Calculate the equilibrium potentials (electrode potentials) Eo for the following electrochemical reactions as a function of the pH of the solution. It is assumed that other soluble ions are present in a concentration of 106 mol/L, and that the gases partial pressure is 1 atm. Nernst's formula is used in the calculation. aA + bB + ze = mM + nN . E = E 0,059 z (M)" (N) log (A) (B) The formula is valid for 25C. The default potential is Eois given for two of the reactions. The values for 50, for the other reactions must be removed from the table. Show the individual steps in the calculation. 1) Fe2+ + 2e = Fe E = -0,44 V = = 2) Fe3O4 + 8 H+ + 8e = 3Fe + 4 H2O E = -0,085 V 3) Fe3O4 + 8 H+ + 2e = 3Fe2+ + 4 H2O 4) 2H+ + 2e = H2 5) O2 + 2H20 + 4e = 40H- Construct a pH-Eo diagram (use the attached diagram or create your own using Excel for example) for the electrochemical reactions. Show the calculations that lie as a basis for drawing the lines in the diagram. It's okay to let the lines cross each other in this chart (will not be a full- fledged Pourbaix chart), however still try to indicate where iron is immune, passive and where it corrodes