Problem 137
Question
The half cell reaction for the corrosion \(2 \mathrm{H}^{+}+1 / 2 \mathrm{O}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}, E^{\circ}=1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Find the \(\Delta \mathrm{G}^{\circ}\) (in \(\mathrm{kJ}\) ) for the overall reaction. (a) \(-76\) (b) \(-322\) (c) \(-161\) (d) \(-152\)
Step-by-Step Solution
Verified Answer
The correct answer is (b) -322 kJ/mol.
1Step 1: Identify the overall cell reaction
The overall cell reaction is the sum of the given half-cell reactions. This involves balancing the electron transfer between the two half reactions:1. Oxidation Reaction: \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s})\) with \(E^{\circ}=-0.44 \mathrm{~V}\)2. Reduction Reaction: \(2 \mathrm{H}^{+}+1/2 \mathrm{O}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}\mathrm{O}\) with \(E^{\circ}=1.23 \mathrm{~V}\)Thus, the overall cell reaction will be \(\mathrm{Fe}^{2+} + 2\mathrm{H}^{+} + \frac{1}{2}\mathrm{O}_{2} \longrightarrow \mathrm{Fe}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}\).
2Step 2: Calculate the cell potential
The standard cell potential \(E^{\circ}_{\text{cell}}\) is the difference between the potential of the cathode and the anode.\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.23 \mathrm{~V} - (-0.44 \mathrm{~V}) = 1.67 \mathrm{~V} \]
3Step 3: Apply the Gibbs Free Energy formula
The change in Gibbs Free Energy \( \Delta G^{\circ} \) is calculated using the formula:\[ \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \]where \( n \) is the number of moles of electrons transferred (2 moles), \( F \) is the Faraday's constant (96485 C/mol).
4Step 4: Calculate \( \Delta G^{\circ} \)
Substitute the known values into the formula:\[ \Delta G^{\circ} = -2 \times 96485 \times 1.67 \]\[ \Delta G^{\circ} = -322114.7 \text{ J/mol} \]Convert \( \Delta G^{\circ} \) into kilojoules:\[ \Delta G^{\circ} = -322.1 \text{ kJ/mol} \]
5Step 5: Determine the correct answer from the options
Matching the calculated \(\Delta G^{\circ}\) with the provided options, the closest and most accurate option is \(b) -322 \text{ kJ/mol}\).
Key Concepts
ElectrochemistryCell PotentialHalf Cell ReactionsFaraday's Constant
Electrochemistry
Electrochemistry is a branch of chemistry that studies the relationship between electricity and chemical reactions. It plays a crucial role in understanding how chemical energy is converted to electrical energy and vice versa. This conversion process is fundamental in various applications, including batteries, corrosion, and electroplating.
In electrochemical reactions, there are always two processes happening simultaneously: oxidation and reduction. Oxidation is the loss of electrons, while reduction is the gain of electrons. These processes often occur in different compartments, known as half-cells, within an electrochemical cell. The flow of electrons via an external circuit generates electricity.
Electrochemical cells can be either galvanic (voltaic) cells, which convert chemical energy to electrical energy, or electrolytic cells, which use electrical energy to drive non-spontaneous chemical reactions. Understanding this concept aids in solving problems related to cell potential and Gibbs Free Energy.
In electrochemical reactions, there are always two processes happening simultaneously: oxidation and reduction. Oxidation is the loss of electrons, while reduction is the gain of electrons. These processes often occur in different compartments, known as half-cells, within an electrochemical cell. The flow of electrons via an external circuit generates electricity.
Electrochemical cells can be either galvanic (voltaic) cells, which convert chemical energy to electrical energy, or electrolytic cells, which use electrical energy to drive non-spontaneous chemical reactions. Understanding this concept aids in solving problems related to cell potential and Gibbs Free Energy.
Cell Potential
Cell potential, also known as electromotive force (emf), is a measure of the voltage or electrical potential difference between two electrodes in an electrochemical cell. It indicates the ability of a cell to drive an electric current through an external circuit.
To calculate the cell potential, you subtract the anode potential from the cathode potential. Each half-cell has a standard electrode potential, denoted as \(E^\circ\). The overall cell potential is determined using the formula: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\).
For instance, in the given exercise, the cell potential is calculated by finding the difference between the reduction potential of the cathode and the oxidation potential of the anode, resulting in 1.67 V. A positive cell potential indicates a spontaneous cell reaction, allowing us to predict whether or not a reaction will occur under standard conditions.
To calculate the cell potential, you subtract the anode potential from the cathode potential. Each half-cell has a standard electrode potential, denoted as \(E^\circ\). The overall cell potential is determined using the formula: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\).
For instance, in the given exercise, the cell potential is calculated by finding the difference between the reduction potential of the cathode and the oxidation potential of the anode, resulting in 1.67 V. A positive cell potential indicates a spontaneous cell reaction, allowing us to predict whether or not a reaction will occur under standard conditions.
Half Cell Reactions
Half cell reactions are the two parts of an overall electrochemical reaction. Each half reaction occurs at separate electrodes called the anode and the cathode. Understanding half cell reactions is fundamental to calculating cell potentials and predicting reaction spontaneity.
In a half cell reaction, one half relates to the reduction process and the other to the oxidation process. The half-cell reaction for oxidation might be represented like this: \(\text{Fe}^{2+} + 2 \text{e}^{-} \rightarrow \text{Fe}(\text{s})\).The reduction half-cell reaction might look like: \(2 \text{H}^{+} + \frac{1}{2} \text{O}_{2} + 2 \text{e}^{-} \rightarrow \text{H}_{2}\text{O}\).
Balancing these half reactions involves ensuring that both the number of electrons and the charge are equal on both sides of the equation. This balance is crucial for calculating the net reaction in an electrochemical cell. The reaction sum gives us the overall chemical equation for the cell.
In a half cell reaction, one half relates to the reduction process and the other to the oxidation process. The half-cell reaction for oxidation might be represented like this: \(\text{Fe}^{2+} + 2 \text{e}^{-} \rightarrow \text{Fe}(\text{s})\).The reduction half-cell reaction might look like: \(2 \text{H}^{+} + \frac{1}{2} \text{O}_{2} + 2 \text{e}^{-} \rightarrow \text{H}_{2}\text{O}\).
Balancing these half reactions involves ensuring that both the number of electrons and the charge are equal on both sides of the equation. This balance is crucial for calculating the net reaction in an electrochemical cell. The reaction sum gives us the overall chemical equation for the cell.
Faraday's Constant
Faraday's Constant is a critical value in electrochemistry that relates to the charge of one mole of electrons. Its value is 96485 Coulombs per mole (C/mol). This constant helps bridge the gap between the macroscopic world of chemistry and the microscopic world of electrons.
In calculations of Gibbs free energy (\(\Delta G^\circ\)) and electrochemical processes, Faraday's Constantis used to quantify how much charge is transferred during a reaction. The formula for Gibbs Free Energy is:\(\Delta G^\circ = -nFE^\circ_{cell}\),where \(n\) is the number of moles of electrons.
For example, the number of electrons transferred in the reaction mentioned in the solution is 2 moles. Using Faraday's Constant allows the conversion of voltage potential into energy change per mole of reaction, quantifying the energy released or needed for a process.
In calculations of Gibbs free energy (\(\Delta G^\circ\)) and electrochemical processes, Faraday's Constantis used to quantify how much charge is transferred during a reaction. The formula for Gibbs Free Energy is:\(\Delta G^\circ = -nFE^\circ_{cell}\),where \(n\) is the number of moles of electrons.
For example, the number of electrons transferred in the reaction mentioned in the solution is 2 moles. Using Faraday's Constant allows the conversion of voltage potential into energy change per mole of reaction, quantifying the energy released or needed for a process.
Other exercises in this chapter
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