Notifications
Clear all
0
30/01/2022 2:30 pm
Topic starter
Consider the following cell reaction:
Cd(s) + Hg2SO4(s) + 9/5H2O(l) ⇌ CdSO4.9/5H2O(s) + 2Hg(l)
The value of E0cell 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]
1 Answer
0
30/01/2022 2:36 pm
ΔG° = -nFE° = ΔH° - TΔS°
= \(\frac{ΔH°+nFE°}{T}\)
= \(\frac{(-825.2 \times 10^3) + (2 \times 96487 \times 4.315)}{298}\)
= \(\frac{-825.2 \times 10^3 + 832.682 \times 10^3}{298}\)
= \(\frac{7.483 \times 10^3}{298}\)
= 25.11 J K-1 mol-1
∴ Nearest integer answer is 25.