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For a cell reaction involving a two elec...

For a cell reaction involving a two electron change, the standrard emf of the cell is found to be `0.295V" at "25^(@)`C. The equilibrium constant of the reaction at `25^(@)`C will be:

A

`1xx10^(-10)`

B

`29.5xx10^(-2)`

C

10

D

`1xx10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
D
`E^(@)=(0.0591)/(n)"log" k( "At" 25^(@)C)`
here n=2 and `E^(@)=0.295V`
`therefore 0.295=(0.0591)/(2)"log"K`
`"log"K=(0.590)/(0.0591)=10`
`K =10^(10)`
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