Consider the following cell reaction :
Cd₍ₛ₎ + Hg₂SO₄₍ₛ₎ + 9/5H₂O₍ₗ₎ ⇌ CdSO₄.9/5H₂O₍ₛ₎ + 2Hg₍ₗ₎
The value of E_(cell)⁰ is 4.315 V at 25°C. If ΔH° = -825.2 kJ mol⁻¹, the standard entropy change ΔS° in J K⁻¹ is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Get ΔG° from the cell voltage, then the Gibbs equation hands you ΔS° once you know ΔH°.

1. Find n. Cd goes from 0 to +2, losing 2 electrons. In Hg₂SO₄ each Hg is +1, and both Hg atoms go to 0, gaining 2 electrons in total. So n = 2.
2. Gibbs energy. Δ G° = -nFE°_(cell) = -2 × 96487 × 4.315 = -832 683 J mol⁻¹, which is about -832.68 kJ mol⁻¹.

   Δ S° = (Δ H° - Δ G°)/T

3. Put in numbers. Work in joules with T = 298 K: Δ S° = (-825 200 - (-832 683))/298 = 7483/298 = 25.1 J K⁻¹.
4. So. Δ S° ≈ 25.1 J K⁻¹, which rounds to 25.

The common slip: -25. Writing Δ S° = (Δ G° - Δ H°)/T flips the sign and gives −25. Rearrange Δ G = Δ H - TΔ S carefully.

The idea to keep. Cell voltage gives ΔG° through -nFE°, and Δ G = Δ H - TΔ S links it to ΔS°. Get n from the oxidation states and put everything in joules before you divide by T.

Answer from the NTA final answer key. Solution by PrepWiser.