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The reducation potential of a `Pt(s)|Cl_(2)(g)|Cl^(-)(aq.)` electrode is found to be `1.42 V` when the pressure olf `Cl_(2)` is 0.25 arm. The concentration of chloride ion in this half cell isA. `0.43 M`B. `0.043 M`C. `0.34 M`D. `0.034 M` |
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Answer» Correct Answer - B The half cell reaction is `Cl_(2)(g) + 2e^(-) hArr 2Cl^(-)(aq.)` and its `E_(Cl_(2)//Cl^(-)` is `1.36 V` The value of `n`, the number of elctrons transferred, is `2`. The expression for `Q` is `C_(Cl^(-)//P_(Cl_(2)))` Substituting these data into the Nernst equation gives `E = E^(@) -(0.0592 v)/(n) log Q` `1.42 V = 1.36 V - (0.0592 V)/(n) "log" (C_(Cl^(-))^(2))/(0.25)` `C_(Cl^(-)) = 0.043 M` |
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