Question

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]\) \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} &\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

Short answer

The potential of the concentration cell is initially calculated to be -0.281 V. Upon adding NH₃ to the left cell compartment, the new potential is found to be -0.915 V.

Step-by-step solution

Calculate the Nernst equation for the cell potential

Using the Nernst equation, we can find the cell potential: E = E0 - (RT/nF) * ln(Q) For the concentration cell with Cu electrodes, the reaction is: Cu2+ (left) + 2e- → Cu (left) Cu (right) → Cu2+ (right) + 2e- Combining these two, we get: Cu2+ (left) → Cu2+ (right) For this reaction, n = 2, and E0 = 0 V (since there is no potential difference between the same metal electrode). Now we need to find the reaction quotient Q. It can be found by the ratio of the concentrations of Cu2+ ions in the right and left compartments. Q = [Cu2+ (right)]/[Cu2+ (left)] Now we can plug in the values to find the cell potential.

Plug in values and find potential

To calculate the potential at 25°C, use R = 8.314 J/(mol·K), T = 298.15 K, n = 2, and F = 96485 C/mol. E = 0 - (8.314 * 298.15 / (2 * 96485)) * ln((1.00) / (1.0 * 10^(-4))) E = -0.0305 * ln(10^4) E = -0.0305 * 9.21 E ≈ -0.281 V Therefore, the potential of the concentration cell is -0.281 V. b. After adding NH3 and forming the complex in the left cell compartment, we need to find how the concentration of Cu2+ changes in equilibrium.

Find the new concentration of Cu2+ in the left compartment

We are given the equilibrium constant K and NH3 concentration in equilibrium, and we can find the new concentration of Cu2+ using the reaction stoichiometry. K = [Cu(NH3)4+2] / ([Cu2+] * [NH3]^4) 1.0 * 10^13 = [Cu(NH3)4+2] / ([Cu2+] * (2)^4) [Cu(NH3)4+2] = (1.0 * 10^13) * ([Cu2+] * 16) Since initially, all Cu2+ is present in the compartment, the final concentration of Cu2+ will be ([Cu2+] - [Cu(NH3)4+2]). Therefore, the total Cu2+ concentration in the left compartment will be: total [Cu2+] (left) = [Cu2+] - [Cu(NH3)4+2] = [Cu2+] - (1.0 * 10^13) * ([Cu2+] * 16) Now we need to find the new potential of the cell in these new conditions.

Calculate the new cell potential

Similar to the first part, we will use the Nernst equation and plug in the new concentration of Cu2+ in the left compartment. E_new = E0 - (RT/nF) * ln(Q_new) Q_new = [Cu2+ (right)] / [total Cu2+ (left)] E_new = 0 - (8.314 * 298.15 / (2 * 96485)) * ln((1.00) / ([Cu2+] - (1.0 * 10^13) * ([Cu2+] * 16))) Since the initial concentration of Cu2+ in the left compartment is much smaller than the equilibrium concentration, it can be neglected in the denominator ([Cu2+initial] = 1.0 * 10^(-4) M). E_new = -0.0305 * ln(10^13) E_new = -0.0305 * 30 E_new ≈ -0.915 V Therefore, after adding NH3 to the left cell compartment, the new cell potential is -0.915 V.

Key concepts

Concentration Cell

A concentration cell uses the same electrodes and solutions but with different concentrations of ions in each half-cell. Here, we're working with copper electrodes in a solution of copper ions (\(\text{Cu}^{2+}\)) at varying concentrations. The core principle is that a difference in ion concentration creates a potential difference, driving the cell reaction from one side to the other.

In this exercise, the left cell has \(1.0 \times 10^{-4}\) M \(\text{Cu}^{2+}\) and the right cell has 1.00 M \(\text{Cu}^{2+}\). These differences lead to a flow of electrons as the system tries to reach equilibrium, generating a measurable voltage.

Cell Potential

The cell potential or electromotive force (EMF) of a concentration cell can be calculated using the Nernst equation. Given by \(E = E^0 - \frac{RT}{nF} \ln(Q)\), this equation helps us determine how the potential is affected by the concentration difference.

For the copper concentration cell, the standard cell potential \(E^0\) is 0 V, since the electrodes are the same. The reaction quotient \(Q\) is the ratio of concentrations of \(\text{Cu}^{2+}\) ions in the two compartments:

• Q = [Cu²⁺(right)] / [Cu²⁺(left)]

By plugging in the known values, we calculate the cell potential using the formula, giving approximately -0.281 V under initial conditions.

Complex Ion Formation

Complex ion formation significantly alters the concentrations in our system. When \(\text{Cu}^{2+}\) ions in the left compartment react with \(\text{NH}_3\) to form \(\text{Cu(NH}_3\text{)}_4^{2+}\), the ion's effective concentration decreases.

This reaction is described by the equilibrium constant \(K = 1.0 \times 10^{13}\). Using the equation:

• \(K = \frac{[\text{Cu(NH}_3\text{)}_4^{2+}]}{[\text{Cu}^{2+}] * [\text{NH}_3]^4}\)

we can determine the concentration of \(\text{Cu}^{2+}\) ions at equilibrium. Having taken part in complex formation, the concentration of free \(\text{Cu}^{2+}\) ions decreases proportionally to the formation of \(\text{Cu(NH}_3\text{)}_4^{2+}\).

Copper Electrode Reaction

Copper serves as both electrode and ion source in our cell. This reaction involves the reduction and oxidation of copper at different concentrations, driving current flow.

For the left side:

• \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\)

For the right side (reverse):

• \(\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-\)

Adding \(\text{NH}_3\) leads to the formation of a complex, shifting the equilibrium.

Incorporating all these reactions, the new cell potential becomes -0.915 V. This illustrates how chemical interactions and ion concentration can influence the behavior of electrodes in an electrochemical system.