol} \), and \( T = 298 \text{ K} \). The reaction quotient \( Q \) can be found from the concentration and pressures: \( Q = \frac{[\mathrm{Fe}^{2+}]^2}{(P_{ ext{O}_2})([\mathrm{H}^+])^4} \). Given \([\mathrm{Fe}^{2+}] = 5 \times 10^{-5} \text{ M} \), \( \text{pH} = 5.9 \rightarrow [\mathrm{H}^+] = 10^{-5.9} \), \( P_{\text{O}_2} = 0.21 \), we calculate \( Q = \frac{(5 \times 10^{-5})^2}{0.21 \times (10^{-5.9})^4} \). Substitute this value into the Nernst equation to find \( E \).

Step 4: Calculate the Potential

Insert the values into the Nernst formula: \[ E = 1.67 - \frac{8.314 \times 298}{4 \times 96485} \ln \left( \frac{(5 \times 10^{-5})^2}{0.21 \times (10^{-5.9})^4} \right) \]. Calculate the value by evaluating the fraction inside the logarithm and substituting to find \( E \).

Key Concepts

Corrosion of Iron

Corrosion of iron is a common electrochemical process that affects materials, causing damage over time. When iron is exposed to oxygen and moisture, it oxidizes, forming rust, which is chemically known as iron (III) oxide. This process can be described by the combination of two half-reactions: oxidation and reduction.
\[\text{Oxidation: } \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2 \mathrm{e}^{-}\]
The iron, in its solid form, loses electrons and converts into iron ions. Now, let's look at the reduction half-reaction:
\[\text{Reduction: } \mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\]
In this reaction, oxygen gas combines with protons and electrons to form water. These two half-reactions together illustrate how corrosion occurs naturally in the presence of oxygen and acid. Balancing these reactions ensures that electrons are neither lost nor gained, maintaining overall charge neutrality. Consequently, the full corrosion reaction can be written as:
\[2 \mathrm{Fe} + \mathrm{O}_2 + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Fe}^{2+} + 2 \mathrm{H}_2 \mathrm{O}\]

Nernst Equation

The Nernst Equation is a vital tool in electrochemistry to determine the cell potential under non-standard conditions. In situations like natural environmental settings, concentrations and pressures differ from standard conditions. The equation is:
\[ E = E^\circ - \frac{RT}{nF} \ln Q\]
To understand it better, each symbol represents important quantities:

• \(E\) is the cell potential under non-standard conditions
• \(E^\circ\) is the standard cell potential
• \(R\) is the universal gas constant (8.314 J/mol·K)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of electrons transferred in the reaction
• \(F\) is Faraday's constant (96485 C/mol)
• \(Q\) is the reaction quotient

The Nernst Equation adjusts the standard potential \(E^\circ\) by considering concentration differences, allowing accurate predictions of cell behavior. Knowing how these variables interact helps in understanding diverse systems, from batteries to natural water systems.
For example, when we apply the Nernst Equation to the corrosion of iron in natural water, we substitute known values such as concentrations, pressures, and pH levels, then calculate the potential \(E\).

Electrode Potentials

Electrode potentials form the backbone of understanding electrochemical reactions. Each half-reaction has a corresponding electrode potential, measured in volts, indicating its ability to either gain or lose electrons. These values help predict the feasibility and extent of reactions.
In standard conditions, the potential for the reduction reaction, \(\mathrm{O}_2 + 4\mathrm{H}^{+} + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O}\), is \(+1.23\;\text{V}\), illustrating a high tendency to occur. Contrary to this, the oxidation half-reaction \(\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^-\) has a potential of \(-0.44\;\text{V}\).
By combining these, we use the difference between these potentials to find the overall cell potential, \(E^\circ_{\text{cell}}\). The equation,
\[E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}\]
yields \(1.67\;\text{V}\) for the corrosion reaction, showing a spontaneous tendency at standard conditions. These principles of electrode potentials, when applied correctly, offer deep insights into system energetics, reaction driving forces, and stability.