Calculate the equilibrium constant for the reaction at 298K 4 Br + O2 + 4H+ → 2 Br2 + 2H2O Given that Ecell = 0.16V​

Explanation

`E_(cell)^(@)=0.16V`
`E=E^(@)-(0.0591)/(n)logQ_(C)`
At equilibrium, `E=0 and Q_(C)=Q_(C)=K_(C)`
`therefore logK_(C)=(nxxE^(@))/(0.0591)`
Cell reactions are,
Oxidation: `4Br^(-) to 2Br_(2)+4e^(-)`
Reduction: `4H^(+)O_(2)+4e^(-) to 2H_(2)O`
Overall: `4H^(+)+O_(2)+4Br^(-) to 2Br_(2)+2H_(2)O`
Here, `n=4`
`therefore logK_(C)=(4xx0.16V)/(0.0591V)=(64)/(5.91)=18.8291`
`K_(C)`=antilog `10.8291=6.747xx10^(10)`