Problem 123 - Chemistry An Atoms First Approach

Question

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{kJ} .\) What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The theoretical maximum voltage this cell can achieve at \(25^\circ \mathrm{C}\) is 1.071 V.

Step 1: Balancing the redox reaction

First, we need to break down the given net reaction into half-reactions to determine the number of electrons being transferred. The given net reaction is: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\) Breaking the net reaction into two half-reactions: Oxidation: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) \rightarrow 2 \mathrm{CO}_{2}(g) + 12 \mathrm{H}^{+}(aq) + 12 \mathrm{e}^-\) Reduction: \(3\mathrm{O}_2(g) + 12\mathrm{H}^{+}(aq) + 12\mathrm{e}^- \rightarrow 6\mathrm{H}_2\mathrm{O}(l)\) In this case, n = 12.

Step 2: Calculating the maximum cell voltage

Now that we have the value of n, we can calculate the maximum cell voltage using the formula: Maximum cell voltage = (\( \dfrac{W_{max}}{n \times F} \)) Plugging in the values, we get: Maximum cell voltage = (\( \dfrac{1.32×10^3 \mathrm{kJ}}{12 \times 9.65×10^4 Cmol^{-1}}\)) Convert 1.32×10^3 kJ to J: \(1.32×10^3 kJ × \dfrac{1000J}{1kJ} = 1.32×10^6 J\) Formula with converted values: Maximum cell voltage = (\( \dfrac{1.32×10^6 \mathrm{J}}{12 \times 9.65×10^4 Cmol^{-1}}\)) Now, calculate the maximum cell voltage: Maximum cell voltage \(= \dfrac{1.32×10^6}{12 \times 9.65×10^4} = 1.071 \mathrm{V}\) The theoretical maximum voltage this cell can achieve at \(25^\circ \mathrm{C}\) is 1.071 V.

Key Concepts

Redox Reactions

Redox reactions are a type of chemical reaction involving the transfer of electrons between two species. The term 'redox' is a shorthand for 'reduction-oxidation.' In these processes, one reactant loses electrons (oxidation) while another gains electrons (reduction). Understanding redox reactions is crucial when studying fuel cells because they are the fundamental processes that allow for the conversion of chemical energy into electrical energy.

For example, the grain alcohol reacting with oxygen in a fuel cell comprises an oxidation reaction for the alcohol, where it loses electrons, and a reduction reaction for the oxygen, where it gains electrons. The electrons donated by the alcohol travel through an external circuit to reach the oxygen, thereby generating an electric current. This makes the mastery of writing and balancing redox reactions vital in predicting the output of a fuel cell.

Half-Reactions in Electrochemistry

Half-reactions are the individual steps that comprise a full redox reaction in electrochemistry. They are split into two parts: the oxidation half-reaction, where electrons are lost, and the reduction half-reaction, where electrons are gained. Each half-reaction occurs at a different electrode within an electrochemical cell.

In a fuel cell, oxidizing the fuel (e.g., grain alcohol) at the anode generates electrons, while at the cathode, oxygen is reduced as it accepts electrons. The separate analysis of these half-reactions is essential to calculate the number of electrons transferred in the full reaction, denoted by 'n' in electrochemical calculations. Understanding half-reactions also helps in measuring the cell's potential and efficiency, which are critical factors in the design and application of fuel cells.

Gibbs Free Energy and Work Relationship

The Gibbs free energy \textbf{(G)} is a thermodynamic property that measures the maximum amount of reversible work that can be done by a system at constant temperature and pressure. It is intimately linked to the concept of work in a fuel cell because the electric work performed by the cell equals the decrease in Gibbs free energy during the chemical reaction.

For an electrochemical reaction, the work done (W) is obtained when charge (q) moves across a potential difference (E), given by the equation \textbf{W = qE}. In the context of a fuel cell, the maximum work \textbf{(W\(_{max}\))} that can be generated corresponds to \textbf{G}, and it can be determined by calculating the difference in free energy between the reactants and products. This relationship is vital for understanding the theoretical maximum voltage that a fuel cell can achieve, as a greater \textbf{G} indicates more work can be extracted, and thus a higher voltage.

Faraday's Constant Applications

Faraday's constant (\textbf{F}) represents the total charge of one mole of electrons, about \textbf{96,485 coulombs per mole}. It is a fundamental value in electrochemistry and plays a crucial role in correlating the amount of chemical substance involved in the reaction with the measured electrical charge that passes through the circuit in a fuel cell.

In voltage calculation for a fuel cell, Faraday's constant is used to determine the charge associated with the mole of electrons transferred. By dividing the maximum work (\textbf{W\(_{max}\)}) by the product of the number of moles of electrons (\textbf{n}) and Faraday's constant (\textbf{F}), you derive the maximum voltage achievable by the cell. This formula is derived from the relationship between Gibbs free energy and work, and it highlights how Faraday's constant is vital for transforming chemical reaction energy into electrical energy within the practical applications of fuel cells.