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Consider the standard galvanic cell based on the following half-reactions: $$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$$ $$\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}$$ The electrodes in this cell are \(A g(s)\) and \(C u(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\operatorname{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) . ]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: Ag' reacts with Cl- to form AgCl(s). ] d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V}$$

Step by step solution

01

(1) Calculate the standard cell potential

To begin, let's first establish the standard cell potential. Since this is galvanic cell with Cu虏鈦�/Cu and Ag鈦�/Ag as electrodes, we can calculate the standard cell potential through: \(E_{cell}^o = E_{reduction, Ag}^o - E_{reduction, Cu}^o\). Given values for \(E_{reduction, Ag}^o = 0.80 V\), and \(E_{reduction, Cu}^o = 0.34 V\). Thus, \(E_{cell}^o = 0.80 - 0.34 = 0.46 V\). Now, let's analyze the effects of the changes in parts a through e. #a. Adding CuSO4 to Cu half-cell compartment#

02

(2a) Effect of CuSO4 on Cu虏鈦� concentration

When we add CuSO4 in the copper half-cell compartment, it dissociates into Cu虏鈦� and SO鈧劼测伝 ions increasing the concentration of Cu虏鈦� ions. This would create more driving force for the reduction of Cu虏鈦� ions.

03

(2b) Calculate the effect on potential

The increase in Cu虏鈦� concentration shifts the Cu虏鈦�/Cu half-reaction according to the Nernst equation: \(E = E^o - \frac{RT}{nF} \ln{Q}\) Since the standard cell potential will remain the same, E_{reduction, Ag}^o, and \(E_{reduction, Cu}^o\) will also remain the same. As the concentration of Cu虏鈦� increases, Q will decrease making the denominator in the equation more negative, thus increasing the cell potential. So, the cell potential will increase in this case. #b. Adding NH3 to Cu half-cell compartment#

04

(3a) Effect of NH3 on Cu虏鈦� concentration

Cu虏鈦� reacts with NH鈧� to form Cu(NH鈧�)鈧劼测伜, which effectively reduces the concentration of Cu虏鈦� in the solution by removing them to form the complex ion.

05

(3b) Calculate the effect on potential

The reduction in Cu虏鈦� concentration as a consequence of the formation of Cu(NH鈧�)鈧劼测伜 will decrease Q value leading to a decrease in cell potential. #c. Adding NaCl to the silver half-cell compartment#

06

(4a) Effect of NaCl on Ag鈦� concentration

NaCl dissociates into Na鈦� and Cl鈦� ions; Cl鈦� ions would slowly remove Ag鈦� concentration by formation of silver chloride AgCl precipitate.

07

(4b) Calculate the effect on potential

The decrease in Ag鈦� concentration raises the log term in the Nernst equation, which will reduce the cell potential. #d. Doubling the volume of both half-cell compartments#

08

(5) Effect of dilution on the cell potential

Diluting both compartments by doubling the volume causes an equal decrease in concentrations of Cu虏鈦� and Ag鈦� ions. Since the change is equal to both half-cell compartments, there will be no effect on the standard cell potential. #e. Replacing the silver electrode with a platinum electrode#

09

(6) Effect of replacing the electrode on potential

Changing the electrode from silver to platinum does not affect the half-reactions occurring in the cell. The platinum electrode does not change the concentrations of the species involved in the half-reactions and thus does not affect the standard cell potential.

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Key Concepts

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

Cell Potential

Cell potential, also known as electromotive force (EMF), is a critical concept in understanding galvanic or voltaic cells. It represents the voltage or the difference in electric potential between two electrodes in an electrochemical cell. The cell potential is determined by the inherent characteristics of the half-reactions occurring at each electrode.

In a standard galvanic cell, the cell potential can be calculated using the standard reduction potentials of the two half-reactions involved. For example, in a cell with copper and silver electrodes, the standard cell potential can be calculated using the formula:

\( E_{cell}^o = E_{reduction, Ag}^o - E_{reduction, Cu}^o \).

The potential difference translates to the cell's ability to do electrical work, and it changes with alterations in concentrations, pressures, or temperatures under non-standard conditions.

Nernst Equation

The Nernst equation is a fundamental tool in electrochemistry that enables us to calculate the cell potential under non-standard conditions. It considers the effect of concentration changes on the cell potential and is especially useful for predicting cell behavior in real practical scenarios.

The Nernst equation is given by:
\[ E = E^o - \frac{RT}{nF} \ln{Q} \] where \( E \) is the cell potential, \( E^o \) 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 transferred, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient representing the concentration ratio of the products to reactants.

Changes in ion concentrations, like increasing \( \mathrm{Cu}^{2+} \) concentration through adding \( \mathrm{CuSO}_{4} \), affect \( Q \) and thereby alter the cell potential, as explained through the Nernst equation.

Half-Reactions

In a galvanic cell, the process of generating electrical energy involves two key reactions at the electrodes known as half-reactions. Each half-reaction occurs in one of the cell's compartments, helping convert chemical energy into electrical energy.

Half-reactions are essentially the breakdown of the overall redox reaction into two parts: one demonstrating oxidation where electrons are lost, and the other showing reduction where electrons are gained. Each half-reaction has its own standard reduction potential, indicating its tendency to gain electrons.

For instance, in a galvanic cell comprising of copper and silver electrodes, the half-reactions are: \( \mathrm{Cu}^{2+}+2 \, \mathrm{e}^{-} \rightarrow \mathrm{Cu} \) and \( \mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \). These half-reactions help in calculating the standard cell potential and analyzing the cell's behavior under varying conditions.

Electrode Reactions

Electrode reactions form the foundation of a galvanic cell's operation where conversion between electrochemical energy and electrical energy occurs.

At each electrode, a separate reaction takes place: the anode hosts the oxidation reaction, and the cathode hosts the reduction reaction. These reactions are driven by differences in standard reduction potentials, creating a flow of electrons through an external circuit.

The materials and reactions at the electrodes define the overall reaction and resulting cell potential. For example, in the standard galvanic cell with copper and silver electrodes, silver acts as the cathode (reduction) and copper as the anode (oxidation). Changes in the chemical environment, such as adding \( \mathrm{NH}_{3} \) which forms complexes with \( \mathrm{Cu}^{2+} \), can significantly affect the electrode reactions and the resulting cell potential. Understanding these reactions allows for prediction and manipulation of galvanic cell behavior.