Question

A concentration cell has the same half-reactions at the anode and cathode, but a voltage results from different concentrations in the two electrode compartments. (a) What is \(x\) in the concentration cell \(\mathrm{Cu}(s) \mid \mathrm{Cu}^{2+}(x \mathrm{M}) \| \mathrm{Cu}^{2+}\) \((0.10 \mathrm{M}) \mid \mathrm{Cu}(s)\) if the measured cell potential is \(0.0965 \mathrm{~V} ?\) (b) A similar cell has \(0.10 \mathrm{M} \mathrm{Cu}^{2+}\) in both compartments. When \(\mathrm{a}\) stoichiometric amount of ethylenediamine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)\) is added to one compartment, the measured cell potential is \(0.179 \mathrm{~V}\). Calculate the formation constant \(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}^{2+}\). Assume there is no volume change.

Short answer

(a) x = 182 M (b) Kf = 1.13 × 10^6

Step-by-step solution

Understand the Concentration Cell

In a concentration cell, the electrodes are identical, but the concentrations of the ions in the solutions differ. The cell potential is generated because of this concentration difference, not because of a difference in electrode materials.

Assign Nernst Equation for the First Part (a)

For a concentration cell, the Nernst equation is given by: \[ E_{cell} = \frac{RT}{nF} \ln \left(\frac{\text{concentration at anode}}{\text{concentration at cathode}}\right) \]At standard temperature (298 K), the equation simplifies for one electron transfer to:\[ E_{cell} = \frac{0.0592}{n} \log \left(\frac{\text{concentration at anode}}{\text{concentration at cathode}}\right) \]Here, \( E_{cell} = 0.0965 \, \text{V} \), \( n = 2 \), the cathode concentration \( = 0.10 \, \text{M} \), and the concentration at the anode is \( x \).

Solve for x in the Nernst Equation

Rearrange the Nernst equation to solve for \( x \):\[ 0.0965 = \frac{0.0592}{2} \log \left(\frac{x}{0.10}\right) \]\[ \log \left(\frac{x}{0.10}\right) = \frac{0.0965 \times 2}{0.0592} \]\[ \log \left(\frac{x}{0.10}\right) = 3.257 \]\[ \frac{x}{0.10} = 10^{3.257} \]\[ x = 0.10 \times 1820 \]\[ x = 182 \, \text{M} \]

Understand the Complex Ion Formation in Part (b)

A similar cell is used with \(0.10 \, \text{M Cu}^{2+}\) in both compartments. The potential difference is due to the formation of a complex ion.

Use Nernst Equation for Complex Ion Formation

The Nernst equation remains valid, but now the concentration change is due to complex formation:\[ E_{cell} = \frac{RT}{nF} \ln K_{f} \]The given \( E_{cell} = 0.179 \, \text{V} \).Use the simplified form:\[ 0.179 = \frac{0.0592}{2} \log K_{f} \]

Solve for Formation Constant Kf

Rearrange to solve for \( K_{f} \):\[ \log K_{f} = \frac{0.179 \times 2}{0.0592} \]\[ \log K_{f} = 6.051 \]\[ K_{f} = 10^{6.051} = 1.13 \times 10^{6} \]

Key concepts

Nernst Equation

The Nernst Equation is a fundamental concept for understanding galvanic cells and, in particular, concentration cells. It helps us determine the electric potential of an electrochemical cell based on the concentrations of reactants and products. In this situation, the Nernst Equation is used to calculate the cell potential for a concentration cell, where two identical electrodes are immersed in solutions of differing concentrations. The equation is:

• \[ E_{cell} = rac{RT}{nF} imes ext{ln} rac{ ext{[Ox]}_{ ext{anode}}}{ ext{[Red]}_{ ext{cathode}}} \]

For our specific problem, involving copper ions, we use a simplified version at room temperature (298 K):

• \[ E_{cell} = rac{0.0592}{n} imes ext{log} rac{ ext{[Cu}^{2+}]_{ ext{anode}}}{ ext{[Cu}^{2+}]_{ ext{cathode}}} \]

In the exercise, students find the unknown concentration at the anode by rearranging this equation, solving for the concentration that gives the observed cell potential of 0.0965 V with known cathode concentration. This step reiterates the importance of understanding how concentration differences drive the potential in a concentration cell.

Complex Ion Formation

Complex ions are formed when transition metals, such as copper in our example, bind with ligands, molecules or ions that donate a pair of electrons. This molecule can significantly influence the electrochemical properties of a system. In our exercise's Part (b), ethylenediamine is added to one compartment, forming a complex ion with copper, affecting the measured cell potential.
The formula for a complex ion, such as \(\text{Cu(NH}_2\text{CH}_2 \text{CH}_2 \text{NH}_2)_2^{2+}\), indicates that two ethylenediamine ligands are bound to a copper ion. The formation constant \( K_f \) indicates the stability of the complex, with higher values suggesting a more stable complex.

• Complex ion formation can:
  • Change the concentration of free metal ions in solution.
  • Result in a different electromotive force.
  • Be calculated using the Nernst equation by integrating \( K_f \).

By applying the Nernst equation, one determines \( K_f \) from the new cell potential, emphasizing the critical role of complex ion formation in electrochemical reactions.

Electrode Potential

Electrode potential is the measure of the ability of a chemical species to gain or lose electrons and thus undergo reduction or oxidation. It's a vital component in understanding how electrochemical cells work. The potential is influenced by factors like temperature, pressure, and importantly in our case, ion concentration.
In a concentration cell, electrode potential arises from differences in concentration. In the exercise, the concentration differences between the copper ions at the anode and the cathode generate a measurable potential difference. This potential difference or voltage is crucial to calculating unknown concentrations using the Nernst Equation.

• Key points on electrode potential include:
  • Electrode potential drives electron flow from higher to lower potential.
  • It reflects the chemical power of a cell to do work.
  • Understanding electrode potential helps predict cell behavior in conditions other than the standard state.

The study of electrode potential is essential, as it directly relates to calculating the potential of an overall cell setup and further solving real-world electrochemical problems.