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`)__

Answer

2

Explanation

`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