Consider a concentration cell similar to the one shown in Exercise 69, except that both electrodes are made of \(\mathrm{Ni}\) and in the left-hand compartment \(\left[\mathrm{Ni}^{2+}\right]=1.0 M .\) Calculate the cell potential at \(25^{\circ} \mathrm{C}\) when the concentration of \(\mathrm{Ni}^{2+}\) in the compartment on the right has each of the following values. a. \(1.0 M\) b. \(2.0 \mathrm{M}\) c. \(0.10 \mathrm{M}\) d. \(4.0 \times 10^{-5} M\) e. Calculate the potential when both solutions are \(2.5 M\) in \(\mathrm{Ni}^{2+}\). For each case, also identify the cathode, anode, and the direction in which electrons flow.

A concentration cell is an electrochemical cell that has the same electrode on both sides but different concentrations of the electrolyte on either side. In this problem, we have a \(\mathrm{Ni}\) concentration cell with varying concentrations of \(\mathrm{Ni}^{2+}\) in both compartments. Left compartment: \(\mathrm{Ni^{2+}}\) with concentration \(1.0 M\) Right compartment: \(\mathrm{Ni^{2+}}\) with different concentrations The cell reaction is: \(\mathrm{Ni^{2+}(right) + 2e^- <=> Ni^{2+}(left)}\)

The Nernst equation is given by: \(E = E_{0} - \frac{RT}{nF} \ln{\frac{Q}}\) where \(E\) is the cell potential, \(E_{0}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. In this case, the standard potential \(E_{0}\) is \(0 \mathrm{V}\) since both compartments have the same electrode. We will calculate the cell potential for each concentration of \(\mathrm{Ni^{2+}}\) in the right compartment.

Temperature \(T\) is given as \(25^{\circ} \mathrm{C}\) which is equal to \(298.15\) K. The Nernst equation becomes: \(E = - \frac{0.0592 \mathrm{V}}{n} \log{\left(\frac{[\mathrm{Ni^{2+}(left)}]}{[\mathrm{Ni^{2+}(right)]}\right)}\) We will calculate the cell potentials for each case: a) When the concentration of \(\mathrm{Ni^{2+}}\) in the right compartment is \(1.0 M\): \(E = - \frac{0.0592}{2} \log{\left(\frac{1}{1}\right)} = 0 \mathrm{V}\) b) When the concentration of \(\mathrm{Ni^{2+}}\) in the right compartment is \(2.0 M\): \(E = - \frac{0.0592}{2} \log{\left(\frac{1}{2}\right)} = 0.0148 \mathrm{V}\) c) When the concentration of \(\mathrm{Ni^{2+}}\) in the right compartment is \(0.10 M\): \(E = - \frac{0.0592}{2} \log{\left(\frac{1}{0.10}\right)} = -0.0296 \mathrm{V}\) d) When the concentration of \(\mathrm{Ni^{2+}}\) in the right compartment is \(4.0 \times 10^{-5} M\): \(E = - \frac{0.0592}{2} \log{\left(\frac{1}{4.0 \times 10^{-5}}\right)} = -0.0992 \mathrm{V}\) e) When both solutions have concentration \(2.5 M\): \(E = - \frac{0.0592}{2} \log{\left(\frac{2.5}{2.5}\right)} = 0 \mathrm{V}\)

For each case: a) The potentials are equal, so there is no electron flow and no cathode or anode. b) The left compartment acts as the cathode, the right compartment acts as the anode, and electrons flow from the right to the left compartment. c) The left compartment acts as the anode, the right compartment acts as the cathode, and electrons flow from the left to the right compartment. d) The left compartment acts as the anode, the right compartment acts as the cathode, and electrons flow from the left to the right compartment. e) The potentials are equal, so there is no electron flow and no cathode or anode.