Question

The cell `Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag` has emf, `E_(298KK) =0`. The electrode potaneial for the reaction `AgI +e^(-) rarr Ag + I^(Theta)` is `-0.151` volt. Calculate the `pH` value:-
A. `3.37`
B. `5.26`
C. `2.56`
D. `4.62`

Answer 1

Correct Answer - C
`(1)/(2)H_(2)AgI rarr H^(+)+Ag+I^(-) E = 0`
`(1)/(2) H_(2)rarr H^(+)+e^(-)` , `E=0.151`
`0.151= -(0.059)/(1)log (H^(+))=0.059xxpH`
`pH = (0.151)/(0.059) = 2.5`