When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by p — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the pH-Dependence of Each Half-Cell Reaction To determine which half-cell reaction is affected by a change in $\text{pH}$, we examine the chemical equations for the half-reactions provided in the passage: 1. **Reduction of free silver ions:** $$\ce{Ag^+(aq) + e^- -> Ag(s)}$$ This reaction does not contain hydrogen ions ($\ce{H^+}$), meaning its potential is not directly defined by a $\text{pH}$ term in its stoichiometry. 2. **Reduction of the silver-ammonia complex:** $$\ce{[Ag(NH3)2]^+(aq) + e^- -> Ag(s) + 2NH3(aq)}$$ This reaction also does not involve $\ce{H^+}$ ions, so its potential is not directly affected by $\text{pH}$. 3. **Oxidation of glucose to gluconic acid:** $$\ce{C6H12O6(aq) + H2O(l) -> C6H12O7(aq) + 2H^+(aq) + 2e^-}$$ This reaction contains hydrogen ions ($\ce{H^+}$) directly on the product side. Since $\text{pH}$ is a measure of the hydrogen ion concentration ($[\ce{H^+}] = 10^{-\text{pH}}$), any change in the $\text{pH}$ of the solution will directly shift the equilibrium and alter the electrode potential of this half-reaction. Therefore, only the **glucose oxidation half-cell reaction** is affected by the $\text{pH}$ of the solution. Step 2 - Apply the Nernst Equation to the Glucose Oxidation Half-Reaction At $298\text{ K}$, the Nernst equation for the oxidation potential of glucose ($E_{\text{oxd}}$) is written as: $$E_{\text{oxd}} = E^\circ_{\text{oxd}} - \frac{2.303 RT}{nF} \log_{10} \left( \frac{[\ce{C6H12O7}][\ce{H^+}]^2}{[\ce{C6H12O6}]} \right)$$ Where: * $E^\circ_{\text{oxd}}$ is the standard oxidation potential of glucose ($E^\circ_{\text{oxd}} = -0.05\text{ V}$). * $n$ is the number of moles of electrons transferred in the balanced reaction ($n = 2$). * The slope parameter is $\frac{2.303 RT}{F} = 0.0592\text{ V}$. Assuming the concentrations of glucose and gluconic acid are maintained under standard-state conditions ($1.0\text{ M}$), the reaction quotient simplifies to $[\ce{H^+}]^2$, and the equation becomes: $$E_{\text{oxd}} = E^\circ_{\text{oxd}} - \frac{0.0592\text{ V}}{2} \log_{10} [\ce{H^+}]^2$$ $$E_{\text{oxd}} = E^\circ_{\text{oxd}} - 0.0592\text{ V} \times \log_{10} [\ce{H^+}]$$ Step 3 - Calculate the Change in Oxidation Potential at pH = 11 Using the definition of pH: $$\text{pH} = -\log_{10} [\ce{H^+}]$$ We substitute $-\text{pH}$ for $\log_{10} [\ce{H^+}]$ into the Nernst relation: $$E_{\text{oxd}} = E^\circ_{\text{oxd}} + 0.0592\text{ V} \times \text{pH}$$ We are given that the addition of ammonia raises the $\text{pH}$ of the solution to $11$. Substituting $\text{pH} = 11$: $$E_{\text{oxd}} = E^\circ_{\text{oxd}} + 0.0592\text{ V} \times 11$$ $$E_{\text{oxd}} = E^\circ_{\text{oxd}} + 0.6512\text{ V}$$ $$E_{\text{oxd}} \approx E^\circ_{\text{oxd}} + \boxed{0.65\text{ V}}$$ This calculation proves that the non-standard oxidation potential of the glucose half-cell ($E_{\text{oxd}}$) **increases by $0.65\text{ V}$** relative to its standard oxidation potential ($E^\circ_{\text{oxd}}$). Step 4 - Evaluate and Explain Each Option * **Option (A) is correct:** As mathematically proven in Step 3, the glucose oxidation half-cell is the only reaction affected, and its oxidation potential ($E_{\text{oxd}}$) increases by a factor of $0.65\text{ V}$ from $E^\circ_{\text{oxd}}$. * **Option (B) is incorrect:** Raising the pH reduces the concentration of product $\ce{H^+}$ ions, which shifts the equilibrium to the right (Le Chatelier's principle), thereby increasing the oxidation potential rather than decreasing it. * **Option (C) is incorrect:** The reduction potentials of the silver half-reactions do not contain a hydrogen ion concentration term in their Nernstian expressions, so they are not directly affected by the pH change. * **Option (D) is incorrect:** The reduction potentials of the silver couples remain unaffected by the direct pH term under these conditions. $$\text{Correct Option: } \boxed{A}$$