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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

`3.13xx10^(-3)`

B

`3.13xx10^(-4)`

C

`6.26xx10^(-3)`

D

`6.26xx10^(-4)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Hg_(2)^(2)+2e^(-)to2Hg,0.789"volt"`
`HgtoHg^(2+)+2e^(-),-0.854"Volt"`
`Hg_(2)^(2+)toHg+Hg^(2+),-0.065"Volt"`
`triangleG=-2xx(-0.065)xx96500=-8.314xx298lnK_(eq),K_(eq)=6.3xx10^(-3)`
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Given : Hg_(2)^(2+) +2 e ^(-)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+) .

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