An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of \(0.10 \mathrm{M} \mathrm{NaOH}\) that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2}\), what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\mathrm{For} \mathrm{Cu}(\mathrm{OH})_{2}, K_{\mathrm{sp}}\right.\) \(\left.=1.6 \times 10^{-19} .\right]\)

In order to calculate cell potential, the balanced half-reactions involving each electrode should be written. For the standard hydrogen electrode (SHE), the half-reaction is: \[\mathrm{2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)}\] For the copper electrode, the half-reaction is: \[\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}\]

To calculate the cell potential, we need to find the concentration of Cu²⁺ ions in the solution. We have the equation \(\mathrm{Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2 OH^{-}(aq)}\) and we know that Ksp = 1.6x10⁻¹⁹ and [OH⁻] = 0.10 M (from NaOH solution). We can write Ksp as: \[K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{OH^{-}}]^{2}\] Replacing the known values, we get: \[\mathrm{1.6 \times 10^{-19} = [Cu^{2+}](0.10)^{2}}\] Now, we can solve for the concentration of Cu²⁺ ions: \[[\mathrm{Cu^{2+}}] = \dfrac{1.6 \times 10^{-19}}{(0.10)^{2}} = 1.6 \times 10^{-17} \mathrm{M}\]

Now that we know the concentration of Cu²⁺ ions, we can use the Nernst equation to find the cell potential. The Nernst equation is given by: \[E_{cell} = E_{\circ} - \dfrac{0.0592}{n}\log Q\] Where \(E_{\circ}\) is the standard potential, n is the number of electrons transferred, and Q is the reaction quotient. For this cell, the standard potential is the difference between the standard reduction potentials of the Cu²⁺/Cu and H⁺/H₂ couple: \(E_{\circ} = E_{\circ}(\mathrm{Cu^{2+}/Cu}) - E_{\circ}(\mathrm{H^{+}/H_{2}}) = 0.34 \mathrm{V} - 0.00 \mathrm{V} = 0.34 \mathrm{V}\) As both half-reactions involve the exchange of 2 electrons, we have n=2. The reaction quotient Q can be written as: \[Q = \dfrac{[\mathrm{Cu^{2+}}]}{[\mathrm{H^{+}}]^{2}}\] Let's assume that the concentration of H⁺ ions is 1 M (neutral solution). Therefore, Q becomes: \[Q=\dfrac{1.6\times10^{-17}}{(1)^2}=1.6\times10^{-17}\] Now, we can plug in the values into the Nernst equation to find the cell potential: \[E_{cell} = 0.34 - \dfrac{0.0592}{2}\log(1.6\times10^{-17})\] \[E_{cell} \approx0.485 \mathrm{V}\] The cell potential at 25°C is approximately 0.485 V.