December 13, 2019, 07:39:05 AM
Forum Rules: Read This Before Posting


Topic: Standard half potential of oxidation of zero-valent sulfur to sulfite  (Read 185 times)

0 Members and 1 Guest are viewing this topic.

Offline lights_that_flash

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
I'm trying to calculate the standard half potential of this reaction:
SO32- + 3 H2O + 4 e-  ::equil:: S0 + 6 OH-
and I would like for someone to verify if I did it right.

I know the following:
ΔGf(S0)=0 kJ/mol
ΔGf(SO32-)=-486.6 kJ/mol
ΔGf(H2O)=-237.14 kJ/mol
ΔGf(OH-)=-157.2 kJ/mol
T=303.15 K
pH=8.5
[SO32-]=1·10-6 M
[OH-]=10-(14-pH)

First, I calculated ΔG0=-ΔGf(SO32-)-3·ΔGf(H2O)+6·ΔGf(OH-)=+254.8 kJ/mol
Then, I calculated E0=(ΔG0/4·F)·1000=-0.660 V
Finally, I calculated E=E0-(R·T/4·F)·ln(([OH-])6/[SO32-])=-0.254V

However, I expected this value to be closer to -0.1V. Did I do something wrong?

Sponsored Links