[lights_that_flash]

I'm trying to calculate the standard half potential of this reaction:
SO₃^2- + 3 H₂O + 4 e^-  S⁰ + 6 OH^-
and I would like for someone to verify if I did it right.

I know the following:
ΔG_f(S⁰)=0 kJ/mol
ΔG_f(SO₃^2-)=-486.6 kJ/mol
ΔG_f(H₂O)=-237.14 kJ/mol
ΔG_f(OH^-)=-157.2 kJ/mol
T=303.15 K
pH=8.5
[SO₃^2-]=1·10^-6 M
[OH^-]=10^-(14-pH)

First, I calculated ΔG⁰=-ΔG_f(SO₃^2-)-3·ΔG_f(H₂O)+6·ΔG_f(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/[SO₃^2-])=-0.254V

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