Question

Given that \(E^{\circ}=0.52 \mathrm{~V}\) for the reduction \(\mathrm{Cu}^{+}(a q)+e^{-}\) \(\longrightarrow \mathrm{Cu}(s),\) calculate \(E^{\circ}, \Delta G^{\circ},\) and \(K\) for the following reaction at \(25^{\circ} \mathrm{C}\) : $$ 2 \mathrm{Cu}^{+}(a q) \rightleftarrows \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s) $$

Short answer

\( E^{\circ} = 0.16 \mathrm{~V} \), \( \Delta G^{\circ} = -15.44 \mathrm{~kJ/mol} \), \( K \approx 506 \).

Step-by-step solution

Write the Half-Reactions

To solve this problem, we first need to write the half-reactions involved. The given reduction half-reaction is \( \mathrm{Cu}^+(aq) + e^- \rightarrow \mathrm{Cu}(s) \) with \( E^\circ = 0.52 \mathrm{~V} \). However, this is only one part of the required process. We should identify the oxidation half-reaction for \( \mathrm{Cu}^+(aq) \rightarrow \mathrm{Cu}^{2+}(aq) + e^- \).

Calculate the Standard Reduction Potential for the Oxidation Half-Reaction

Given the complete cell reaction \( 2\mathrm{Cu}^{+}(aq) \rightleftarrows \mathrm{Cu}^{2+}(aq) + \mathrm{Cu}(s) \), we note its decomposition into \( \mathrm{Cu}^+(aq) + e^- \rightarrow \mathrm{Cu}(s) \) and \( \mathrm{Cu}^{2+}(aq) + e^- \rightleftarrows \mathrm{Cu}^+(aq) \). The potential for \( \mathrm{Cu}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cu}(s) \) is known to be \( 0.34 \mathrm{~V} \). Therefore, for the oxidation \( \mathrm{Cu}^+(aq) \rightarrow \mathrm{Cu}^{2+}(aq) + e^- \), \( E^\circ = 0.34 \mathrm{~V} - 0.52 \mathrm{~V} = -0.18 \mathrm{~V} \).

Calculate the Standard Electrode Potential for the Full Reaction

This reaction ocurss by combining the resulting oxidation and reduction reactions. Since we have identified two half reactions: \( \mathrm{Cu}^+(aq) + e^- \rightarrow \mathrm{Cu}(s) \) and \( \mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(aq) + 2e^- \), the full cell reaction potential \( E^\circ = 0.34 \mathrm{~V} + (-0.18) \mathrm{~V} = 0.16 \mathrm{~V} \).

Calculate the Standard Free Energy Change \(\Delta G^\circ\)

From the Nernst equation, \( \Delta G^\circ = -nFE^\circ \), where \( n = 1 \) (the number of moles of electrons transferred), \( F = 96485 \mathrm{~C \cdot mol^{-1}} \) (Faraday's constant), and \( E^\circ = 0.16 \mathrm{~V} \). Thus, \( \Delta G^\circ = -1 \times 96485 \times 0.16 = -15437.6 \mathrm{~J/mol} \) or \( -15.4376 \mathrm{~kJ/mol} \).

Calculate the Equilibrium Constant \(K\)

Using the relationship \( \Delta G^\circ = -RT\ln K \), where \( R = 8.314 \mathrm{~J/mol \cdot K} \) and \( T = 298 \mathrm{~K} \), rearrange to find \( \ln K = \frac{\Delta G^\circ}{-RT} = \frac{-15437.6}{-8.314 \times 298} \approx 6.23 \). Thus, \( K \approx e^{6.23} \approx 506 \).

Key concepts

Standard Electrode Potential

In electrochemistry, the standard electrode potential is a fundamental concept used to predict the direction of electron flow in a redox reaction. It is denoted as \( E^\circ \) and represents the potential difference between a reduction half-reaction under standard conditions compared to the standard hydrogen electrode, which is set at zero volts. Standard conditions imply solutions are 1 M, gases at 1 atm pressure, and 25°C (298 K). Understanding the standard electrode potential allows us to determine if a half-cell will act as an oxidizing or reducing agent.

• A positive \( E^\circ \) indicates the half-cell can gain electrons, thus it will easily undergo reduction.
• A negative \( E^\circ \) suggests the half-cell tends to lose electrons, making it likely to be oxidized.

In the given exercise, we calculated \( E^\circ \) for the reaction \( 2 \mathrm{Cu}^{+}(aq) \rightleftarrows \mathrm{Cu}^{2+}(aq) + \mathrm{Cu}(s) \) to be 0.16 V. This value indicates the tendency for the given redox reaction to proceed under standard conditions.

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) is a critical concept connecting thermodynamics with electrochemistry. It provides insights into the spontaneity of a chemical process. The equation \( \Delta G^\circ = -nFE^\circ \) shows the relationship between Gibbs free energy change and standard electrode potential.Here:

• \( n \) is the number of moles of electrons transferred in the reaction.
• \( F \) is Faraday's constant, approximately 96485 C/mol.

When \( \Delta G^\circ < 0 \), the reaction is spontaneous. If \( \Delta G^\circ > 0 \), it is non-spontaneous. In the exercise, we found that \( \Delta G^\circ \) for the reaction is \(-15.4376 \mathrm{~kJ/mol}\), indicating that the process is spontaneous since energy is released.

Nernst Equation

The Nernst equation allows us to calculate the cell potential at any concentrations, not just under standard conditions. It is a versatile tool that incorporates reaction quotient \( Q \) and temperature effects. The Nernst equation is given by:\[E = E^\circ - \frac{RT}{nF} \ln Q\]In this equation:

• \( E \) is the cell potential under non-standard conditions.
• \( R \) is the universal gas constant (8.314 J/mol·K).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons transferred.
• \( Q \) is the reaction quotient, reflecting the ratio of concentrations of products to reactants.

By using the Nernst equation, electrochemists can predict how the potential of a cell will change as concentrations change, providing real-world applicability in sensor design and battery technology.

Equilibrium Constant

The equilibrium constant \( K \) is another significant concept in electrochemistry, linking the extent of the reaction to thermodynamic data. The relationship is given by:\[\Delta G^\circ = -RT \ln K\]From this, the equation for \( K \) becomes:\[K = e^{-\frac{\Delta G^\circ}{RT}}\]A larger \( K \) value indicates a greater extent of reaction completion favoring products at equilibrium. In our exercise, we calculated \( K \approx 506 \), signifying that, at equilibrium, the concentration of products \( \mathrm{Cu}^{2+}(aq) \) and \( \mathrm{Cu}(s) \) is significantly higher compared to reactants, leaning towards the completion of the reaction.

Half-Reaction

In electrochemistry, a half-reaction is a part of the overall redox reaction and focuses on either the oxidation or reduction process. Each half-reaction can be written separately to identify electron balance and the specific redox changes occurring.For the given exercise, the half-reactions involved are:

• Reduction: \( \mathrm{Cu}^+(aq) + e^- \rightarrow \mathrm{Cu}(s) \)
• Oxidation: \( \mathrm{Cu}^{2+}(aq) + e^- \rightarrow \mathrm{Cu}^+(aq) \)

By balancing these half-reactions, we determine how electrons are transferred and compute the final cell potential. Understanding half-reactions is essential as they allow precise calculation of the changing reduction states of elements within an electrochemical cell.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between chemical species. One substance gets oxidized (loses electrons), while the other gets reduced (gains electrons).Redox reactions are fundamental in electrochemistry and are often harnessed in batteries, corrosion, and electroplating.Key aspects include:

• Oxidizing agent: Species that gets reduced.
• Reducing agent: Species that gets oxidized.

In the problem statement, the reaction is \( 2 \mathrm{Cu}^{+}(aq) \rightleftarrows \mathrm{Cu}^{2+}(aq) + \mathrm{Cu}(s) \). It involves copper changing oxidation states, an example of a typical redox process where electron management facilitates electrochemical reactions.