Question

For a cell reaction, Pb(s) + Hg$_2$Cl$_2$(s) → PbCl$_2$(s) + 2Hg(l), $\left( \frac{\partial E^\circ}{\partial T} \right)_P$ is 1.45 × 10$^{-4}$ V K$^{-1}$. The entropy change (in J mol$^{-1}$ K$^{-1}$) for the reaction is ..........

Hint:

The temperature dependence of cell potential directly gives entropy change.

Answer 1

Correct Answer: 27.9

Step 1: Use thermodynamic relation.
For an electrochemical cell:
ΔS = nF $\left( \frac{\partial E^\circ}{\partial T} \right)_P$
Step 2: Determine n (electrons transferred).
Hg$_2^{2+}$ + 2e$^-$ → 2Hg(l)
Hence, n = 2.
Step 3: Substitute values.
ΔS = 2 × 96500 × (1.45 × 10$^{-4}$)
= 2 × 96500 × 0.000145
= 27.985 ≈ 28 J mol$^{-1}$ K$^{-1}$.
Note: The reaction is written as reduction of Hg$_2^{2+}$, but overall sign becomes negative depending on direction.
Step 4: Conclusion.
Thus, the entropy change is approximately −28 J mol$^{-1}$ K$^{-1}$.