Question

A A cell is constructed using the following half-reactions: $$ \begin{array}{c} \mathrm{Ag}^{+}(\mathrm{aq})+e^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \\ \mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s})+2 e^{-} \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{array} $$ \(E^{\bullet}=0.653 \mathrm{V}\) (a) What reactions should be observed at the anode and cathode? (b) Calculate the solubility product constant, \(K_{\mathrm{zp}}\) for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\)

Short answer

Cathode: \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \); Anode: \( \mathrm{Ag}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{Ag}^{+} + \mathrm{SO}_{4}^{2-} \); \(K_{sp} = 4s^3\).

Step-by-step solution

Identify the Cathode Reaction

In a galvanic cell, the reaction with the higher standard potential occurs at the cathode. Here, the reduction of \( \mathrm{Ag}^{+} \) to \( \mathrm{Ag} \) with \( E^\bullet = 0.653 \mathrm{V} \) is known from standard tables to have a higher potential than the reduction of \( \mathrm{Ag}_2\mathrm{SO}_4 \). Thus, the cathode reaction is:\[ \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \]

Identify the Anode Reaction

The oxidation reaction occurring at the anode usually involves the substance with the lower potential in its reduction form. Given the half-reaction: \( \mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) + 2 e^{-} \rightarrow 2 \mathrm{Ag}(\mathrm{s}) + \mathrm{SO}_{4}^{2-}(\mathrm{aq}) \), we understand this implies the dissolution of \( \mathrm{Ag}_2\mathrm{SO}_4 \) and it can act as the anode.

Write the Overall Cell Reaction

By combining the half-reactions, the overall cell reaction can be written. Since we have: - Cathode: \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \) - Anode: \( \mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{SO}_{4}^{2-}(\mathrm{aq}) \) The overall reaction becomes: \[ \mathrm{Ag}_{2}\mathrm{SO}_{4}(\mathrm{s}) \rightarrow 2\mathrm{Ag}(\mathrm{s}) + 2\mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{SO}_{4}^{2-} \]

Calculate the Solubility Product Constant

The solubility product constant \(K_{sp}\) is related to the solubility of \(\mathrm{Ag}_2\mathrm{SO}_4\) in water. We denote its dissociation in water as: \[ \mathrm{Ag}_{2}\mathrm{SO}_{4} \rightleftharpoons 2\mathrm{Ag}^{+} + \mathrm{SO}_{4}^{2-} \]. Let \(s\) be the solubility of \( \mathrm{Ag}_{2}\mathrm{SO}_{4} \), then \([\mathrm{Ag}^{+}] = 2s\) and \([\mathrm{SO}_{4}^{2-}] = s \). Therefore, \(K_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{SO}_{4}^{2-}] = (2s)^2(s) = 4s^3\). We need \(E^\bullet\) or experimental values to solve for \(s\) and \(K_{sp}\).

Key concepts

Cell potential

In the realm of electrochemistry, the term 'cell potential' is crucial. It reflects the voltage, or the electric potential difference, generated by an electrochemical cell. In simpler terms, it indicates how much energy is available to drive reactions within the cell. Cell potential depends on the redox reactions occurring at the electrodes. Each half-reaction comes with a standard cell potential (Eº), available from electrochemical tables.
For any galvanic cell, the overall cell potential is calculated by subtracting the anode potential from the cathode potential using the formula:\[E_{ ext{cell}} = E_{ ext{cathode}} - E_{ ext{anode}} \]
The cathode reaction will always have a larger (more positive) cell potential than the anode. Therefore, identifying both the cathode and anode reactions determines the overall potential. Understanding cell potential helps predict the feasibility and direction of the chemical reaction.

Half-reaction

Half-reactions are an integral part of understanding electrochemical cells. They represent the individual parts of a redox reaction—either the oxidation or reduction process. Recognizing these reactions is essential to grasp how electrons move in an electrochemical circuit.
Each half-reaction consists of two main elements: a reductant and an oxidant. In the cathode, reduction occurs, which involves the gain of electrons. At the anode, oxidation involves the loss of electrons.
For instance, we might have half-reactions like:

• Reduction: \( ext{Ag}^{+} + e^{-} ightarrow ext{Ag}( ext{s}) \)
• Oxidation: \( ext{Ag}_2 ext{SO}_4( ext{s}) + 2e^{-} ightarrow 2 ext{Ag}( ext{s}) + ext{SO}_4^{2-}( ext{aq}) \)

Balancing these equations is crucial since they ensure that charge and mass are conserved. Ultimately, half-reactions provide insight into the electron flow and the atomic transformations occurring within the cell.

Solubility product constant

The solubility product constant, often denoted as \( K_{sp} \), is vital when discussing slightly soluble compounds in solutions. It is a measure of the degree to which a compound can dissolve in water. The larger the \( K_{sp} \), the more soluble the compound is in the solution.
For a compound like \( ext{Ag}_2 ext{SO}_4 \), the equation for dissociation is:\[ ext{Ag}_2 ext{SO}_4(s) ightleftharpoons 2 ext{Ag}^{+}(aq) + ext{SO}_4^{2-}(aq)\]
When calculating \( K_{sp} \), we consider the concentrations of the ions formed in the equilibrium. If \( s \) represents the solubility of \( ext{Ag}_2 ext{SO}_4 \):

• \( [ ext{Ag}^{+}] = 2s \)
• \( [ ext{SO}_4^{2-}] = s \)

Thus, \( K_{sp} = [ ext{Ag}^{+}]^2[ ext{SO}_4^{2-}] = 4s^3 \).
Determining the actual \( K_{sp} \) requires experimental data or known solubility values for the compound in question.

Galvanic cell

Galvanic cells, also known as voltaic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. They are the cornerstone of many modern batteries. The structure of a galvanic cell includes two electrodes immersed in electrolyte solutions, connected by a salt bridge, facilitating ionic movement, and completing the electrical circuit.
Each electrode houses a particular half-reaction—reduction occurs at the cathode, and oxidation happens at the anode. These reactions collectively generate electrical energy, evident in the form of cell potential.
Consider a typical galvanic cell setup:

• The cathode might involve a reaction like \( ext{Ag}^{+} + e^{-} \rightarrow ext{Ag}( ext{s}) \).
• Whereas, the anode might involve \( ext{Ag}_2 ext{SO}_4( ext{s}) \rightarrow 2 ext{Ag}^{+}(aq) + ext{SO}_4^{2-}(aq) \).

Understanding the role of a galvanic cell aids in comprehending how batteries operate, facilitating devices we use daily, from phones to cars.