The half cell reaction for the corrosion: \(2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+\overline{2} \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}\) Find the \(\Delta \mathrm{G}^{\circ}\) (in \(\mathrm{kJ}\) ) for the overall reaction. (a) \(-76\) (b) \(-322\) (c) \(-161\) (d) \(-152\)

Corrosion chemistry involves the study of materials, typically metals, deteriorating due to chemical reactions with their environment. A classic example of this is the rusting of iron when it reacts with moisture in the air. The process is essentially a series of electrochemical reactions leading to the loss of electrons by the metal, known as oxidation.

For iron, one of the half-cell reactions would be represented by \(\mathrm{Fe}^{2+} + 2\mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) \), where iron loses electrons to form iron ions. The other half-cell involves the reduction of oxygen, which is common in corrosion processes and can be represented by \(2 \mathrm{H}^{+} + \frac{1}{2} \mathrm{O}_{2} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}\).

Understanding the details of these reactions allows us to calculate the overall energy change and to design measures to protect materials from corrosion.

The electrochemical cell potential, often represented by \(E^\circ_{\text{cell}}\), is a measure of the voltage produced by an electrochemical cell when no current is being drawn from the cell. It essentially reflects the tendency of the cell components to gain or lose electrons, a key element in the corrosion chemistry.

Calculated as the difference between the reduction potentials of the cathode and the anode, the cell potential can predict whether a chemical reaction will occur spontaneously. If \(E^\circ_{\text{cell}}\) is positive, the reaction is spontaneous, which in the context of corrosion, means the metal is more likely to corrode under standard conditions.

Each electrochemical reaction can be broken down into two half-cell reactions, each representing either the oxidation or reduction process that occurs at one of the electrodes.

In the context of the corrosion of iron, the oxidation half-cell reaction sees iron turning into iron ions while losing electrons, \(\mathrm{Fe}(\mathrm{s})\longrightarrow \mathrm{Fe}^{2+} + 2\mathrm{e}^{-}\). The reduction half-cell reaction is the acceptance of electrons by oxygen to form water, \(2\mathrm{H}^{+} + \frac{1}{2}\mathrm{O}_{2} + 2\mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}\mathrm{O}\). For calculating the overall cell potential and Gibbs free energy, we combine these two reactions to get the overall cell reaction.

Faraday's constant, denoted by \(F\), is the amount of electric charge per mole of electrons. It has a value of approximately 96,485 Coulombs per mole. During the calculation of the Gibbs free energy change for an electrochemical reaction, \(F\) is used to convert the electrochemical cell potential from volts into an energy value.

The formula \(\Delta G^\circ = -nFE^\circ_{\text{cell}}\) highlights the use of Faraday's constant, where \(n\) is the number of moles of electrons exchanged in the reaction and \(E^\circ_{\text{cell}}\) is the standard electrochemical cell potential. The negative sign indicates that a spontaneous reaction (positive \(E^\circ_{\text{cell}}\)) will result in energy being released, hence a decrease in Gibbs free energy.