The redox reaction is 2 SmCl3(aq) + 6 Li(aq) → 2 Sm(s) + 6 LiCl(aq) Calculate the following for the reaction at 298.0 K .A) The standard Gibbs free energy change of the reaction @ 298.0 K B) The equilibrium constant for the reaction C) The standard cell potential

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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The redox reaction is

2 SmCl3(aq) + 6 Li(aq) → 2 Sm(s) + 6 LiCl(aq)

Calculate the following for the reaction at 298.0 K

.A) The standard Gibbs free energy change of the reaction @ 298.0 K

B) The equilibrium constant for the reaction

C) The standard cell potential

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Follow-up Question

this aprt of calculation seems wrong 

ΔG° = -RTln(K) 

=>-3.9 × 10^5 J/Mol.  = - 8.314 J/K .Mol × 298K ln (K)

By Solving K = 9.77× 10^68

I think it is 

(-3.9 × 10^5  J/Mol) = (- 8.314 J/K Mol × 298K x ln(K)

ln (K)=  (-3.9 × 10^5 J/Mol) / (-8.314 J/K Mol × 298)

ln (K)= 157.4

  K = e^157.4

K = 2.3 x 10^68


are you right or me 

 

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Follow-up Question

2 SmCl3(aq) + 6 LiCl(aq) + 6 e- → 2 Sm(s) + 6 Li+(aq) + 6 Cl-(aq) how di you get this reaction , i provided a different one 

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