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

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.

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.