A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)M` hydrogen ions. The `EMF` of the cell is `0.118V` at `25^(@)C`. Calculate the concentration of hydrogen ions at the positive electrode.

Explanation

Anode `H_(2) rarr 2H^(+) + 2e`
(negative polarity) `[H^(+)] = 10^(-6)M`
Cathode: `2H^(+) + 2e rarr H_(2)`
(positive polarity) `[H^(+)] rarr aM`
`:. E_(cell) = E_(OP_(H^(+)//H)) + E_(RP_(H^(+)//H))`
`E_(OP_(H^(+)//H))^(@) - (0.059)/(2)log_(10)[H^(+)]_("Anode")^(2)`
`+ E_(RP_(H^(+)//H))^(@) + (0.059)/(2)log_(10)[H^(+)]_("Cathode")^(2)`
`= (0.059)/(2)log_(10).([H^(+)]_("Cathode")^(2))/([H^(+)]_("Anode")^(2))`
`0.118= (0.059)/(2)log_(10).([H^(+)]_("Cathode")^(2))/(10^(-6))^(2)`
`= (0.059)/(1)log_(10).([H^(+)]_("Cathode")^(2))/(10^(-6))`
`:. [H^(+)]_("Cathode") = 10^(-4) M`