Consider the following cell reaction: 

C d ( s ) + H g 2 S O 4 ( s ) + 9 5 H 2 O ( I ) + C d S O 4 9 5 H 2 O ( s ) + 2 H g ( I )

The value of E c e l l 0 is 4.315V at 25°C. If Δ H o = -825.2 kJ mol-1, the standard entropy change   Δ S °  in JK-1 is_________. (Nearest integer)

           [Given: Faraday constant = 96487 C mol-1]

3 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
R
6 months ago

Change in oxidation number = 2

Δ G ° = n E c e l l o F

Δ G ° = n E c e l l o F

=-2 * 4.315 * 96487 Jmol-1

Δ G ° = Δ H ° T Δ S °

Δ S ° = Δ H ° Δ G ° T

= 8 2 5 . 2 * 1 0 0 0 ( 2 * 4 . 3 1 5 * 9 6 4 8 7 ) 2 9 8 . 1 5 J K 1

= 7 4 8 2 . 8 1 2 9 8 . 1 5 = 2 5 . 0 9 J K 1

25.1 JK-1

Ans. = 25

 

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Similar Questions for you

ΔG° = –RT * 2.303 log K

–nFE° = +RT * 2.303 log K

2 * 96500 * 0.295 = 8.314 * 298 * 2.303 log10 K

10 = log10 K = 1010

It has chiral centre and differently di substituted double bonded carbon atoms.

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Thirteen 2025

Chemistry NCERT Exemplar Solutions Class 12th Chapter Thirteen 2025

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