4 error in equation balance

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $$\ce{OH-}$$ according to the reaction $$\ce{O2 + H2O + 4e- -> 4OH-}$$$$\ce{O2 + 2H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{4OH-}$$$$\ce{2OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + H2O -> 4OH-}$$$$\ce{2O- + 2H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $$\ce{OH-}$$ according to the reaction $$\ce{O2 + H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{4OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $$\ce{OH-}$$ according to the reaction $$\ce{O2 + 2H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{2OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + 2H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

3 Removed all-caps from title, added more MathJax, slight reformatting of chem eqns.

# partial Partial reactions in REDOXredox formation of rust

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form OH-$$\ce{OH-}$$ according to the reaction $$\ce{O2 + H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{4OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + H2O -> 4OH-}$$$$\ce{2O- + H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

# partial reactions in REDOX formation of rust

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form OH- according to the reaction $$\ce{O2 + H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{4OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

# Partial reactions in redox formation of rust

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $$\ce{OH-}$$ according to the reaction $$\ce{O2 + H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{4OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

2 deployment of \mhchem

In the iron oxidation that results in rust, the first step is to steal 2 electrons from Fe$$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form OH- according to the reaction O2 + H2O + 4e- → 4OH-$$\ce{O2 + H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically 4OH- $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and 4OH-$$\ce{4OH-}$$. Is it that, first, O2$$\ce{O2}$$ steals the two electrons from the Fe$$\ce{Fe}$$ lattice becoming 2O-$$\ce{2O-}$$ and then, later, each O-$$\ce{O-}$$ takes an H$$\ce{H}$$ from H2O$$\ce{H2O}$$ to form 4OH-$$\ce{4OH-}$$? That does not seem right to me, because this means that what is happening is:

O2 + 4e- → 2O-$$\ce{O2 + 4e- -> 2O-}$$

2O- + H2O → 4OH-$$\ce{2O- + H2O -> 4OH-}$$

But the reaction "O2 + 4e- → 2O-"$$\ce{O2 + 4e- -> 2O-}$$" implies that 2O-$$\ce{2O-}$$ is more stable than O2$$\ce{O2}$$. Is that right?

In the iron oxidation that results in rust, the first step is to steal 2 electrons from Fe. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form OH- according to the reaction O2 + H2O + 4e- → 4OH-

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically 4OH- comes out of the collision. I imagine that there must be some steps between the reactants and 4OH-. Is it that, first, O2 steals the two electrons from the Fe lattice becoming 2O- and then, later, each O- takes an H from H2O to form 4OH-? That does not seem right to me, because this means that what is happening is:

O2 + 4e- → 2O-

2O- + H2O → 4OH-

But the reaction "O2 + 4e- → 2O-" implies that 2O- is more stable than O2. Is that right?

In the iron oxidation that results in rust, the first step is to steal 2 electrons from $$\ce{Fe}$$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form OH- according to the reaction $$\ce{O2 + H2O + 4e- -> 4OH-}$$.

What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically $$\ce{4 OH-}$$ comes out of the collision. I imagine that there must be some steps between the reactants and $$\ce{4OH-}$$. Is it that, first, $$\ce{O2}$$ steals the two electrons from the $$\ce{Fe}$$ lattice becoming $$\ce{2O-}$$ and then, later, each $$\ce{O-}$$ takes an $$\ce{H}$$ from $$\ce{H2O}$$ to form $$\ce{4OH-}$$? That does not seem right to me, because this means that what is happening is:

$$\ce{O2 + 4e- -> 2O-}$$

$$\ce{2O- + H2O -> 4OH-}$$

But the reaction "$$\ce{O2 + 4e- -> 2O-}$$" implies that $$\ce{2O-}$$ is more stable than $$\ce{O2}$$. Is that right?

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