InterviewSolution
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InterviewSolution
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The dissociation constant of `[M (NH_(3))_(2)]^(+)` complex into `M^(+)` and Ammonia is `1xx10^(-17)`. The standard reduction poential of the same metal electrode `[E_(M^(+),M)^(@)]` is `1.02 V`. Hence `E^(@)` of the half cell reaction : `[M(NH_(3))_(2)]^(+)+e^(-)rarrM(s)+2NH_(3)` is found to be `chi xx 10^(-2)`. Express the value of x to the nearest integer ? (Given R = 8.3 and 2.3 log x = `log_(e )x`)__ |
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Answer» Correct Answer - 2 `Eq(A)rarr[M(NH_(3))_(2)]^(+)underset(larr)rarr M^(+)+2NH_(3)` `Delta G_(1)^(@)=-(kJ.2.303 log 10^(-17))/(96,500)=1.0F` `Eq(B)rarr M^(+)+e^(-)underset(larr)rarr M(s)` `Delta G_(2)^(@)=-1.02F` on adding the two `[M(NH_(3))_(2)]^(+)+e^(-)rarr M(S)+2NH_(3)` `Delta G_(3)^(@)=-nFE^(@)` `Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(3)^(@)=-0.02F-1xxFxxE^(@)=-0.02F` `E^(@)=2xx10^(-2)` x = 2 |
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