[Single-digit Integer] The standard reduction potential for Cu^2+ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Half-Cell Reaction and Nernst Equation The reduction half-reaction occurring at the copper electrode is: $$\ce{Cu^2+(aq) + 2e^- -> Cu(s)}$$ Here, the number of moles of electrons transferred is $n = 2$. The standard reduction potential for this electrode is given as: $$E^\circ_{\ce{Cu^2+/Cu}} = +0.34\text{ V}$$ The non-standard reduction potential of the electrode is given as: $$E_{\ce{Cu^2+/Cu}} = 0.31\text{ V}$$ Using the Nernst equation at $298\text{ K}$ ($25^\circ\text{C}$): $$E_{\ce{Cu^2+/Cu}} = E^\circ_{\ce{Cu^2+/Cu}} - \frac{2.303 RT}{nF} \log_{10} \left( \frac{1}{[\ce{Cu^2+}]} \right)$$ This can be rewritten as: $$E_{\ce{Cu^2+/Cu}} = E^\circ_{\ce{Cu^2+/Cu}} + \frac{2.303 RT}{2F} \log_{10} [\ce{Cu^2+}]$$ Using the standard approximation for the Nernstian slope coefficient at $298\text{ K}$: $$\frac{2.303 RT}{F} \approx 0.06\text{ V}$$ Substituting this value into the equation yields: $$E_{\ce{Cu^2+/Cu}} = E^\circ_{\ce{Cu^2+/Cu}} + \frac{0.06\text{ V}}{2} \log_{10} [\ce{Cu^2+}]$$ Step 2 - Calculate the Concentration of Copper Ions ($[\ce{Cu^2+}]$) Now, we substitute the given potentials into our Nernst relation: $$0.31\text{ V} = 0.34\text{ V} + 0.03\text{ V} \times \log_{10} [\ce{Cu^2+}]$$ Subtract $0.34\text{ V}$ from both sides to isolate the logarithmic term: $$-0.03\text{ V} = 0.03\text{ V} \times \log_{10} [\ce{Cu^2+}]$$ Divide both sides by $0.03\text{ V}$: $$\log_{10} [\ce{Cu^2+}] = \frac{-0.03\text{ V}}{0.03\text{ V}} = -1$$ Taking the antilogarithm on both sides: $$[\ce{Cu^2+}] = 10^{-1}\text{ M} = 0.1\text{ M}$$ Step 3 - Set up the Solubility Product Equilibrium for Copper(II) Hydroxide Copper(II) hydroxide, $\ce{Cu(OH)2}$, is a sparingly soluble base that establishes the following solubility equilibrium in water: $$\ce{Cu(OH)2(s) <=> Cu^2+(aq) + 2OH^-(aq)}$$ The solubility product constant ($K_{\text{sp}}$) expression for this equilibrium is: $$K_{\text{sp}} = [\ce{Cu^2+}][\ce{OH^-}]^2$$ Step 4 - Calculate the Hydroxide Ion Concentration ($[\ce{OH^-}]$) and $\text{pOH}$ We are given that $K_{\text{sp}}$ of $\ce{Cu(OH)2}$ is $10^{-19}$ at $25^\circ\text{C}$. Substituting the values of $K_{\text{sp}}$ and $[\ce{Cu^2+}]$ into the expression: $$10^{-19} = (10^{-1}\text{ M}) \times [\ce{OH^-}]^2$$ Solving for $[\ce{OH^-}]^2$: $$[\ce{OH^-}]^2 = \frac{10^{-19}}{10^{-1}} = 10^{-18}\text{ M}^2$$ Taking the square root on both sides: $$[\ce{OH^-}] = \sqrt{10^{-18}}\text{ M} = 10^{-9}\text{ M}$$ Now, we calculate the $\text{pOH}$ of the solution: $$\text{pOH} = -\log_{10} [\ce{OH^-}]$$ $$\text{pOH} = -\log_{10} (10^{-9}) = 9$$ Step 5 - Reconcile the Question Phrasing and State the Final Answer The question text mentions a "solution of $\text{pH} = 2$". However, in a saturated solution of copper hydroxide where the reduction potential is $0.31\text{ V}$, the concentration of $\ce{OH^-}$ is thermodynamically constrained by the $K_{\text{sp}}$ equilibrium to be $10^{-9}\text{ M}$ ($\text{pOH} = 9$), which corresponds to a $\text{pH}$ of $5$. Therefore, the thermodynamic value of $\text{pOH}$ based on the copper concentration in equilibrium with $\ce{Cu(OH)2}$ is indeed $9$. $$\text{Final Single-digit Integer Answer: } \boxed{9}$$