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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. $$ \begin{array}{r} \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})+\mathrm{Zn}(\mathrm{s}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \\ E_{\mathrm{cell}}^{\circ}=1.10 \mathrm{~V} \end{array} $$

Short Answer

Expert verified
As the reaction proceeds, \( E_{\text{cell}} \) decreases, \( \Delta_{r} G \) becomes less negative, and \( K_{c} \) remains constant.

Step by step solution

01

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 \( E^{\circ}_{\text{cell}} = 1.10 \, \text{V} \).
02

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.
03

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.
04

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 \( E_{\text{cell}} = 0 \).

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Key Concepts

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

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.
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:
  • \( E^{\circ}_{\text{cell}} \) 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.
As the reaction in the voltaic cell proceeds, the concentration of reactants decreases and the concentration of products increases. Consequently, the value of \( Q \) rises, leading to a decrease in the value of \( E_{\text{cell}} \).
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.
  • 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.
Understanding Gibbs Free Energy helps determine how long a cell can produce electrical energy before a reaction reaches 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.
Understanding \( K_{c} \) provides context for how far a reaction will proceed, as well as the maximal voltage a cell can produce before reaching equilibrium. This concept is critical for predicting the efficiency and limits of electrochemical cells.

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Most popular questions from this chapter

Consider two different electrolytic cells; one cell contains aqueous \(\mathrm{Zn}^{2+}\) and the other contains \(\mathrm{Cr}^{3+}\). The initial metal ion concentration is the same in each cell and the metal ions are reduced to the metal during the electrolysis. Each cell operates at the same current. Without doing calculations, predict which cell has the greater mass of metal deposited after 5 min. Explain your prediction.

In this reaction, assign an oxidation number to each atom in reactants and products. Identify which substance is oxidized and which is reduced. Identify the oxidizing agent and the reducing agent. $$ \begin{aligned} 8 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq})+& 5 \mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \\ & 5 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Mn}^{2+}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned} $$

This reaction occurs in a cell with \(\mathrm{H}_{2}(\mathrm{~g})\) pressure of \(1.0 \mathrm{~atm}\) and (conc. \(\left.\mathrm{Cl}^{-}\right)=1.0 \mathrm{M}\) at \(25^{\circ} \mathrm{C} ;\) the measured \(E_{\text {cell }}=0.34 \mathrm{~V}\). Calculate the \(\mathrm{pH}\) of the solution. $$ \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{AgCl}(\mathrm{s}) \longrightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) $$

The standard cell potential for the oxidation of \(\mathrm{Mg}\) by $$ \begin{aligned} \mathrm{Br}_{2} \text { is } 3.42 \mathrm{~V} \\ & \mathrm{Br}_{2}(\ell)+\mathrm{Mg}(\mathrm{s}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \end{aligned} $$ (a) Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) for this reaction. (b) Suppose the equation is written with all coefficients doubled. Determine \(\Delta_{\mathrm{r}} G^{\circ}\) and \(E^{\circ}\) for this new equation.

To measure the \(\mathrm{Ag}^{+}\) concentration, \(25.00 \mathrm{~mL}\) of a silvercontaining solution is titrated with \(0.015 \mathrm{M} \mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) by using a silver electrode immersed in the test solution and the electrical potential measured against a standard hydrogen electrode. It required \(16.7 \mathrm{~mL}\) of the KI solution to reach the equivalence point, where the potential was \(0.325 \mathrm{~V}\). (a) Calculate the molarity of \(\mathrm{Ag}^{+}\) in the solution. (b) Calculate the \(K_{\mathrm{sp}}\) of \(\mathrm{AgI}\).

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