Correct Answer - A
(a) In the cell,
At anode : `H to H^(+)+e^(-)`
At cathode : `H^(+)+e^(-) to H`
we know that,
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([H^(+)]Anode)/([H^(+)]Cathode)`
`0.118=0-(0.0591)/(1)"log"(10^(-6))/([H^(+)]Cathode)`
or `"log"10^(-6)/([H^(+)]Cathode)=-(0.118)/(0.0591)=-2`
`(10^(-6))/([H^(+)]Cathode)=10^(-2)`
`[H^(+)]_(Cathode)=(10^(-6))/(10^(-2))=10^(-4)M`