The dissociation constant for CH3COOH is 1.8 × 10^-5 at 298 K. The electrode potential for the half- — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Half-Cell Notation and Chemical Reactions The given half-cell is represented as: $$\text{Pt} \mid \ce{H2(1 bar)} \mid \ce{H+(aq)} \text{ (from } 0.5\text{ M }\ce{CH3COOH}\text{)}$$ This notation conventionally represents an **oxidation half-cell (anode)**, where the oxidation of hydrogen gas occurs: $$\ce{\frac{1}{2} H2(g) -> H+(aq) + e^-}$$ The corresponding reduction reaction is: $$\ce{H+(aq) + e^- -> \frac{1}{2} H2(g)}$$ We will evaluate both the standard reduction potential (as per standard IUPAC definition) and the oxidation potential (as represented by the cell notation) to provide a complete understanding. Step 2 - Calculate the Concentration of Hydrogen Ions ($[\ce{H+}]$) Acetic acid ($\ce{CH3COOH}$) is a weak monobasic acid that dissociates reversibly in aqueous solution: $$\ce{CH3COOH(aq) <=> CH3COO^-(aq) + H^+(aq)}$$ The concentration of hydrogen ions produced by a weak acid is given by the formula: $$[\ce{H+}] = \sqrt{K_a \cdot C}$$ Where: * $K_a = 1.8 \times 10^{-5}$ is the acid dissociation constant of acetic acid. * $C = 0.5\text{ M}$ is the molar concentration of the acetic acid solution. Substituting the given values into the formula: $$[\ce{H+}] = \sqrt{1.8 \times 10^{-5} \times 0.5\text{ M}}$$ $$[\ce{H+}] = \sqrt{9.0 \times 10^{-6}\text{ M}^2}$$ $$[\ce{H+}] = 3.0 \times 10^{-3}\text{ M}$$ Step 3 - Calculate the pH of the Solution The $\text{pH}$ is defined as the negative logarithm of the hydrogen ion concentration: $$\text{pH} = -\log[\ce{H+}]$$ Substituting $[\ce{H+}] = 3.0 \times 10^{-3}\text{ M}$: $$\text{pH} = -\log(3.0 \times 10^{-3})$$ $$\text{pH} = 3 - \log 3$$ Using the given value $\log 3 = 0.48$: $$\text{pH} = 3 - 0.48$$ $$\text{pH} = 2.52$$ Step 4 - Calculate the Electrode Potential using Nernst Equation * **Case 1: Standard Reduction Potential (SRP)** By standard IUPAC convention, "electrode potential" refers to the reduction potential ($E_{\text{red}}$): $$E_{\text{red}} = E^\circ_{\ce{H+|H2}} - \frac{2.303RT}{F} \log \left(\frac{P_{\ce{H2}}^{1/2}}{[\ce{H+}]}\right)$$ Given $E^\circ_{\ce{H+|H2}} = 0\text{ V}$, $P_{\ce{H2}} = 1\text{ bar}$, and $\frac{2.303RT}{F} = 0.06\text{ V}$: $$E_{\text{red}} = 0 - 0.06 \log \left(\frac{1}{[\ce{H+}]}\right)$$ $$E_{\text{red}} = -0.06 \times \text{pH}$$ $$E_{\text{red}} = -0.06 \times 2.52 = -0.1512\text{ V}$$ * **Case 2: Oxidation Potential (OP)** If we calculate the potential based on the explicit anode/oxidation representation of the half-cell ($\text{Pt} \mid \ce{H2} \mid \ce{H^+}$): $$E_{\text{ox}} = E^\circ_{\text{ox}} - 0.06 \log \left(\frac{[\ce{H+}]}{P_{\ce{H2}}^{1/2}}\right)$$ $$E_{\text{ox}} = 0 - 0.06 \log[\ce{H+}]$$ $$E_{\text{ox}} = +0.06 \times \text{pH}$$ $$E_{\text{ox}} = +0.06 \times 2.52 = +0.1512\text{ V}$$ Step 5 - Explanation of Options * **Option (A)** represents $-0.3024\text{ V}$, which would occur if the number of electrons in the Nernst equation was incorrectly substituted or if the pH was doubled. * **Option (B)** represents $-0.1512\text{ V}$, which is the standard reduction potential of the electrode. Under general IUPAC guidelines, this is the default electrode potential. * **Option (C)** represents $+0.3024\text{ V}$, which is numerically incorrect. * **Option (D)** represents $+0.1512\text{ V}$, which represents the oxidation potential of the half-cell corresponding directly to the written cell notation $\text{Pt} \mid \ce{H2} \mid \ce{H^+}$. Depending on whether reduction potential (IUPAC default) or oxidation potential (as per cell notation) is prioritized: - As per the written notation of the cell, the potential is $+0.1512\text{ V}$, which corresponds to **Option (D)**. - As per standard IUPAC default definition of reduction potential, the potential is $-0.1512\text{ V}$, which corresponds to **Option (B)**. $$\text{Correct Answer (based on cell notation): } \boxed{\text{D}}$$ $$\text{Correct Answer (based on IUPAC reduction potential): } \boxed{\text{B}}$$