# Balancing disproportionation reactions

How to balance the following disproportionation reaction $$\ce {XeF2 + H2O → Xe + XeO3 + HF + O2}$$ My attempt: I tried by making total number of reductions equal to total number of oxidation but I couldn't do because Xe is both oxidised and reduced also the solvent is water so what should I do to balance protons and oxygen on both sides?. Please help by solving it on the basis of same procedure that is making oxidation equal to reduction

• since acid is in the product side balancing should be done as in acidic medium
– JM97
Feb 17 '16 at 6:49

It can't be balanced unambiguously, for this is not a single reaction, but a mechanical sum of two reactions, and these can be formally mixed in arbitrary proportion.

The first reaction is indeed a disproportionation. To balance it, use the common trick: consider these two xenons different elements for a while. $$\ce{\underbrace{XeF2}_{for\;reduction} + \underbrace{XeF2}_{for\;oxidation} +H2O->Xe + XeO3 + HF}$$ The second is a typical redox: $$\ce{XeF2 + H2O -> Xe + O2 + HF}$$

• Nertin how do you know that it is a sum of two reactions? Could you please suggest any good textbook for it
– JM97
Feb 17 '16 at 8:25
• This is a sum of two reactions because it has two independent oxidized products: $\ce{XeO3}$ and $\ce{O2}$. As for the book, I don't think there is one; the matter is too trivial to deserve a whole book on its own. Feb 17 '16 at 8:32
• Nertin how did you figure out what reactants and products would be in each reaction?
– JM97
Feb 17 '16 at 8:45
• Why, the OP specified them all in the original question. I just figured that the oxidized products are independent, so there would be only one of them in each reaction. Feb 17 '16 at 9:15

The original equation cannot be solved using mathematics because it gives the coefficient of $$\ce{XeO3}$$ zero. The reason for that is exact reason given by Ivan Neretin (see else where). Yet, I'd like to show a way of solving a balancing problem of true chemical equation by using mathematics. First, assume the compound is $$\ce{XeF4}$$, which undergoes disproportion in water. The chemical equation for that disproportion can be written as:

$$\ce{a XeF4 + b H2O -> c Xe + d XeO3 + e HF + f O2 \tag 1}$$

So the atom balance of equation $$(1)$$ yields:

$$\text{Balancing } \ce{Xe}: \quad a = c + d \tag 2$$ $$\text{Balancing } \ce{F}: \quad 4a = e \tag 3$$ $$\text{Balancing } \ce{H}: \quad 2b = e \tag 4$$ $$\text{Balancing } \ce{O}: \quad b = 3d + 2f \tag 5$$

Now you have four equations relate to the six unknown coefficients $$a$$ through $$f$$, each coefficient is in general, an integer greater than zero. Yet, keep in mind that, a fraction of coefficients are allowed in chemistry if at least one coefficient is an integral. Since there are only 4 equations for 6 variables, we can give an arbitrary value for at least one variable. Since $$\ce{H}$$ and $$\ce{F}$$ are appeared in only one compound in either side, we can give our arbitrary value to either $$b$$ or $$e$$, respectively. I'd prefer to give $$e = 4$$ for the sake of argument. Hence, we can proceed as follows:

Assuming $$e = 4$$ gives $$a = 1$$ from the equation $$(3)$$ and $$b = 2$$ from the equation $$(4)$$. However this makes mathematics complicated because in right hand side of equation $$(1)$$, there are two $$\ce{Xe}$$ atoms and that makes $$c$$ and $$d$$ fractions (since $$a = c + d$$). Thus, we assume $$e = 8$$, doubling the original assumption. Then, from equation $$(3)$$, $$a = \frac14e =\frac14 \times 8 = 2$$. Similarly, from equation $$(4)$$, $$b = \frac12e =\frac12 \times 8 = 4$$.

Applying $$a = 2$$ and $$b = 4$$ in equations $$(1)$$ and $$(5)$$ respectively give:

$$2 = c + d \tag 6$$ $$4 = 3d + 2f \tag 7$$

By subtracting the equation $$(7)$$ from the equation $$(6) \times 3$$ gives you:

$$4 = 3c - 2f = 2 \tag 8$$

The smallest values to satisfy the equation $$(8)$$ are $$c = 1$$ and $$f = \frac12$$, which is allowed since other coefficients are integers. From the equation $$(2)$$, $$d = a - c = 2 - 1 = 1$$. From the equation $$(4)$$, $$e = 2b = 8$$. Hence, all numerical values of coefficients are: $$a = 2$$, $$b = 4$$, $$c = 1$$, $$d = 1$$, $$e = 8$$, and $$f = \frac12$$. Therefore, the equation would be:

$$\ce{2 XeF4 + 4 H2O -> 1 Xe + 1 XeO3 + 8 HF + 1/2 O2 \tag 9}$$

If you multiply whole equation $$(9)$$ by 2, it becomes:

$$\ce{4 XeF4 + 8 H2O -> 2 Xe + 2 XeO3 + 16 HF + 1 O2} \tag {10}$$

The equation $$(10)$$ is the correct balance equation, using mathematics.

The given equation in the question is not true redox equation. Yet, it is a combination of two redox reactions, one of which is a disproportionation reaction (of $$\ce{XeF2}$$) while the other is regular oxidation-reduction of two compounds $$\ce{XeF2}$$ and $$\ce{H2O}$$. This complex mixture can be solved by step-by-step analysis:

As Ivan Neretin correctly pointed out, the given reaction is sum of two reactions: The first reaction is redox reaction where one molecule of $$\ce{XeF2}$$ is oxidized while another $$\ce{XeF2}$$ molecule is reduced. Let's look at oxidation first:

$$\ce{XeF2 <=> XeO3 + 2 F-}$$

Balance oxygen by $$\ce{H2O}$$, hydrogen by $$\ce{H+}$$, and charges by $$\ce{e-}$$:

$$\ce{XeF2 + 3H2O <=> XeO3 + 2 F- + 6H+ + 4e-} \tag{11}$$

The reduction of $$\ce{XeF2}$$ is:

$$\ce{XeF2 <=> Xe + 2 F-}$$

Just balance charges by $$\ce{e-}$$:

$$\ce{XeF2 + 2 e- <=> Xe + 2 F- } \tag{12}$$

Now add the equations $$(11)$$ and $$(12)$$ in order to cancel $$e^-$$s ($$(11) + (12) \times 2$$):

$$\ce{3XeF2 + 3H2O -> XeO3 + 2 Xe + 6 F- + 6H+} \tag{13}$$

or

$$\ce{3XeF2 + 3H2O -> XeO3 + 2 Xe + 6 HF} \tag{14}$$

This is the first redox reaction, now balanced. The second redox reaction, which is correctly given by Ivan Neretin:

$$\ce{XeF2 + H2O -> Xe + O2 + HF} \tag{15}$$

To balance this equation, follow above methodology:

$$\text{[Ox]}: \quad \ce{2H2O <=> O2 + 4H+ + 4e-}\tag{16}$$ $$\text{[Red]}: \quad \ce{XeF2 + 2e- <=> Xe + 2F- }\tag{12}$$

Add the equations $$(12)$$ and $$(16)$$ in order to cancel $$e^-$$s ($$(16) + (12) \times 2$$):

$$\ce{2XeF2 + 2H2O -> 2Xe + O2 + 4H+ + 4F-} \tag{17}$$ or $$\ce{2XeF2 + 2H2O -> 2Xe + O2 + 4HF} \tag{18}$$

Now, these two equations, $$(14)$$ and $$(18)$$ can be add together in any proportion to get the sought equation of OP's. For example, add one to one proportion gives:

$$\ce{5XeF2 + 5H2O -> XeO3 + 4 Xe + O2 + 10 HF} \tag{19}$$

This is one of balanced equation in the equation given in OP's question. There are many of them can be written by different combinations of equations $$(14)$$ and $$(18)$$.

• Excellent! The circle is now complete! ;)
– Ed V
May 31 '20 at 2:51