A cell contain two hydrogen electrodes. The negatove electrode is in contact with a solution of `pH = 5.5`. The emf of the cell is `0.118 V` at `25^(@)C`. Calculate the `pH` of solution positive electrode. (assume pressure of `H_(2)` in the both electrondes `= 1` bar )

Options

A. `3 . 5`
B. `7 . 5`
C. `4.5`
D. `6.5`

Answer

A

Explanation

Anode `(-) rArr underset(1 atm) (H_(2))(g) rarr underset(10^(-55))(2H^(+)) (aq)+2e^(-)`
Cathode `(+) rArr underset(xM)(2H^(+)) (aq) + 2e^(-) rarr underset((1 atm)) (H_(2))(g)`
`0.118V = 0-(0.059)/(2)log_(10). (10^(-55))^(2)/((x)^(2)) rArr x=10^(-35)M`
`rArr pH= 3.5`