Correct Answer - C
`{:("Anode":, H_(2) rarr 2H^(+) +2e^(-)),("Cathode":,CI_(2)+2e^(-) rarr 2CI^(-)):}`
`E = E^(@) - 0.0591 log [H^(+)] [CI^(-)]`
on increasing concentration by 10 E wil increase by a factor of `-0.0591log 100 =- 0.0591 xx 2`
`=- 0.1182 V`