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

Options

A. `1xx10^(-10)`
B. `29.5xx10^(-2)`
C. 10
D. `1xx10^(10)`

Answer

D

Explanation

`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)`