Consider the following half cell reaction ) ( O H ) aq ( Cr ) aq ( H e O Cr 3 7 The reaction was con — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Write the Nernst equation for the half-cell reaction** For the dichromate/chromium(III) half-cell: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O The Nernst equation is: $$E = E° - \frac{0.059}{n} \log Q$$ where n = 6 electrons, and Q is the reaction quotient. **Step 2: Set up the reaction quotient** $$Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}}$$ Given ratio: [Cr₂O₇²⁻]/[Cr³⁺] = 2/3, so [Cr³⁺]/[Cr₂O₇²⁻] = 3/2 $$Q = \frac{(3/2)^2}{[H^+]^{14}} = \frac{9/4}{[H^+]^{14}}$$ **Step 3: Apply condition for zero EMF** When EMF = 0: $$0 = 1.33 - \frac{0.059}{6} \log Q$$ $$\log Q = \frac{1.33 × 6}{0.059} = 135.25$$ **Step 4: Solve for [H⁺]** $$\log\left(\frac{9/4}{[H^+]^{14}}\right) = 135.25$$ $$\log(2.25) - 14\log[H^+] = 135.25$$ $$0.352 - 14\log[H^+] = 135.25$$ $$\log[H^+] = -9.63$$ **Step 5: Calculate pH** $$pH = -\log[H^+] = 9.63 ≈ 10$$ Therefore, the answer is **10