An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at \(25^{\circ} \mathrm{C}\) is \(0.195 \mathrm{V}\). What is \(\left[\mathrm{Cu}^{2+}\right] ?\) (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

First, you need to write the half-cell reactions for the standard hydrogen electrode (SHE) and the copper electrode. The SHE half-cell reaction is defined as: \[ 2H^{+} (aq) + 2e^{-} \rightleftharpoons H_{2} (g) \ \ \ \ \ \ E^{0} = 0V \] For the copper electrode, we have the following half-cell reaction: \[ Cu^{2+} (aq) + 2e^{-} \rightleftharpoons Cu (s) \ \ \ \ \ \ E^{0} = +0.337V \] Now, we will proceed to calculate the cell potential for part (a).

Now that we have the half-cell reactions, we need to calculate the potential for part (a). We will be using the Nernst equation, but in order to do that, we must write the overall cell reaction, which is the sum of the two half-cell reactions: \[ Cu^{2+} (aq) + H_{2} (g) \rightleftharpoons Cu (s) + 2H^{+} (aq) \] For case (a): \(\left[ \mathrm{Cu}^{2+} \right] = 2.5 \times 10^{-4} \mathrm{M}\) The Nernst equation for the cell reaction is: \[ E = E^{0} - \frac{RT}{nF} \times \ln{Q} \] where - \(E\) is the cell potential - \(E^{0}\) is the standard cell potential, which is the sum of the standard half-cell potentials - \(R\) is the gas constant (8.314 J/mol·K) - \(T\) is the temperature in Kelvin (for 25°C, T=298K) - \(n\) is the number of electrons transferred in the reaction (in this case, 2) - \(F\) is the Faraday's constant (96485 C/mol) - \(Q\) is the reaction quotient, which is: \[ Q=\frac{[\mathrm{H}^{+}]^{2}}{[\mathrm{Cu}^{2+}]}\] Now, calculate the cell potential for part (a).

First, we need to find \(E^{0}\) for the overall cell reaction by adding the standard reduction potentials of both half-cell reactions: \[ E^{0} = 0V + 0.337V = 0.337V \] Now, we can plug in the Nernst equation values: \[ E = 0.337V - \frac{8.314 \times 298}{2 \times 96485} \times \ln{\left(\frac{1}{2.5 \times 10^{-4}}\right)} \] Calculating the values, we get: \[ E \approx 0.34V \] Therefore, the cell potential at \(25^{\circ}\mathrm{C}\) for case (a) is approximately \(0.34V\). Now we can proceed to solve for case (b).

For case (b), we are given that the measured cell potential is \(0.195V\). We have to find the concentration of \(\mathrm{Cu}^{2+}\) ions in the solution. We'll use the Nernst equation again: \[ E = E^{0} - \frac{RT}{nF} \times \ln{Q} \] But this time, we will solve for \([\mathrm{Cu}^{2+}]\) by substituting the given values: \[ 0.195V = 0.337V - \frac{8.314 \times 298}{2 \times 96485} \times \ln{\left(\frac{1}{[\mathrm{Cu}^{2+}]}\right)} \]

We can now calculate the concentration of \(\mathrm{Cu}^{2+}\) ions by rearranging the equation: \[0.142V = \frac{8.314 \times 298}{2 \times 96485} \times \ln{\left(\frac{1}{[\mathrm{Cu}^{2+}]}\right)}\] Now, we can isolate the concentration term by raising \(e\) to the power of both sides of the equation: \[[\mathrm{Cu}^{2+}] = \frac{1}{e^{\frac{2 \times 96485 \times 0.142V}{8.314 \times 298}}}\] Calculating the values, we get: \[[\mathrm{Cu}^{2+}] \approx 6.5 \times 10^{-4} \mathrm{M}\] Thus, the concentration of \(\mathrm{Cu}^{2+}\) ions in case (b) is approximately \(6.5 \times 10^{-4} \mathrm{M}\).