Question

In the presence of oxgyen and acid, two half. reactions responsible for the corrosion of iron are $$ \begin{array}{c} \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 e^{-} \\ \mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) \end{array} $$ Calculate the the standard potential, \(E^{\circ},\) and decide whether the reaction is product-favored at equilibrium. Will decreasing the pH make the reaction less thermodynamically product-favored at equilibrium?

Short answer

The reaction is product-favored with a positive cell potential (1.67 V). Lowering pH makes it more product-favored.

Step-by-step solution

Identify the half-reactions

The given half-reactions for the corrosion of iron in acidic conditions are:1. The oxidation of iron: \( \mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2 e^{-} \).2. The reduction of oxygen: \( \mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \).

Find the standard electrode potentials

Consult a standard reduction potential table to find the standard potentials for the half-reactions:1. \( \mathrm{Fe}^{2+} + 2 e^{-} \rightarrow \mathrm{Fe} \): \( E^{\circ} = -0.44 \text{ V} \).2. \( \mathrm{O}_{2} + 4 \mathrm{H}^{+} + 4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \): \( E^{\circ} = +1.23 \text{ V} \).

Calculate the cell potential

To find the standard cell potential, \( E^{\circ}_{\text{cell}} \), calculate the difference between the reduction and oxidation potentials:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.23 \text{ V} - (-0.44 \text{ V}) = 1.67 \text{ V}.\]

Determine product-favored conditions

Since the standard cell potential \( E^{\circ}_{\text{cell}} = 1.67 \text{ V} \) is positive, the reaction is thermodynamically product-favored at equilibrium. This means the forward reaction (corrosion of iron) is favored under standard conditions.

Analyze the effect of decreasing pH

Decreasing the pH increases the concentration of \( \mathrm{H}^{+} \) ions. Since \( \mathrm{H}^{+} \) is a reactant in the reduction half-reaction, its increased concentration shifts the equilibrium towards the products, thus increasing the \( E^{\circ}_{\text{cell}} \) value. Therefore, the reaction becomes more thermodynamically product-favored as the pH decreases.

Key concepts

Electrode Potentials

Electrode potentials are a crucial concept in understanding chemical reactions that involve electrons. In the corrosion of iron, two half-reactions take place. One for oxidation, where iron loses electrons, and another for reduction, where oxygen gains electrons. Each of these half-reactions has an associated electrode potential which signifies how easily a species gains or loses electrons.

Electrode potentials are measured in volts (V) and are retrieved from standard reduction potential tables. These potentials indicate a substance's likelihood to be reduced. The more positive the potential, the greater the substance's tendency to gain electrons. In the corrosion problem, the electrode potential for iron's oxidation is negative, while the potential for oxygen's reduction is positive, suggesting that oxygen readily accepts electrons, driving the corrosion process.

Half-Reactions

Understanding half-reactions is fundamental when analyzing redox processes, like corrosion. Each half-reaction represents either oxidation or reduction. In the case of iron corrosion:

• For oxidation: Iron (\[\text{Fe}\] ) turns into iron ions (\[\text{Fe}^{2+}\] ) and releases electrons.
• For reduction: Oxygen interacts with hydrogen ions and electrons to form water.

Each half-reaction shows either a gain or loss of electrons, and these two processes must balance in a full redox reaction. By breaking reactions into these smaller steps, it's easier to calculate associated potentials and understand how substances transform. Recognizing the role of electrons helps us predict whether the entire reaction will proceed spontaneously.

Cell Potential Calculation

Calculating the cell potential is pivotal to determining whether a chemical reaction will occur spontaneously. In corrosion, we calculate the cell potential (\[E^{\circ}_{\text{cell}}\] ) by finding the difference between the reduction potential of the cathode and the oxidation potential of the anode:

\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]

For corrosion, the cathode is the reduction of oxygen, with a potential of +1.23 V, and the anode is the oxidation of iron, with a potential of -0.44 V. Calculating their difference gives a cell potential of 1.67 V. Because this value is positive, the corrosion reaction is spontaneous under standard conditions. This spontaneity is why rusting is such a common problem in everyday life, as it indicates a natural tendency for materials to degrade.

pH Effect on Corrosion

The pH level of the environment dramatically impacts the corrosion of iron. pH measures the concentration of hydrogen ions (\[\text{H}^{+}\] ). A lower pH means more acidic conditions with a higher \[\text{H}^{+}\] concentration.

In the reduction half-reaction, \[\text{H}^{+}\] ions are needed. Thus, more \[\text{H}^{+}\] will shift the equilibrium of the reaction toward the products, enhancing the likelihood of reduction.

• Decreasing the pH makes the environment more favorable for corrosion.
• More \[\text{H}^{+}\] ions drive the reaction forward, increasing the oxidation rate of iron.

As a result, corrosion accelerates in acidic conditions, a crucial consideration for protecting metal structures exposed to varied environments.