Two concentration cells are prepared, both with \(90.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) and a Cu bar in each half-cell. (a) In the first concentration cell, \(10.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NH}_{3}\) is added to one half-cell; the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) forms, and \(E_{\text {cell }}\) is \(0.129 \mathrm{~V}\). Calculate \(K_{\mathrm{f}}\) for the formation of the complex ion. (b) Calculate \(E_{\text {cell }}\) when an additional \(10.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NH}_{3}\) is added. (c) In the second concentration cell, \(10.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) is added to one half-cell; the precipitate \(\mathrm{Cu}(\mathrm{OH})_{2}\) forms \(\left(K_{\mathrm{sp}}=\right.\) \(\left.2.2 \times 10^{-20}\right) .\) Calculate \(E_{\mathrm{cell}}^{0}\) (d) What would the molarity of NaOH have to be for the addition of \(10.0 \mathrm{~mL}\) to result in an \(E_{\text {cell }}^{0}\) of \(0.340 \mathrm{~V} ?\)

Step by step solution

01

- Calculate concentrations after dilution (Parts a, b)

Determine the new concentration after diluting 90.0 mL with 10.0 mL of either 0.500 M NH₃ or NaOH. This gives you \[ C_{\text{new}} = \frac{C_{\text{initial}} \times V_{\text{initial}}}{V_{\text{total}}}\]

02

- Setup the formula for Kf using the given cell potential (Part a)

Given: \[ E_{\text{cell}} = 0.129\text{ V}\]The formation constant of the complex ion can be found from the Nernst equation and the relationship between Kf and E: \[ E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0592}{n} \text{log Kf}\]

03

- Plug in the values to solve for Kf (Part a)

Rearrange: \[ 0.129\text{ V} = - \frac{0.0592}{2} \text{log Kf}\]Then solve for Kf: \[ \text{log Kf} = -\frac{0.129}{0.0592/2}\]Kf = 2.9×10¹³.

04

- Calculate E_cell after adding extra NH3 (Part b)

The concentration of NH₃ is doubled, calculate the new E_cell: Using the Nernst equation again, \[ E_{\text{cell}} = 0.0592\text{ V} \times \text{log}(\text{first E_cell}) + 0.0592 \times log (\text{concentration ratio})\]

05

- Determine the standard cell potential E_cell^0 with given Ksp for precipitation (Part c)

Calculate the reaction quotient for Cu(OH)₂ formation. Setup for Ksp reaction using: \[ E_{\text{cell}}^{0} = - \frac{0.0592}{2} \text{log} \frac{1}{K_{\text{sp}}}\]

06

- Calculate NaOH concentration for a given E_cell^0 (Part d)

Finding the required NaOH molarity so that E_cell^0 matches: \[ E_{\text{cell}}^{0} = 0.0592 \times\text{log} (\text{concentrations into the equation})\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concentration Cells

Concentration cells are a unique type of electrochemical cells. They generate an electric current by exploiting the difference in the concentration of ions in two half-cells. Every concentration cell involves the same chemical species in both half-cells, but in different concentrations. In the exercise above, copper(II) nitrate \(\text{Cu(NO}_3)_2\) is used in both half-cells, but additional substances like ammonia (NH₃) and sodium hydroxide (NaOH) are added to alter the ion concentration.

Cell Potential

The cell potential \(E_{\text{cell}}\) is crucial in understanding how an electrochemical cell operates. It represents the driving force behind the electric current generated by the cell. For concentration cells, \(E_{\text{cell}}\) is calculated based on the concentrations of ions in the half-cells. The Nernst equation is often used to determine the cell potential, which can be influenced by factors such as temperature and ion concentration. In the exercise, \( E_{\text {cell}} = 0.129 \text{ V} \) when NH₃ is added, showing how changes in concentration impact the potential.

Formation Constant (Kf)

The formation constant \(K_f\) is a measure of the stability of a complex ion in solution. It is particularly important in reactions where complex ions form, as seen with \(\text{Cu(NH}_3)_4^{2+}\) in the exercise. The formation constant can be found using the Nernst equation and the given cell potential. For instance, in the exercise, the cell potential \( E_{\text{cell}} = 0.129 \text{ V} \). By rearranging and solving, \( K_f \) for \( \text{Cu(NH}_3)_4^{2+} \) is calculated to be approximately \( 2.9 \times 10^{13} \).

Nernst Equation

The Nernst equation is a key tool in electrochemistry. It relates the cell potential to the concentrations of the reactants and products in an electrochemical cell. The Nernst equation is given by: \[ E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0592}{n} \log Q \] where:
\( E_{\text{cell}} \) is the observed cell potential,
\( E_{\text{cell}}^0 \) is the standard cell potential,
\( n \) is the number of electrons transferred, and
\( Q \) is the reaction quotient.
By manipulating this equation, one can determine cell potentials under varying conditions. For example, in the exercise, it is used to calculate \(E_{\text{cell}} \) after the addition of NH₃.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine to form an insoluble product, called a precipitate. In the context of the exercise, adding NaOH to a solution containing Cu²⁺ ions leads to the formation of copper(II) hydroxide \( \text{Cu(OH}_2) \), a solid precipitate. The solubility product constant (Ksp) indicates the extent to which a compound will dissolve. When solving for the standard cell potential, the Nernst equation can be adapted to consider Ksp in the electrochemical reaction, helping to predict if and how a precipitate will form in the reaction.