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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» 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` |
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