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Given : Hg(2)^(2+) rightarrow 2Hg, E^(...

Given :
`Hg_(2)^(2+) rightarrow 2Hg`, `E^(@) = 0.789 V` and `Hg^(2+) + 2e^(-) rightarrow Hg`, `E^(@) = 0.854V`
Calculate the equilibrium consant for `Hg_(2)^(2+) rightarrow Hg + Hg^(2+)`.

A

89

B

82.3

C

79

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`E^(@)` for the given reaction
`=0.855-0.799=0.056V`
`E_(cell)^(@)=(0.0591)/(2)"log"K_(c)`
or `logK_(c)=(0.056xx2)/(0.059)=1.8983`
`K_(c)`=Antilog `1.8983=79:12=79`
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