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 0.118 V at `25^(@)C` calculate the concentration of hydrogen ions at the positive electrode.

Explanation

The reaction is : `H^(+)+e^(-)rarr1//2H_(2)(g)`
`E_(cell)=(0.0591)/(1)"log"(C_(1))/(C_(2))`
`0.118 V=(0.0591)/(1)"log"(C_(1))/(C_(2))`

`"log"(C_(1))/(C_(2))=(0.118 Vxx1)/(0.0591)=2 or(C_(1))/(C_(2))="Antilog "2=10^(2)`
`C_(1)=C_(2)xx10^(2)=10^(-6)xx10^(2)=10^(-4) M`