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

💡 Solution & Explanation

$[H^+] = \sqrt{K_a \cdot C} = \sqrt{1.8 \times 10^{-5} \times 0.5} = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3}$ M. pH = $-\log(3 \times 10^{-3}) = 3 - \log 3 = 3 - 0.48 = 2.52$. The cell notation Pt \| H2 \| H+ represents an oxidation half-cell. Oxidation potential = $0.06 \times \text{pH} = 0.06 \times 2.52 = +0.1512$ V.