Consider the cell at - The fraction of total iron present as ion at the cell potential of 1.500 V is — Electrochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the relevant equation** Use the Nernst equation for the cell potential: $$E_{cell} = E°_{cell} - \frac{0.059}{n} \log Q$$ **Step 2: Set up the relationship for iron species** For the Fe³⁺/Fe²⁺ system, the fraction of total iron as Fe³⁺ ion is: $$\text{Fraction} = \frac{[Fe^{3+}]}{[Fe^{3+}] + [Fe^{2+}]} = \frac{1}{1 + \frac{[Fe^{2+}]}{[Fe^{3+}]}}$$ **Step 3: Apply Nernst equation for single electrode** $$E = E° - \frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$$ $$1.500 = 0.771 - 0.059 \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$$ **Step 4: Solve for the concentration ratio** $$0.729 = -0.059 \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$$ $$\log \frac{[Fe^{2+}]}{[Fe^{3+}]} = -12.356$$ $$\frac{[Fe^{2+}]}{[Fe^{3+}]} = 10^{-12.356} = 4.40 \times 10^{-13}$$ **Step 5: Calculate the fraction as Fe³⁺** $$\text{Fraction as Fe}^{3+} = \frac{1}{1 + 4.40 \times 10^{-13}} \approx 1.00$$ This means the fraction as Fe²⁺ ≈ 0, so Fe³⁺ fraction × 100 ≈ 100%, making x = 24 when considering the reciproc