Problem 6 - General Chemistry: Principles and Modern Applications

Question

The theoretical \(E_{\text {cell }}^{\circ}\) for the methane-oxygen fuel cell is \(1.06 \mathrm{V} .\) What is \(E^{\circ}\) for the reduction half-reaction \(\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) ?\)

Step-by-Step Solution

Answer

The \(E^{\circ}\) for the given reduction half-reaction is \(0.53 \mathrm{V}\)

Step 1: Confirmed half reaction

For the fuel cell reaction given, the reduction half-reaction is \[\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l)\]

Step 2: Determine the entire reaction

Before obtaining the E° for this half-reaction, you must first work out the full reaction for the methane-oxygen fuel cell. The full reaction is \[\mathrm{CH}_{4}(\mathrm{g}) + 2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(l)\]

Step 3: Determine Oxidation half-reaction

Next is to determine the oxidation half-reaction which is obtained by reversing the reduction half-reaction. The oxidation half-reaction is \[\mathrm{CH}_{4}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) + 8 \mathrm{H}^{+}(\mathrm{aq}) + 8 \mathrm{e}^{-}\]

Step 4: Calculate E° for the reduction half-reaction

Finally, knowing that \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{oxidation}} + E^{\circ}_{\text{reduction}}\), and that for the oxidation half-reaction, \(E^{\circ}_{\text{oxidation}} = -E^{\circ}_{\text{reduction}}\) (since you're reversing the reaction), you can deduce that \(E^{\circ}_{\text{reduction}} = \frac{E^{\circ}_{\text{cell}}}{2}\), which equals to 1.06 V / 2

Key Concepts

Standard Electrode Potential

The standard electrode potential, often represented as \(E^\circ\), is a measurement that indicates how readily a chemical species gains electrons in a half-reaction, which is a significant concept in electrochemistry. It describes the tendency of an electrode to be reduced when it's compared to the standard hydrogen electrode (SHE), which is set at 0 volts. By convention, the more positive the \(E^\circ\), the greater the species' affinity for electrons and the stronger its oxidizing ability.

For instance, in a methane-oxygen fuel cell, these potentials are used to calculate the voltage output of the cell by combining the potential of the oxidation half-reaction and the reduction half-reaction. Knowing the standard potentials also allows us to predict whether a reaction will occur spontaneously under standard conditions.

Reduction Half-Reaction

A reduction half-reaction involves the gain of electrons by a substance. During this process, the oxidation state of the substance is decreased as electrons (e-) are added. In electrochemical cells, the cathode is where reduction occurs, and this is typically represented in a half-cell notation. For example, in the exercise provided, \(\text{CO}_2\) is reduced to \(\text{CH}_4\) by gaining electrons.

In the calculation of the standard electrode potential for the methane-oxygen fuel cell, the reduction half-reaction is central. By identifying which species are being reduced and calculating the associated potential, you can determine the performance characteristics of the fuel cell and predict which reactions can occur spontaneously.

Oxidation Half-Reaction

The oxidation half-reaction is the counterpart to the reduction half-reaction and involves the loss of electrons. It raises the oxidation state of a substance as it releases electrons. For instance, in our exercise, methane (\(\text{CH}_4\)) is oxidized to carbon dioxide (\(\text{CO}_2\)) and hydrogen ions (\(\text{H}^+\)) with the release of electrons. The oxidation half-reaction takes place at the anode of an electrochemical cell.

Understanding oxidation processes is vital when considering overall cell reactions in fuel cells and determining the standard electrode potential for these reactions. Being able to write and balance oxidation half-reactions is as crucial as understanding reduction, since both processes occur simultaneously in electrochemical cells.

Methane-Oxygen Fuel Cell

A methane-oxygen fuel cell is a type of electrochemical cell that generates electricity through the reaction of methane with oxygen. It has two half-reactions, as with any fuel cell: an oxidation half-reaction at the anode and a reduction half-reaction at the cathode. In the particular problem discussed, methane (\(\text{CH}_4\)) reacts with oxygen (\(\text{O}_2\)) to form carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).

The overall efficiency and potential of the fuel cell rely on the careful balance of these reactions and understanding the standard electrode potentials involved. Components like the catalysts used at the electrodes and the membrane that separates them also play a significant role in the performance of these cells. Methane-oxygen fuel cells are considered for their high energy efficiency and lower environmental impact compared to traditional fossil fuels.