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What is the value of the equilibrium constant for the cell reaction below at 25 degree...
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We can use the expression relating the equilibrium and the delta G:
`0 = E^o _(cell) - (RT)/(nF) lnK`
`(RT)/(nF) lnK =E^o _(cell)`
`lnK = (E^o _(cell) * nF)/(RT) `
`K ` = equilibrium constant
`E^o _(cell)` = is the standard cell potential at the temperature of interest = 0.61
`n` = 6
`F` = 96485 C/mol
`R` = 8.31447 J/mol-K
`T` = 25 + 273.15 = 298.15 K
`lnK = (0.61*6*96485)/(8.31447*298.15) `
`lnK = 142.453 `
`K = e^(142.453) `
`K = 7.35e^61 answer`
Posted by jerichorayel on July 10, 2013 at 5:59 AM (Answer #1)
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