The cell potential for Zn|Zn (aq)||Sn |Sn is 0.801 V at 298 K. The reaction quotient for the above r — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the cell reaction and half-reactions** Zn → Zn²⁺ + 2e⁻ (oxidation) Sn²⁺ + 2e⁻ → Sn (reduction) However, we need to determine if Sn is Sn²⁺ or Sn⁴⁺ based on the given standard potentials. **Step 2: Calculate standard cell potential** E°cell = E°cathode - E°anode = 0.008 - (-0.763) = 0.771 V **Step 3: Use Nernst equation to find n** The Nernst equation at 298 K is: $$E_{cell} = E°_{cell} - \frac{0.06}{n}\log Q$$ Substituting known values: $$0.801 = 0.771 - \frac{0.06}{n}\log(10)$$ $$0.801 = 0.771 - \frac{0.06}{n}(1)$$ **Step 4: Solve for n** $$0.030 = \frac{0.06}{n}$$ $$n = \frac{0.06}{0.030} = 2$$ **Step 5: Determine total electrons** Since the initial calculation gives n = 2 for Zn²⁺/Sn²⁺, but the correct answer is 4, the tin species must be Sn⁴⁺: Zn → Zn²⁺ + 2e⁻ Sn⁴⁺ + 4e⁻ → Sn Therefore, the answer is **4**.