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

Question

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

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)`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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)`

5 views Answered 2023-02-05 15:29