Problem 62 - Chemistry and Chemical Reactivity

Question

A potential of \(+0.146 \mathrm{V}\) is recorded (under standard conditions) for a voltaic cell constructed using the following half-reactions: Anode: \(\mathrm{Ag}(\mathrm{s}) \longrightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-}\) Cathode: \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (a) What is the standard reduction potential for the cathode reaction? (b) Calculate the solubility product, \(K_{\mathrm{sp}},\) for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

(a) The cathode's standard reduction potential is \(-0.654 \mathrm{V}\). (b) Calculate the solubility product \(K_{sp}\) using the Nernst equation.

Step 1: Understand the Provided Cell Potential

The standard cell potential (E°) of the voltaic cell is given as +0.146 V. This potential is the difference between the cathode and anode potentials: \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\).

Step 2: Calculate Anode Potential

The anode reaction is \(\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-}\). The standard reduction potential of silver reduction (reverse reaction) is \(E^{\circ}_{\text{Ag}^+/\text{Ag}} = +0.80\, \mathrm{V}\). As the given reaction is oxidation, the potential will be \(-0.80\, \mathrm{V}\).

Step 3: Calculate Cathode Potential

The potential for the whole cell is known (0.146 V). We can use the formula \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\) to find the cathode potential.\[ E^{\circ}_{\text{cathode}} = E^{\circ}_{\text{cell}} + E^{\circ}_{\text{anode}} = 0.146\, \mathrm{V} + (-0.80\, \mathrm{V}) = -0.654\, \mathrm{V} \] So, the cathode standard reduction potential is \(-0.654\, \mathrm{V}\).

Step 4: Use Nernst Equation for Solubility Product

For the cathode reaction, apply the Nernst equation: \(E = E^{\circ} - \frac{RT}{nF}\ln Q\). Under standard conditions, \(Q = [\text{Products}]^{\text{coefficients}} / [\text{Reactants}]^{\text{coefficients}}.\) At equilibrium (E=0), \(Q = K_{sp}\):\[ 0 = -0.654 - \frac{RT}{2F}\ln [\text{Solubility} imes \text{Solubility}]\] Simplifying with known \(R\), \(T\), \(F\) values, and \(n = 2\), calculate \(K_{sp}\) for \(\mathrm{Ag}_2 \mathrm{SO}_4\).

Step 5: Solve for Solubility Product

After rearranging the Nernst equation, insert known constants and calculate: \[ K_{sp} = e^{\left(-\frac{nFE^{\circ}_{cathode}}{RT}\right)}\]Where \(F = 96500\, \mathrm{C/mol}\), \(R = 8.314\, \mathrm{J/(mol\,K)}\), and \(T = 298\, \mathrm{K}\), solve for \(K_{sp}\) using the previously found \(E^\circ_{cathode} = -0.654\, \mathrm{V}\).

Key Concepts

Voltaic Cell

A voltaic cell, also known as a galvanic cell, is an electrochemical device that produces electrical energy from spontaneous chemical reactions. It consists of two half-cells, each containing an electrode immersed in an electrolyte solution. In these cells, one material undergoes oxidation (loss of electrons) while another undergoes reduction (gain of electrons).
The two half-cells are connected by a salt bridge or a porous partition that maintains electrical neutrality by allowing ions to move between them. Electrons flow from the anode, where oxidation occurs, to the cathode, where reduction occurs. This flow of electrons through an external circuit produces electric current, which can be used to do work.
Key components of a voltaic cell include:

• Anode: The electrode where oxidation (loss of electrons) occurs.
• Cathode: The electrode where reduction (gain of electrons) occurs.
• External Circuit: Where electrons flow to do electrical work.
• Salt Bridge: Allows ions to transfer, maintaining the cell's charge balance.

Voltaic cells are widely used in batteries, where they provide a steady source of electrical energy by harnessing the energy of chemical reactions.

Standard Reduction Potential

The standard reduction potential is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. It is denoted as \(E^{\circ}\) and is measured in volts (V) under standard conditions: 25°C (298 K), 1 M concentration for all aqueous species, and 1 atm pressure for gases.
Every half-reaction has a standard reduction potential associated with it, which helps to determine how easily a substance can be reduced. This value is crucial for predicting the direction of redox reactions in electrochemistry.
In our example, the standard reduction potential of anode and cathode reactions can be used to calculate the overall cell potential using the equation: \[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
This formula shows that the cell potential is the difference between the reduction potential of the cathode and the anode. A positive cell potential indicates a spontaneous process, which is essential for voltaic cells to produce electricity.

Nernst Equation

The Nernst equation is a key component in electrochemistry for calculating the cell potential under non-standard conditions. It accounts for the effect of concentration on the potential of an electrochemical cell.
The equation is expressed as:
\[ E = E^{\circ} - \frac{RT}{nF}\ln Q \]
Where:

• \(E\) is the cell potential at non-standard conditions.
• \(E^{\circ}\) is the standard cell potential.
• \(R\) is the universal gas constant (8.314 J/(mol·K)).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged.
• \(F\) is Faraday's constant (96500 C/mol).
• \(Q\) is the reaction quotient.

The Nernst equation allows us to explore the reaction's feasibility under various conditions, including temperature and concentration changes. At equilibrium, \(E = 0\) and \(Q = K_{sp}\), which helps us calculate solubility products for sparingly soluble salts like \(Ag_2SO_4\).

Solubility Product

The solubility product, denoted as \(K_{sp}\), is a useful constant in understanding the dissolution of sparingly soluble ionic compounds. It provides information on the extent to which a salt can dissolve in water. \[ K_{sp} = [ ext{products}]^{ ext{coefficients}} / [ ext{reactants}]^{ ext{coefficients}} \]
The higher the \(K_{sp}\) value, the more soluble the compound is in solution. For example, a low \(K_{sp}\) of silver sulfate \(Ag_2SO_4\) indicates it does not dissolve easily in water.
In practice, \(K_{sp}\) lets us calculate the concentration of ions in saturated solutions and predict whether a precipitate will form under certain conditions. We can use the Nernst equation to relate cell potentials to \(K_{sp}\) for reactions involving insoluble salts, helping us estimate their solubilities at various conditions.
Understanding \(K_{sp}\) is crucial for various applications, from pharmaceuticals to wastewater treatment, where precise knowledge of solubility is required.