Question

Why is this not possible? 2Cl2 + 4NaOH -> 4NaCl + 2H20 + O2

2Cl2 + 4NaOH -> 4NaCl + 2H20 + O2

Why isn't this reaction possible? Because the oxygen in [OH] will have to undergo oxidation from [-2] to [zero], and there's no powerful oxidizer around?

The reaction from the textbook is:

I tried to come up with an alternative reaction with no NaClO3 arising.

Answer 1

`2"Cl"_2 + 4"NaOH" -> 4"NaCl" + 2"H"_2"O" + "O"_2`

First, let's assume we don't know the products. At that point we should know that chlorine gas is reacting in base. If that is the case, it is not a simple acid/base reaction, because there is no acid.

That, along with the single-replacement form of the reaction, suggests that it is a redox reaction (like you know), and we can't immediately expect that we would make an equimolar amount of `"NaCl"` as we have `"NaOH"`, as we would have had it been an actual acid/base reaction.

(Perhaps you put `"O"_2` because it looked like a simplistic single-replacement reaction, so that's understandable.)

Another way you could look at it is to notice that oxygen changed oxidation state from `-2` to both `-2` AND `0` (in `"O"_2`), but in addition, chlorine went from a `0` state (in `"Cl"_2`) to a `-1` state.

That means that somehow, you had oxygen get... oxidized. More often than not, oxygen or oxygen-containing compounds (like oxohalogen anions, or peroxides) would oxidize, not get oxidized.

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WHEN IN DOUBT, USE YOUR TEXTBOOK REFERENCE TABLES

Now, I would start looking at what kinds of redox reactions in base (or not in acid) that `"Cl"_2` could undergo. Only one is shown here.

`"Cl"_2(g) + 2e^(-) -> 2"Cl"^(-)` (you could have predicted this one)

But that doesn't utilize `"NaOH"`. Since we have examined `"Cl"_2` reacting on its own, it makes sense to now consider reactions that have both `"OH"^(-)` and some form of chlorine on the same side.

Two of the reactions in the reference table qualify, but since `"Cl"_2` is a reactant, we should reverse these reactions to get an eventual cancellation of `"Cl"^(-)` on the reactants side only:

`"Cl"^(-)(aq) + 2"OH"^(-)(aq) -> "ClO"^(-)(aq) + "H"_2"O" + 2e^(-)`

`"Cl"^(-)(aq) + 6"OH"^(-)(aq) -> "ClO"_3^(-)(aq) + 3"H"_2"O" + 6e^(-)`

As you can see, neither of these yield `"O"_2`. :)

DETERMINING THE FINAL REACTION

Now let's suppose we went with the first option as a half reaction.

`"Cl"_2(g) + cancel(2e^(-)) -> 2"Cl"^(-)`
`"Cl"^(-)(aq) + 2"OH"^(-)(aq) -> "ClO"^(-)(aq) + "H"_2"O" + cancel(2e^(-))`
`"-------------------------------------------------"`
`"Cl"_2(g) + "Cl"^(-)(aq) + 2"OH"^(-)(aq) -> "ClO"^(-)(aq) + 2"Cl"^(-)(aq) + "H"_2"O"`

Now, our only cation is sodium, so if we add back the sodium spectator ions, we get:

`color(blue)("Cl"_2(g) + 2"NaOH"(aq) -> "NaClO"(aq) + "NaCl"(aq) + "H"_2"O")`

Here, `"Cl"` got oxidized AND reduced, from `0` to both `+1` and `-1`. So this is a plausible reaction. Now let's see the other one.

`3("Cl"_2(g) + cancel(2e^(-)) -> 2"Cl"^(-))`
`"Cl"^(-)(aq) + 6"OH"^(-)(aq) -> "ClO"_3^(-)(aq) + 3"H"_2"O" + cancel(6e^(-))`
`"-------------------------------------------------"`
`3"Cl"_2(g) + 6"OH"^(-)(aq) -> "ClO"_3^(-)(aq) + 5"Cl"^(-)(aq) + 3"H"_2"O"`

Add back the spectator sodium:

`color(blue)(3"Cl"_2(g) + 6"NaOH"(aq) -> "NaClO"_3(aq) + 5"NaCl"(aq) + 3"H"_2"O")`

Here, `"Cl"` got oxidized AND reduced, as before, from `0` to both `+5` and `-1`. So this is a plausible reaction.

Both of these are possible reactions. They are called disproportionation reactions.

So I'm guessing that you were given ahead of time that you are making `"NaClO"_3`. Otherwise, you have two possible reactions.