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The standard emf of a galvanic cell invo...

The standard emf of a galvanic cell involving cell reaction with `n = 2` is found to be `0.295 V` at `25^(@)C`. The equilibrium constant of the reaction would be (Given`F=96,500 C mol^(-1), R = 8.314 JK^(-1) mol^(-1)`):

A

(a) `2.0xx10^(11)`

B

(b) `4.0xx10^(12)`

C

(c) `1.0xx10^(2)`

D

(d) `1.0xx10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) By Nernst equation,
`E_(cell)=E_(cell)^(@)-(2.303RT)/(nF) log_(10)K`
At equilibrium, `E_(cell)=0`
Given that,
`R=8.314JK^(-1)mol^(-1)`
`T=25^(@)C+273=298K`
`F=96500C and n=2`
`:. E_(cell)^(@)=(2.303xx8.314xx298)/(2xx96500)log_(10)K`
`=(0.0591)/(2) log_(10)K`
Given that `E_(cell)^(@)=0.295 V`
`:. 0.295=(0.0591)/(2) log_(10)K`
`log_(10)K=(0.295xx2)/(0.0591)=10`
antilog `log_(10)` K= antilog 10
`K=1xx10^(10)`
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