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

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.

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

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.

a. To identify the substances oxidized and reduced in the given corrosion reaction, we need to determine the changes in oxidation states of the elements involved.

In the reactants, the iron (Fe) is in its elemental state, which has an oxidation state of 0. In the products, however, iron is oxidized to Fe^2+, which has an oxidation state of +2. This indicates that iron is being oxidized.

On the other hand, oxygen (O2) in the reactants has an oxidation state of 0, while in the products, it forms hydroxide ions (OH-) with an oxidation state of -1. This indicates that oxygen is being reduced.

b. To write the balanced redox equation using half reactions, we divide the overall reaction into two half reactions, one for oxidation and one for reduction.

Oxidation half reaction: Fe(s) -> Fe^2+(aq) + 2e-
Reduction half reaction: O2(g) + 2H2O(l) + 4e- -> 4OH-(aq)

To balance the number of electrons transferred, the reduction half reaction is multiplied by 2:

2O2(g) + 4H2O(l) + 8e- -> 8OH-(aq)

To balance the number of OH- ions on both sides, we multiply the oxidation half reaction by 4:

4Fe(s) -> 4Fe^2+(aq) + 8e-

Finally, adding the two half reactions gives the balanced redox equation:

4Fe(s) + 2O2(g) + 4H2O(l) -> 4Fe^2+(aq) + 8OH-(aq)

c. To prove that the reaction is spontaneous at standard conditions, we can calculate the standard cell potential (Ecell) of the electrochemical cell using the standard reduction potentials of the half reactions involved.

The standard reduction potential for the reduction half reaction, 2O2(g) + 4H2O(l) + 8e- -> 8OH-(aq), can be found in a standard reduction potential table. The standard reduction potential for the oxidation half reaction, Fe(s) -> Fe^2+(aq) + 2e-, is considered to be 0V since it is the reference half reaction.

By subtracting the reduction potential of the oxidation half reaction from the reduction potential of the reduction half reaction, we can calculate the standard cell potential:

Ecell = [Reduction potential of reduction half reaction] - [Reduction potential of oxidation half reaction]

If the calculated value of Ecell is positive, then the reaction is spontaneous at standard conditions.

Note: The reduction potentials used should be from the same reference table to ensure accuracy.