What is the value of the equilibrium constant for the reaction between following species at 25 degree C? Fe(s) and Cr^3+ (aq)

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llltkl | College Teacher | (Level 3) Valedictorian

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The relation between `E^0` and the equilibrium constant K is given by


`rArr lnK=nFE^0/RT`

Here, relevant `E^0` values are:

`Fe(s) rarr Fe^(2+)(aq.) +2e- ( E^0=0.44 V)` *3

`rArr 3Fe(s) rarr 3Fe^(2+)(aq.) +6e- ( E^0=0.44 V)`

and `Cr^(3+)(aq.) + 3e- rarr Cr(s) (E^0=-0.41 V)` *2

`rArr 2Cr^(3+)(aq.) + 6e- rarr 2Cr(s) (E^0=-0.41 V)`


`E^0(overall) = 0.44+(-0.41) = 0.03 V`

Plugging in the values in the relation of K,


`lnK=(6*96485*0.03)/(8.314*298) `


`rArr K=1.107*10^3`

Therefore, the value of equilibrium constant for the reaction between Fe(s) and Cr^(3+)(aq.) is 1.107*10^3.


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