Chemistry - Balancing disproportionation reactions

Solution 1:

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

Solution 2:

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 $(18)$.