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 \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

Step by step solution

01

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.

02

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.

03

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.

04

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.

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