Question

(a) In a cell reaction, equilibrium constant L is less than one. Is `E^(@)` for the cell positive or negative ?
(b) What will be the value of K if `E_(cell)^(@)=0` ?

Answer 1

(a) For a cell , `E^(@)=(0.0591)/(n)logK`
Let the value of K be 0.01 (less than one)
`E^(@)=(0.0591)/(n)log0.01=(0.0591)/(n)log(10^(-2))=(-(2)xx(0.0591))/(n)=-"ve"`.
(b) If `E_(cell)^(@)=0, "then" 0=(0.0591)/(n)log K" or " log K=0,K="Antilog"(0)=1`.