​ Consider the following cell reaction : Cd(s)+Hg2SO4(s)+ H2O(l) ⇌ CdSO4. H2O(s) + 2Hg(l) ​

Question

Consider the following cell reaction :

Cd_(s)+Hg₂SO_4(s)+  \(\frac{9}{5}\)H₂O_(l) ⇌ CdSO₄^. \(\frac{9}{5}\) H₂O_(s) + 2Hg_(l)

The value of E⁰_cell 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]

Answer 1

Answer is 25

∴ Nearest integer answer is 25

Comments

Bhai apne n =2 kaise liya hai please bta do

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Bro cadmium ka compund ban rhaaa hai it moves from charge 0 to 2 so n= no of electrons involved which gives 2