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If the equilibrium constant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{-5},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Short Answer

Expert verified
\( \Delta G^\circ \) is 26.556 kJ/mol and \( E^\circ \) is approximately -0.275 V.

Step by step solution

01

Recalling the Relationship between Equilibrium Constant and Gibbs Free Energy

To begin, we use the equation that relates the standard Gibbs free energy change \( \Delta G^\circ \) to the equilibrium constant \( K \), which is: \[ \Delta G^\circ = -RT \ln K \]. Here, \( R \) is the universal gas constant (8.314 J/mol·K), \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant.
02

Calculating the Natural Logarithm of the Equilibrium Constant

We need the natural logarithm of the equilibrium constant \( K = 2.2 \times 10^{-5} \). \( \ln(2.2 \times 10^{-5}) \approx -10.72 \).
03

Substituting Values into the Gibbs Free Energy Equation

Substitute the known values into the equation: \( R = 8.314 \) J/mol·K, \( T = 298 \) K, and \( \ln K = -10.72 \): \[ \Delta G^\circ = - (8.314 \text{ J/mol·K})(298 \text{ K})(-10.72) \].
04

Calculating \( \Delta G^\circ \)

Compute \( \Delta G^\circ \) using the substituted values: \[ \Delta G^\circ \approx 26556 \text{ J/mol} = 26.556 \text{ kJ/mol} \].
05

Recalling the Relationship between Gibbs Free Energy and Cell Potential

Next, use the relationship \( \Delta G^\circ = -nFE^\circ \), where \( n \) is the number of moles of electrons transferred (1 for this reaction), \( F \) is the Faraday constant (96485 C/mol), and \( E^\circ \) is the standard cell potential.
06

Solving for the Standard Cell Potential \( E^\circ \)

Rearrange the equation \( 26.556 \text{ kJ/mol} = -(1)(96485 \text{ C/mol})(E^\circ) \) to find \( E^\circ \): \[ E^\circ = \frac{-26.556 \times 10^3 \text{ J/mol}}{96485 \text{ C/mol}} \approx -0.275 \text{ V} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, often denoted as \( K \), is a key concept in chemical reactions and thermodynamics. It provides a measure of the extent to which a chemical reaction reaches equilibrium. For a given reaction, it's defined by the ratio of the concentrations of products to reactants, each raised to the power of their respective stoichiometric coefficients.

This constant is crucial in predicting the direction in which a reaction will proceed to reach equilibrium. If \( K \) is much greater than 1, the reaction favors the formation of products; conversely, if \( K \) is much smaller than 1, the reactants are favored and very little product is formed. In our example, the equilibrium constant \( K = 2.2 \times 10^{-5} \) indicates a strong favor towards the reactants.
  • The equilibrium constant is temperature-dependent.
  • A small \( K \) value signifies that at equilibrium, the concentration of reactants is much larger than that of the products.
Understanding \( K \) allows chemists to manipulate conditions to drive reactions toward more desired outcomes.
Gibbs Free Energy
Gibbs free energy, represented by \( Delta G \), is a thermodynamic quantity critical for predicting whether a reaction can occur spontaneously. The change in Gibbs free energy, \( Delta G^\circ \), relates directly to the equilibrium constant through the equation:

\[ Delta G^\circ = -RT \ln K \]

Here, \( R \) is the universal gas constant (8.314 J/mol·K), \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant. When \( Delta G^\circ \) is negative, the reaction proceeds spontaneously in the forward direction. A positive \( Delta G^\circ \) indicates a non-spontaneous reaction under standard conditions.
  • \( Delta G^\circ \) incorporates both enthalpy (\( Delta H^\circ \)) and entropy (\( Delta S^\circ \)) changes in the reaction.
  • The equation \( Delta G^\circ = Delta H^\circ - TDelta S^\circ \) provides a deeper insight into what dominates the spontaneity: heat exchange or disorder.
In our exercise, the calculation of \( Delta G^\circ \) showcases how low equilibrium constants result in positive \( Delta G^\circ \), indicating non-spontaneous reactions.
Standard Cell Potential
The standard cell potential, \( E^\circ \), is a measure of the voltage difference between two half-cells in a galvanic cell under standard conditions (1 atm, 1 M concentration, and 298 K). It reflects the ability of a redox reaction to perform electrical work.

\( E^\circ \) is directly related to Gibbs free energy through the equation:

\[ Delta G^\circ = -nF E^\circ \]

where \( n \) is the number of moles of electrons transferred during the reaction, and \( F \) is the Faraday constant (96485 C/mol). The sign of \( E^\circ \) indicates the spontaneity of a reaction: positive values suggest a reaction can occur spontaneously, while negative values suggest non-spontaneity.
  • This relationship provides a bridge between electrochemistry and thermodynamics.
  • Given \( E^\circ \) and \( Delta G^\circ \), one can determine the efficacy of a cell to produce electrical energy.
In our solved task, we find \( E^\circ = -0.275 \text{ V} \), indicating the reaction in the standard state isn't able to do work without additional input of energy.

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Most popular questions from this chapter

A voltaic cell utilizes the following reaction: \(4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=\) \(0.010 M, P_{\mathrm{O}_{2}}=50.7 \mathrm{kPa}\), and the \(\mathrm{pH}\) of the solution in the cathode half-cell is \(3.50 ?\)

Indicate whether each statement is true or false: (a) The cathode is the electrode at which oxidation takes place. (b) A galvanic cell is another name for a voltaic cell. (c) Electrons flow spontaneously from anode to cathode in a voltaic cell.

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) \(14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)\) $$ \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q) $$ (b) \(2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) $$ \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q) $$ (c) \(\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

You may have heard that "antioxidants" are good for your health. Is an "antioxidant" an oxidizing agent or a reducing agent? [Sections 20.1 and 20.2\(]\)

(a) Write the half-reaction that occurs at an oxygen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell. (b) Write the half-reaction that occurs at an oxygen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard oxygen electrode?

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