What is the equilibrium concentration of Zn2+ at 25C if [Cr3+]=3.7x10-3M for the reaction: 3Zn + 2Cr3+ ->3Zn2+ + 2Cr, E=+0.02V 

Expert Answers
mwmovr40 eNotes educator| Certified Educator

The relationship between the energy released in a oxidation-reduction reaction is

`0=DeltaG + RTlnK`

For this reaction the change in Gibbs Free energy is just the electrical energy o 0.02V

Solving for K (the equilibrium constant) gives:

`RTlnK = -DeltaG`

`lnK = -(DeltaG)/(RT)`

`K = e^[-(DeltaG)/(RT)]`

the Equilibrium constant gives  you the following relationship between the concentrations of the reactants and products

`K = ([Zn^(+2)]^3)/([Cr^(+3)]^2)`

` `Thus

`([Zn^(+2)]^3)/([Cr^(+3)]^2) = e^[-(DeltaG)/(RT)]`

Solving for the concentration of zinc ions gives

`[Zn^(+2)] = ([Cr^(+3)]^2*e^[-(DeltaG)/(RT)])^(1/3)`


In this equation 

The concentration of `Cr^(+2) =3.7times10^(-3)(mol)/L`

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` ` `DeltaG = 0.02V`

R is the Universal Gas Constant in units of `((V)*(C))/(mol*K)`

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` `V is volt, C is Coulomb (or together = joules)

T is temperature in units of kelvin

`[Zn^(+2)] = (3.7times10^(-3)(mol)/L^2*e^[-(0.02)/(8.33*(298))])^(1/3)`

`[Zn^(+2)] = 1.55times10^(-1) M`


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