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. `29.5 xx 10^(-2)`
B. `10`
C. `1 xx 10^(10)`
D. `2.95 xx 10^(-10)`

Answer

C

Explanation

Using the relation ,
`E_(cell)^(@) = (2.303 RT)/(nF) "log" K_(c) = (0.0591)/(n) "log" K_(c)`
`therefore 0.295 V = (0.0591)/(2) "log" K_(c)`
or `log K_(c) = (2 xx 0.295)/(0.0591) = 10`
or `K_(c) = 1 xx 10^(10)`