Reaction |
Eo / V |
Nernst Equation E = Eo– (2.303RT/ nF) * log{ [Products] / [Reactants]} |
1) Fe 2+ + 2e = Fe |
-0.440 |
-0.440 + 0.0295 log [ a Fe2+] |
2) Fe3 O4 + 8H + + 8e - = 3Fe + 4 H 2O |
-0.085 |
-0.085 – 0.0591 pH |
3) Fe3 O4 + 8H + + 2e - = 3Fe2+ + 4 H 2O |
0.980 |
0.980 – 0.2364 pH - 0.0886 log (a Fe 3+ ) |
4) Fe3+ + e- = Fe2+ |
0.771 |
0.771 + 0.0591 log ( a Fe 3+/ a Fe 2+ )
|
5) 2Fe3+ + 3H2O = Fe2O3 + 6H+ |
Use (aH+)6/(aFe3+)2 =100 (This is not an electrode reaction) |
pH = { - log (a Fe3+ ) - 1} / 3 |