# 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.

##### 1 Answer
Feb 8, 2016

$2 {\text{Cl"_2 + 4"NaOH" -> 4"NaCl" + 2"H"_2"O" + "O}}_{2}$

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 $\text{NaCl}$ as we have $\text{NaOH}$, as we would have had it been an actual acid/base reaction.

(Perhaps you put ${\text{O}}_{2}$ 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 ${\text{O}}_{2}$), but in addition, chlorine went from a $0$ state (in ${\text{Cl}}_{2}$) 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.

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 ${\text{Cl}}_{2}$ could undergo. Only one is shown here.

${\text{Cl"_2(g) + 2e^(-) -> 2"Cl}}^{-}$ (you could have predicted this one)

But that doesn't utilize $\text{NaOH}$. Since we have examined ${\text{Cl}}_{2}$ reacting on its own, it makes sense to now consider reactions that have both ${\text{OH}}^{-}$ and some form of chlorine on the same side.

Two of the reactions in the reference table qualify, but since ${\text{Cl}}_{2}$ is a reactant, we should reverse these reactions to get an eventual cancellation of ${\text{Cl}}^{-}$ on the reactants side only:

$\text{Cl"^(-)(aq) + 2"OH"^(-)(aq) -> "ClO"^(-)(aq) + "H"_2"O} + 2 {e}^{-}$

$\text{Cl"^(-)(aq) + 6"OH"^(-)(aq) -> "ClO"_3^(-)(aq) + 3"H"_2"O} + 6 {e}^{-}$

As you can see, neither of these yield ${\text{O}}_{2}$. :)

DETERMINING THE FINAL REACTION

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

${\text{Cl"_2(g) + cancel(2e^(-)) -> 2"Cl}}^{-}$
$\text{Cl"^(-)(aq) + 2"OH"^(-)(aq) -> "ClO"^(-)(aq) + "H"_2"O} + \cancel{2 {e}^{-}}$
$\text{-------------------------------------------------}$
$\text{Cl"_2(g) + "Cl"^(-)(aq) + 2"OH"^(-)(aq) -> "ClO"^(-)(aq) + 2"Cl"^(-)(aq) + "H"_2"O}$

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

$\textcolor{b l u e}{\text{Cl"_2(g) + 2"NaOH"(aq) -> "NaClO"(aq) + "NaCl"(aq) + "H"_2"O}}$

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

$3 \left({\text{Cl"_2(g) + cancel(2e^(-)) -> 2"Cl}}^{-}\right)$
$\text{Cl"^(-)(aq) + 6"OH"^(-)(aq) -> "ClO"_3^(-)(aq) + 3"H"_2"O} + \cancel{6 {e}^{-}}$
$\text{-------------------------------------------------}$
$3 \text{Cl"_2(g) + 6"OH"^(-)(aq) -> "ClO"_3^(-)(aq) + 5"Cl"^(-)(aq) + 3"H"_2"O}$

Add back the spectator sodium:

$\textcolor{b l u e}{3 \text{Cl"_2(g) + 6"NaOH"(aq) -> "NaClO"_3(aq) + 5"NaCl"(aq) + 3"H"_2"O}}$

Here, $\text{Cl}$ 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 ${\text{NaClO}}_{3}$. Otherwise, you have two possible reactions.