Question

A galvanic cell is based on the following half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{array}$$ In this cell, the silver compartment contains a silver electrode and excess \(\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. Assuming \(1.0 \mathrm{L}\) of \(2.0 \mathrm{M}\space \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) & \rightleftharpoons \ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$

Short answer

In this galvanic cell with specified half-reactions, the potential at 25°C is 0.46 V. To obtain a cell potential of 0.52 V, approximately 0.017 moles of NH₃ should be added to the copper compartment.

Step-by-step solution

Determine the balanced overall cell reaction

First, we need to balance the two half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & (1) \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & (2) \end{array}$$ Multiply reaction (1) by 2, and then subtract the resulting reaction from reaction (2) to get the overall balanced reaction: 2Ag⁺ + 2e⁻ → 2Ag(s) (multiply by 2) Cu²⁺ + 2e⁻ → Cu(s) --------------------------------------- 2Ag⁺ + Cu → 2Ag(s) + Cu²⁺

Calculate cell potential using standard electrode potentials

We can now calculate the cell potential (E_cell) using the standard electrode potentials: E_cell = Eº_reduced - Eº_oxidized In this case, Ag⁺ + e⁻ → Ag(s) is the reduction reaction and Cu²⁺ + 2e⁻ → Cu(s) is the oxidation reaction. So, we have: E_cell = 0.80 V - 0.34 V = 0.46 V The potential for this cell at 25°C is 0.46 V. #b. Moles of NH₃ needed for specific cell potential#

Calculate the [Cu(NH₃)₄²⁺] needed for the desired cell potential

We have to use the Nernst equation to find the required concentration of Cu(NH₃)₄²⁺: E_cell = Eº_cell - \(\frac{RT}{nF}\) ln(Q) Here, E_cell is 0.52 V (given), Eº_cell is 0.46 V (from previous step), R is the gas constant (8.314 J/mol·K), T is the temperature (25°C = 298 K), n is the number of electrons transferred (2), F is Faraday's constant (96485 C/mol), and Q is the reaction quotient. 0.52 V = 0.46 V - \(\frac{8.314 \times 298}{2 \times 96485}\) ln(Q) Now, let's find Q. We can write it as: Q = \(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}] \times [\mathrm{Ag}^+]^2}{[\mathrm{Cu}^{2+}]}\) The standard Gibbs free energy change for dissolution of AgCl is: ΔGº = -RT ln(K_sp) = -nFEº_AgCl From the information given in the problem, we can find Eº_AgCl: Eº_AgCl = -\(\frac{ΔGº}{nF}\) = -\(\frac{RT \times ln(K_{sp})}{nF}\) = -\(\frac{8.314 \times 298 \times ln(1.6 \times 10^{-10})}{96485}\) ≈ 0.225 V We also know that Eº_cell = Eº_Ag - Eº_AgCl, so: Eº_Ag = Eº_cell + Eº_AgCl = 0.46 V + 0.225 V ≈ 0.685 V The concentration of Ag⁺ ions is: [Ag⁺] = \(\sqrt{\frac{K_{sp}}{[\mathrm{Ag}]}} = \sqrt{\frac{1.6 \times 10^{-10}}{1}}\) ≈ 4.0 × 10⁻⁶ M Now we can find the required [Cu(NH₃)₄²⁺] to achieve 0.52 V: 0.52 V = 0.46 V - \(\frac{8.314 \times 298}{2 \times 96485}\) ln(\(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}] \times (4.0 \times 10^{-6})^2}{2.0}\)) On solving the above equation, we get: [Cu(NH₃)₄²⁺] ≈ 1.91 M

Calculate the moles of NH₃ required

To find the moles of NH₃ required, we'll use the given expression for K: K = \(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}]}{[\mathrm{Cu}^{2+}] \times [\mathrm{NH}_3]^{4}}\) Rearranging the equation and substituting the values: [\mathrm{NH}_3]⁴ = \(\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}]}{K \times [\mathrm{Cu}^{2+}]}\) = \(\frac{1.91}{1.0 \times 10^{13} \times 2.0}\) [\mathrm{NH}_3] ≈ 0.017 M Now, considering this problem asked for the moles of NH₃ needed and a 2.0 M solution of Cu²⁺ in 1.0 L: Moles of NH₃ required = (0.017 mol/L) × 1.0 L ≈ 0.017 mol The moles of NH₃ that would have to be added to give a cell potential of 0.52 V at 25°C are approximately 0.017 mol.

Key concepts

Understanding Standard Electrode Potentials

Standard electrode potentials, expressed as E° values, are measures of the inherent tendency of redox species to gain or lose electrons. These values serve as a benchmark telling us how likely an element is to be reduced under standard conditions, which are typically 25°C (298 K), 1 atm pressure, and 1M concentration for each ion. When comparing two half-reactions in a galvanic cell, the reaction with the higher E° acts as the cathode (reduction), while the one with the lower E° acts as the anode (oxidation).

In our exercise, the standard electrode potentials of silver and copper are given as +0.80V and +0.34V, respectively. This implies that under standard conditions, silver ions are more likely to gain electrons and form solid silver compared to copper ions, which makes silver the cathode and copper the anode.

Interpretation of E° values:

• A higher E° value indicates a greater likelihood of reduction.
• A negative E° value signifies a tendency to undergo oxidation.
• Standard electrode potentials are crucial for calculating the overall cell voltage by subtracting the anode E° from the cathode E°.

Understanding these potentials is essential as they are foundational elements of electrochemistry and crucial to determining the direction in which an electrochemical reaction will proceed.

The Nernst Equation and Its Application

The Nernst equation is a relationship that allows us to calculate the actual cell potential under non-standard conditions, taking into account the concentration (or activities) of the reactants and products. It's a crucial formula that bridges the gap between real-world scenarios and standard conditions.

General Form of the Nernst Equation:

E = E° - (RT/nF) ln(Q)

Here, E is the cell potential, E° 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, and Q is the reaction quotient reflecting the concentrations of the reactants and products.

Adjusting Cell Potential:

• For reaction spontaneity, E needs to be positive. This, at times, requires modifying reactant or product concentrations.
• The logarithmic relationship in the Nernst equation means that even small changes in concentration can have a significant impact on cell potential.

In our exercise, the Nernst equation is used to calculate the required concentration of complex ions to adjust the cell potential to a specified value, which reflects the versatility of this equation in practical electrochemistry.

Chemical Equilibrium In A Galvanic Cell

Chemical equilibrium is a state where the rates of the forward and reverse reactions are equal, resulting in no net change of the reactants and products over time. In a galvanic cell, establishing equilibrium involves balancing redox reactions, ensuring charge neutrality, and often involves complex ion formation and dissolution processes influencing the cell's dynamics.

How Equilibrium Affects Cell Potential:

• As a reaction approaches equilibrium, the cell potential moves towards zero because there's less driving force for the reaction to proceed in either direction.
• Variations in reactant or product concentrations shift the equilibrium position and consequently alter the cell potential.

In the context of our exercise, the chemical equilibrium of the silver chloride dissolution and copper-ammonia complex formation are pivotal to calculating the final cell potential and the amount of ammonia needed to achieve a specific cell potential.

Complex Ion Formation and Its Impact on Galvanic Cells

Complex ion formation is an equilibrium process where central metal ions bond with surrounding ligands to form charged species. This process can significantly alter the concentrations of participating ions in a solution, thus impacting the reaction quotient, Q, in the Nernst equation and ultimately affecting the cell potential.

Role in Electrochemistry:

• Complexation can change the solubility of salts (like AgCl), affecting ion availability for the redox reactions.
• Formation constants (Kf) of complex ions indicate their stability and help in calculating equilibrium concentrations.

The exercise guides us through the formation of a copper-ammonia complex, utilizing the formation constant (K) to calculate the moles of ammonia needed to alter the cell potential. Complexation reactions are thus integral to the understanding of electrochemical systems.