The redox reaction involving samarium is as below. 2 Sm(s) + 6 HCl(aq) → 2 SmCl3(aq) + 3 H2(g) The oxidation half-reaction is: Sm - 3e- → Sm3+
Hi, this is extremely important to me. I have done the calculations. Please check if it is right. I really would appreciate it if you check each calculation. Please also check E of half-reactions from the tables to be sure. The values bolded are the one that needed to be found.Thanks
The
2 Sm(s) + 6 HCl(aq) → 2 SmCl3(aq) + 3 H2(g)
The oxidation half-reaction is:
Sm - 3e- → Sm3+
E° = - 2.304 V
The reduction half-reaction is:
H+ + e- → H
E° = 0 V
E° = E°(cathode) - E°(anode)
E° = 0 - (- 2.304)
E° = 2.304 V
ΔG° = -nFE°
E° = 2.304 V
ΔG°= -(6 x 2.304 x 96,485)
ΔG°= -1,333,808 .6 J
ΔG° = -RTln(K)
(-1,333,808 .6 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)= 538.35
K = e157.4
K = 6.35 X 10^233
Calculate the free energy change and the cell potential for the reaction at 298.0 K if all reactants are at 2.00 M and/or 2.00 atm and all products are at 0.0100 M and/or 0.0100 atm. Make sure to write half reactions while calculating.
E = E° - (RT/nF)ln(Q)
At non-standard conditions, the concentrations are:
[Sm] = 0 (solid)
[HCl] = 2.00M
[SmCl3] = 0.0100 M
[H2] = 0.0100 M
Q = {[SmCl3]^2[H2]^3 }/ [HCl ]^6
= (0.0100)^2 (0.0100)^3/ (2.00)^6
= 1.5625 X 10^-12
Now
ΔG = ΔG° + RTlnQ
= -1,333,808 .6 J + (8.314 x 298) × ln(1.5625 X 10^-12)
= -1401160.736 J
= -1401.2 KJ
Substituting the values into the Nernst equation, the new cell potential (E) can be calculated as follows:
E = E° - (RT/nF)ln(Q)
E = 2.304 - ((8.314 × 298 )/(6 × 96,485)) × ln(1.5625 X 10^-12)
E = 2.403V
Step by step
Solved in 4 steps
