Question

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) 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 electrochemical potential of the concentration cell is 0.1028 V. After adding NH3, the new cell potential is 0.1310 V.

Step-by-step solution

Part a: Calculate the cell potential using the Nernst equation

To determine the electrochemical potential of the concentration cell, we will use the Nernst equation, given by: \(E = E^0 - \frac{RT}{nF} \ln Q\) Where \(E^0\) is the standard electrode potential, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. For a concentration cell, \(E^0 = 0\), since the same redox reaction takes place at both the anode and cathode. The reaction is: \(\mathrm{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \mathrm{Cu}(s)\) First, convert the given temperature to Kelvin: \(25^{\circ} \mathrm{C} + 273.15 = 298.15 \: \mathrm{K}\) Then apply the Nernst equation: \(E = 0 - \frac{8.314 \times 298.15}{2 \times 96485} \ln \frac{[1.0 \times 10^{-4}]}{[1.0]}\) \(E = -0.0257 \: \mathrm{V} \ln 10^{-4}\) \(E = -0.0257 \: \mathrm{V} \times (-4)\) \(E = 0.1028 \: \mathrm{V}\) The electrochemical potential of the concentration cell is 0.1028 V.

Part b: Add NH3, determine equilibrium, and calculate the new cell potential

We need to calculate the new cell potential after adding enough NH3 to the left cell compartment such that at equilibrium, \([NH_{3}]=2.0\: M\). The equilibrium constant, K, is given as \(1.0 \times 10^{13}\). The equilibrium expression for the reaction of Cu2+ with NH3 is: \(K = \frac{\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{NH}_{3}\right]^4}\) We know the equilibrium concentrations of NH3 and Cu2+, and the equilibrium constant. We can now solve for the equilibrium concentration of Cu(NH3)42+: \(1.0 \times 10^{13} = \frac{\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\right]}{(1.0 \times 10^{-4})(2.0)^4}\) \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\right] = 8 \times 10^{-6} M\) Now apply the Nernst equation to find the new cell potential after adding NH3: \(E' = 0 - \frac{8.314 \times 298.15}{2 \times 96485} \ln \frac{[8 \times 10^{-6}]}{[1.0]}\) \(E' = -0.0257 \: \mathrm{V} \ln 8\times10^{-6}\) \(E' = -0.0257 \: \mathrm{V} \times (-5.10)\) \(E' = 0.1310 \: \mathrm{V}\) The new cell potential after adding NH3 is 0.1310 V.

Key concepts

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry that allows us to calculate the electrochemical potential of a cell under non-standard conditions. It takes into account the temperature, the number of electrons involved in the redox reaction, and the ratio of the concentrations of reactants and products.
The equation is written as:\[ E = E^0 - \frac{RT}{nF} \ln Q \]Here,

• \(E\) is the cell potential.
• \(E^0\) is the standard electrode potential. In a concentration cell, \(E^0 = 0\) because the same electrochemical reaction occurs at both the anode and cathode.
• \(R\) is the ideal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is the Faraday constant (96485 C/mol).
• \(Q\) is the reaction quotient, defined as the ratio of the activities (or concentrations) of the products to the reactants.

Calculating the cell potential gives us insight into the driving force of an electrochemical reaction. A positive cell potential indicates a spontaneous reaction under the given conditions.

Equilibrium Constant

The equilibrium constant \( K \) is a measure of the extent to which a chemical reaction proceeds at equilibrium. In the context of the reaction that forms \( \mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+} \), it tells us how much product will be formed from the reactants at equilibrium.
The reaction equation is:\[ \mathrm{Cu}^{2+}(aq) + 4 \mathrm{NH}_3(aq) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}(aq) \]The equilibrium constant is represented as:\[ K = \frac{[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4} \]In problems involving concentration cells, when a new species such as \( \mathrm{NH}_3 \) is added, it shifts the equilibrium according to Le Chatelier's Principle. The large \( K \) value ( \( 1.0 \times 10^{13} \)) indicates a strong tendency for the formation of the product complex \( \mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+} \), significantly affecting the equilibrium concentrations and thus the cell potential.

Electrochemical Potential

Electrochemical potential is the measure of the ability of an electrochemical cell to drive an electric current through an external circuit. It reflects the energy released or absorbed when a reaction occurs.
In simpler terms, it is the voltage output of a cell, which results from the difference in the electrochemical potentials of the oxidizing and reducing agents involved in the reaction.
In a concentration cell, the electrochemical potential arises due to concentration differences between the two half-cells. Electrons will flow from the area of lower concentration to higher concentration, creating a measurable potential difference. The concentration cell we are considering operates because of the concentration gradient of \( \mathrm{Cu}^{2+} \) ions.

• For example, a gradient from \( 1.0 \times 10^{-4} \) M to 1.00 M in the original setup.
• Adding \( \mathrm{NH}_3 \) alters this concentration gradient, thus affecting the cell potential.

Understanding these concepts is crucial in predicting how changes in conditions, such as added reagents, can affect the performance of an electrochemical cell.