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The cell Pt(H(2))(1atm) |H^(+) (pH =?) T...

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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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