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.

Answer

`10^(-4)M`

Explanation

`Pt,H_(2)|underset(pH=6)(H^(o+)(10^(-6)M))||underset(pH=?)(H^(o+)(xM))|H_(2),Pt`
`E_(cell)=-0.059(pH_(c)-pH_(a))`
`0.118V=-0.059(pH_(c)-6)`
`:. pH_(c)=(-0.118)/(0.0590+6`
`=-2+6=4`
`:. [H^(o+)]_(c)=10^(-4)M`