4 error in equation balance
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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.
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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
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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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