Problem 44
Question
Consider a voltaic cell with the reaction given below. As the cell reaction proceeds, what happens to the values of \(E_{\text {cell }}, \Delta_{r} G,\) and \(K_{\mathrm{c}} ?\) Explain your answers. $$ \begin{array}{r} \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \\ E_{\mathrm{cell}}^{\circ}=1.10 \mathrm{~V} \end{array} $$
Step-by-Step Solution
Verified Answer
As the reaction proceeds, \( E_{\text{cell}} \) decreases, \( \Delta_{r} G \) becomes less negative, and \( K_{c} \) remains constant.
1Step 1: Initial Understanding of the Reaction
This voltaic cell involves the reduction of Cu²⁺ to Cu and the oxidation of Zn to Zn²⁺. Initially, when the reaction starts, concentrations of reactants and products are at their standard states (1 M each), and the cell potential \( E_{\text{cell}} \) is equal to the standard cell potential \( E^{\circ}_{\text{cell}} = 1.10 \, \text{V} \).
2Step 2: Understanding Cell Potential \(E_{\text{cell}}\)
As the cell reaction proceeds, Cu²⁺ is consumed, and Zn²⁺ is produced, altering their concentrations away from 1 M. According to the Nernst equation, \( E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF}\ln Q \), where \( Q \) is the reaction quotient. Since \( Q \) will increase as the reaction progresses, \( E_{\text{cell}} \) will decrease.
3Step 3: Understanding Gibbs Free Energy Change \(\Delta_{r} G\)
The relationship between Gibbs free energy change and cell potential is given by \( \Delta_{r} G = -nFE \). As \( E_{\text{cell}} \) decreases, \( \Delta_{r} G \) becomes less negative, indicating the reaction becomes less spontaneous.
4Step 4: Understanding Equilibrium Constant \(K_{c}\)
\( K_{c} \) is a constant and represents the position of equilibrium. While the reaction proceeds towards equilibrium, \( K_{c} \) does not change because it is only dependent on temperature, which is constant unless otherwise specified. The cell reaction continues until \( Q = K_{c} \) at equilibrium, making \( E_{\text{cell}} = 0 \).
Key Concepts
ElectrochemistryNernst EquationGibbs Free EnergyEquilibrium Constant
Electrochemistry
Electrochemistry is a field that studies the interplay between electricity and chemical reactions. Central to electrochemistry are voltaic cells, or galvanic cells, which convert chemical energy into electrical energy. These cells consist of two half-reactions, with oxidation occurring at the anode and reduction occurring at the cathode.
This process results in the flow of electrons through an external circuit, generating electric current. In our reaction, copper ions are reduced while zinc is oxidized, creating a potential difference which can be measured as the cell potential \( E_{\text{cell}} \).
Understanding how these potentials change as the reaction proceeds is fundamental in electrochemistry, as it provides insight into the feasibility and direction of the reaction.
This process results in the flow of electrons through an external circuit, generating electric current. In our reaction, copper ions are reduced while zinc is oxidized, creating a potential difference which can be measured as the cell potential \( E_{\text{cell}} \).
Understanding how these potentials change as the reaction proceeds is fundamental in electrochemistry, as it provides insight into the feasibility and direction of the reaction.
Nernst Equation
The Nernst Equation is a critical tool in electrochemistry that links the cell potential \( E_{\text{cell}} \) to the concentrations of the reactants and products. It is defined as:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF}\ln Q \]
Where:
This equation thus allows us to predict how the cell potential changes with changing concentrations of ions in the solution.
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF}\ln Q \]
Where:
- \( E^{\circ}_{\text{cell}} \) is the standard cell potential.
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons exchanged.
- \( F \) is Faraday’s constant.
- \( Q \) is the reaction quotient.
This equation thus allows us to predict how the cell potential changes with changing concentrations of ions in the solution.
Gibbs Free Energy
Gibbs Free Energy, denoted by \( \Delta_{r} G \), is a thermodynamic quantity that reflects the amount of usable energy in a chemical reaction. It is directly related to the cell potential in an electrochemical cell via the equation:
\[ \Delta_{r} G = -nFE \]
Here, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( E \) is the cell potential.
\[ \Delta_{r} G = -nFE \]
Here, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday's constant, and \( E \) is the cell potential.
- When \( \Delta_{r} G \) is negative, the reaction is spontaneous under the given conditions.
- As \( E_{\text{cell}} \) decreases due to the progression of the reaction, \( \Delta_{r} G \) becomes less negative.
- This indicates the spontaneity of the reaction decreases as it moves towards equilibrium.
Equilibrium Constant
The Equilibrium Constant, \( K_{c} \), represents the ratio of concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced chemical equation. For the voltaic cell reaction, \( K_{c} \) can be related to the Gibbs Free Energy change by:\[ \Delta_{r} G^{\circ} = -RT\ln K_{c} \]This equation tells us:
- \( K_{c} \) remains constant for a reaction at a given temperature.
- The reaction proceeds until \( Q = K_{c} \), at which point the system reaches equilibrium.
- Once equilibrium is reached, \( E_{\text{cell}} = 0 \), indicating no net electron flow.
Other exercises in this chapter
Problem 37
The standard cell potential for the oxidation of \(\mathrm{Mg}\) by $$ \begin{aligned} \mathrm{Br}_{2} \text { is } 3.42 \mathrm{~V} \\ & \mathrm{Br}_{2}(\ell)+
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The standard cell potential, \(E^{\circ},\) for the reaction of \(\mathrm{Zn}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{~g})\) is \(2.12 \mathrm{~V}\). Write
View solution Problem 49
Calculate the cell potential of a concentration cell that contains two hydrogen electrodes if the cathode contacts a solution with \(\mathrm{pH}=7.8\) and the a
View solution Problem 50
Calculate the potential of a cell with one electrode made from zinc metal immersed in a solution where (conc. \(\left.\mathrm{Zn}^{2+}\right)=0.010 \mathrm{M}\)
View solution