Question

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) 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: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) 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. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

Short answer

In summary: a) When CuSO4(s) is added to the copper half-cell, the cell potential increases. b) When NH3(aq) is added to the copper half-cell, the cell potential decreases. c) When NaCl(s) is added to the silver half-cell, the cell potential increases. d) When water is added to both half-cell compartments, the cell potential remains the same. e) When the silver electrode is replaced with a platinum electrode, the cell potential remains the same.

Step-by-step solution

Write the overall cell reaction

First, we will write the overall cell reaction by combining the copper and silver half-reactions: \[ Cu^{2+}(aq) + 2Ag^+(aq) \longrightarrow Cu(s) + 2Ag(s) \] Now, we can analyze each change in the exercise and determine its effect on the cell potential.

a. Adding CuSO4(s) to the copper half-cell

When CuSO4 is added to the copper half-cell compartment, it will dissolve and increase the concentration of Cu²⁺ ions. According to Le Chatelier's principle, the reaction will shift in the direction that reduces the stress caused by this change, namely, towards the reduction of Cu²⁺ ions. Therefore, the cell potential will increase.

b. Adding NH3(aq) to the copper half-cell

When NH3 is added to the copper half-cell, it will react with Cu²⁺ ions to form Cu(NH3)₄²⁺ complex ions. This reaction will decrease the concentration of free Cu²⁺ ions in the solution. According to Le Chatelier's principle, the reaction will shift to restore the concentration of Cu²⁺ ions, favoring the oxidation of Cu(s). Therefore, the cell potential will decrease.

c. Adding NaCl(s) to the silver half-cell

When NaCl(s) is added to the silver half-cell compartment, it will dissolve and increase the concentration of Cl⁻ ions. Cl⁻ ions react with Ag⁺ ions to form AgCl(s), decreasing the concentration of free Ag⁺ ions in the solution. According to Le Chatelier's principle, the reaction will shift to restore the concentration of Ag⁺ ions, favoring the reduction of Ag⁺ ions. Therefore, the cell potential will increase.

d. Doubling the volume of the solution in both half-cells

When water is added to both half-cell compartments, diluting the solution, the concentrations of Cu²⁺ and Ag⁺ ions will decrease. According to the Nernst equation, the cell potential depends on the concentrations of the ions involved in the reaction. However, since both half-cells are diluted by the same factor, the ratio of the concentrations of Cu²⁺ and Ag⁺ ions will remain the same, and the cell potential will remain the same.

e. Replacing the silver electrode with a platinum electrode

Changing the electrode material does not affect the overall cell reaction or the redox processes taking place at the electrodes, as long as the new electrode (in this case, platinum) is a chemically inert conductor. In this situation, the platinum electrode will only serve as an electron conductor and will not participate in the redox reactions directly. Therefore, the cell potential will remain the same. In summary: a) Cell potential increases b) Cell potential decreases c) Cell potential increases d) Cell potential remains the same e) Cell potential remains the same

Key concepts

Galvanic Cells

Galvanic cells, also known as voltaic cells, are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two separate compartments called half-cells. Each half-cell contains an electrode and an electrolyte solution, where one species undergoes oxidation and the other undergoes reduction.

• The anode is the electrode where oxidation occurs. Electrons are released during this process.
• The cathode is the electrode where reduction happens. Electrons are gained here.

In a galvanic cell, electrons flow from the anode to the cathode through an external circuit, creating an electric current. This current can be harnessed to perform work like turning on a light bulb.
The salt bridge is a crucial part of these cells, allowing ions to move between the half-cells. This movement helps maintain electrical neutrality and enables continuous electron flow.
Each half-cell in a galvanic cell is characterized by a specific half-reaction and an associated electrode potential. The difference in these potentials between the two half-cells gives rise to the cell potential or cell voltage.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in chemistry that helps predict how a system at equilibrium responds to various changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to partially counteract the change and reach a new equilibrium. In electrochemical contexts, this principle aids in understanding how alterations in concentration or pressure affect cell reactions.

When changes are introduced to a galvanic cell, such as adding a reactant or a product, the reaction equilibrium shifts to reduce the impact of that change.

• Increasing the concentration of reactants causes the equilibrium to shift towards products, potentially increasing cell potential.
• Conversely, increasing the concentration of products shifts the equilibrium towards reactants, possibly decreasing cell potential.

Le Chatelier's Principle helps explain why, for instance, adding \( ext{CuSO}_4\) to a copper half-cell increases the cell potential, as the reaction leans towards the copper's reduction to consume excess ions. Understanding this principle is fundamental to predicting the behavior of galvanic cells under different environmental changes.

Half-Reactions

In the study of electrochemistry, half-reactions are an important concept. They split an overall redox reaction into two parts: oxidation and reduction. Each half-reaction represents what happens at one electrode of a galvanic cell.

• The oxidation half-reaction occurs at the anode and indicates the loss of electrons by a species.
• The reduction half-reaction occurs at the cathode and shows the gain of electrons by another species.

By examining half-reactions, one can determine which species is oxidized and which is reduced, as well as the direction of electron flow in the cell.
For instance, in a standard galvanic cell where copper and silver are involved, the half-reactions might be written as:
\[ \text{Copper: } Cu^{2+} + 2e^{-} \rightarrow Cu \] \[ \text{Silver: } Ag^{+} + e^{-} \rightarrow Ag \] Understanding half-reactions is critical because it helps calculate the cell potential and shows how changes affect the cell's function.

Cell Potential

Cell potential, also called electromotive force (EMF), is a measure of the voltage difference between two electrodes in a galvanic cell. It indicates the cell's ability to drive electric current through an external circuit. The standard cell potential \( E^0_{cell} \) is calculated using the standard reduction potentials of the cathode and anode.

The formula for calculating cell potential is:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]
This equation means the cell potential is the difference in reduction potentials between the cathode and anode. A positive cell potential indicates a spontaneous reaction, essential for a functionating galvanic cell.

Modifications like changing ion concentrations can alter cell potential. For example, adding reactants at the cathode usually increases it, and adding them at the anode often decreases it. The Nernst Equation further helps in understanding how real-life cell potentials deviate from standard conditions due to concentration changes:
\[ E = E^0_{cell} - \frac{RT}{nF} \ln Q \] where \ R \ is the gas constant, \ T \ is the temperature in Kelvin, \ n \ is the number of electrons exchanged per mole of reaction, \ F \ is Faraday's constant, and \ Q \ is the reaction quotient.
Understanding cell potential is crucial for assessing cell performance and reaction spontaneity.