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$M \mathrm{NaOH}\( that is saturated with \)\mathrm{Cu}(\mathrm{OH})_{2},$ what is the cell potential at $25^{\circ} \mathrm{C} ?\left[\text { For } \mathrm{Cu}(\mathrm{OH})_{2}\right.\( \)K_{\mathrm{sp}}=1.6 \times 10^{-19} . ]$

The dissolution equation for Copper(II) hydroxide in water can be written as: \[ \mathrm{Cu}(\mathrm{OH})_{2}(s) \rightleftarrows \mathrm{Cu}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq) \]

Using the given \(K_{sp}=1.6 \times 10^{-19}\), we can set up the following solubility equilibrium expression: \[K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^{2}\] Since the solution contains 0.10 M of NaOH, we know the initial concentration of OH- ions is 0.10 M. Let \(x\) be the concentration of \(\mathrm{Cu}^{2+}\) ions in equilibrium, the concentration of \(\mathrm{OH}^{-}\) will then be \(0.10+x\). So, the equilibrium expression becomes: \[1.6 \times 10^{-19} = x(0.10+x)^2\]

We can approximate that \(x << 0.10\), and thus neglect \(x\) in the term \(0.10+x\). Therefore, we have: \[1.6 \times 10^{-19} = x(0.10)^2\] \[x = \frac{1.6 \times 10^{-19}}{(0.10)^2}\] \[x = 1.6 \times 10^{-17} \] So, the concentration of \(\mathrm{Cu}^{2+}\) ions in equilibrium is approximately \(1.6 \times 10^{-17} \mathrm{M}\).

The two half-cell reactions are: Cathode (Reduction): Cu2+ + 2e- → Cu (E° = +0.34 V) Anode (Oxidation): 2H+ + 2e- → H2 (Standard Hydrogen Electrode, E° = 0.00 V)

Now we can use the Nernst equation to determine the cell potential at 25°C (298.15 K): \[E_{cell} = E_{cathode}° - E_{anode}° - \frac{RT}{nF} \ln\frac{[\mathrm{H}^{+}]^{2}}{[\mathrm{Cu}^{2+}]}\] where R = 8.314 J/mol·K (gas constant), T = 298.15 K (temperature), n = 2 (moles of electrons exchanged), and F = 96485 C/mol (Faraday's constant). Remember that \(E_{anode}° = 0.00 \, \mathrm{V}\) since the anode is the standard hydrogen electrode. We are given \(E_{cathode}° = +0.34 \, \mathrm{V}\), and we know that pH of a 0.10 M NaOH solution is \(pH = 13\), so the concentration of \(\mathrm{H}^{+}\) ions can be calculated as \([\mathrm{H}^{+}] = 10^{-pH} = 10^{-13}\,\mathrm{M}\). Inserting these values into the Nernst equation and solving for \(E_{cell}\): \[E_{cell} = 0.34 - \frac{8.314 \cdot 298.15}{2 \cdot 96485} \ln{\frac{(10^{-13})^2}{1.6 \times 10^{-17}}}\] \[E_{cell} = 0.34 + 0.0285\] \[E_{cell} = 0.3685 \, \mathrm{V}\] Therefore, the cell potential of the electrochemical cell is approximately 0.3685 V at 25°C.