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

1 Answer
Feb 8, 2016

#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.


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.