What is the value of the equilibrium constant for the cell reaction below at 25 degree C? E degree cell = 0.61

2Cr(s) +3Pb^2+(aq) ====> 3Pb(s)+2Cr^3+(aq)

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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) `

Where:

`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`

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