InterviewSolution
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InterviewSolution
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`K_(d)` for dissociation of `[Ag(NH_(3))_(2)]^(+)` into `Ag^(+)` and `NH_(3)` is `6 xx 10^(-8)`. Calculae `E^(@)` for the following half reaction. `AG(NH_(3))_(2)^(+) +e^(-) rarr Ag +2NH_(3)` Given `Ag^(+) +e^(-) rarr Ag, E^(@) = 7.99V` |
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Answer» Correct Answer - `0.373V` `Ag rarr Ag^(+) +e^(-) E_(RP) = 0.799V` `Ag(NH_(3))_(2)^(+) +e rarr Ag +2 NH_(3) E_(RP) = ?` `Ag(NH_(3))_(2)^(+) hArr Ag^(+) +2 NH_(3)` `E_(cell) = E_(cell)^(@) +(0.0591)/(1)log_(10).([Ag(NH_(3))_(2)^(+)])/([Ag^(+)][NH_(3)]^(2))` `0 =E_(cell)^(@) +(0.0591)/(1)log(6 xx 10^(-8)) rArr E_(cell)^(@) =- 0.426` `E_(cell)^(@) = E_(C)^(@) - E_(A)^(@)` `-0.426 =E_(C)^(@) - 0.799 rArr E_(C)^(@) = 0.373 V` |
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