Calculate the equilibrium constant for the cell reaction :
`4Br^(-)+O_(2)+4H^(+)rarr2Br_(2)+2H_(2)O."Given "E_(cell)^(@)=0.16 V`

Explanation

`E_(cell)^(Theta)` is related to equilibrum constant , `K_(c)` as :
`E_(cell)^(Theta) = (0.059)/(n) "log" K_(c)` or `log K_(c) = (n E_(cell) ^(Theta))/(0.059)`
In this reaction , four electrons are transferred so that n = 4
`log K_(c) = ( 4 xx 0.16 V)/(0.059 V) = 10.847`
`K_(c) = 7.03 xx 10^(10)`.
The value of equilibrium constant shows that the equilibrium position lies for the right and therefore , large amount of products are formed .