Question

the corrosion of iron is similar to the reactions that occur in an electrochemical cell. Oxidation and reduction occur at separate places on the metal and the circuit is completed by an electrolyte in solution one possible corrosion is Fe(s)+O2(g) + H2O(l) --> Fe^2+(aq) + OH-(aq)
a. identify the substances oxidized and the substance reduced?

b. using the half reactions write the balanced redox equation

c. prove the reaction is spontaneous at standard conditions

Answer 1

The substance oxidized is the one that loses electrons. The substance reduced is the one that gains electrons. It looks like Fe changes from zero to +2 and oxygen changes from zero to -2.

Fe ==> Fe^2+ + 2e
O2 +4e + 2H2O ==> 4OH^-
Multiply #1 by 2 and #2 by 1 and add them for the complete redox reaction.

Look up Eo values (be careful with the signs) and add the oxidation half to the reduction half. Add the E values for each and it should come out a positive number.

Answer 2

Eo cell- Eo 1 + Eo 2 = -0.44+ +1.23= 0.79V?
how do i explain that its spontaneous because delta H is greater then O

Answer 3

The reaction is spontaneous because Ecell is +. I looked up the Eo values. E for Fe==> Fe^2+ is 0.44
O2==> OH^- is +0.401 Ecell = 0.841 which is about the same as you obtained but with different numbers.