An oxidation-reduction reaction in which 3 electrons are transferred has a of 17.37 kJ mol at 25ºC. — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the relationship between ΔG° and E°cell** Use the fundamental equation: $$\Delta G° = -nFE°_{cell}$$ where: - ΔG° = 17.37 kJ/mol = 17,370 J/mol - n = 3 electrons transferred - F = 96,500 C/mol - E°cell = unknown **Step 2: Rearrange to solve for E°cell** $$E°_{cell} = \frac{-\Delta G°}{nF}$$ **Step 3: Substitute values** $$E°_{cell} = \frac{-17,370}{3 \times 96,500}$$ $$E°_{cell} = \frac{-17,370}{289,500}$$ **Step 4: Calculate E°cell** $$E°_{cell} = -0.06 \text{ V}$$ **Step 5: Express in the required format** The question asks for the answer as _____ × 10⁻² V $$-0.06 \text{ V} = -6.0 \times 10^{-2} \text{ V}$$ Taking the absolute value: 6.0 × 10⁻² Therefore, the answer is 6.00.