How do I balance this redox reaction?

Question

The reaction given was $\ce{FNO3 -> O2 + F^- + NO3^-}$ (in basic medium)

I've noticed that the atoms which have their oxidation numbers changed are the oxygen and the nitrogen atom (as the oxidation number in fluorine in $\ce{FNO3}$ should be $-1$, right?)

I'm having troubles at dissecting it into half reactions:

$\ce{FNO3 -> O2 + F-}$ (This one doesn't have a nitrogen atom on the right hand side)

$\ce{FNO3 -> NO3- + F-}$

Answer 1

Unfortunately, the OP's starting reaction expression is simply wrong: the left hand side has only a neutral compound while the right hand side has anions among the product species. From the wikipedia article on fluorine nitrate, the unbalanced reaction of fluorine nitrate and water is

$$\ce{a FNO3 + b H2O -> c O2 + d OF2 + e HF + f HNO3 \tag 1}$$

For the same reason that this problem had infinitely many correctly balanced solutions, this one does as well. So here is the minimal coefficient solution.

Atom balance yields:

$$a=2d + e \tag 2$$ $$a=f \tag 3$$ $$3a + b = 2c + d + 3f \tag 4$$ $$2b = e + f \tag 5$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:

1. From equation $(2)$, $a \ge 3 $, since d and e cannot be zero. From equation $(3)$, $a = f$. From equations $(3)$ and $(4)$, $b = 2c + d$, so $b \ge 3 $, because c and d cannot be zero.

2. Assume $a = 3 = f$. If $e = 1$, then $d = 1$ and $b = 2$, which is impossible. If $e = 2$, then $d = 0.5$, which is impossible.

3. Assume $a = 4 = f$. If $e = 1$, then $d = 1.5$, which is impossible. If $e = 2$, then $d = 1$ and $2b = 2 + 4 = 6$, so $b = 3$. Then $c = 1$.

Almost final result: $a = 4, b = 3, c = 1, d = 1, e = 2,$ and $f = 4$. Hence

$$\ce{4FNO3 + 3H2O -> O2 + OF2 + 2HF + 4HNO3 \tag 6}$$

There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is $5a + 3b$. In the present case, $5a + 3b = 29$.

But what about the basic medium mentioned by the OP? Add 6 hydroxides to both sides and cancel 3 waters from both sides.

$$\ce{4FNO3 + 6OH- -> O2 + OF2 + 2F- + 4NO3- + 3H2O \tag 7}$$

Two other answers assume no oxygen difluoride, hence $d = 0$ in equation $(1)$. Then the atom balance equations reduce to $a = b = 2c = e = f$, so let $c = 1$ and the others are 2. Hence

$$\ce{2FNO3 + 2H2O -> O2 + 2HF + 2HNO3 \tag 8}$$

Then add 4 hydroxides to each side and cancel 2 waters on each side:

$$\ce{2FNO3 + 4OH- -> O2 + 2F- + 2NO3- + 2H2O \tag 9}$$

Answer 2

As noted elsewhere, the reaction as given is not balanced. Assuming that other products such as $\ce{OF2}$ are ignored, apply the following rules for basic solution:

1. Use hydroxide ion to allow for balanced charges.

2. Use water to allow balancing hydrogen and oxygen atoms.

Here, hydroxide ions should he added on the left by Rule 1 and then water is needed on the right to balance atoms by Rule 2. Thus, properly (prior to balancing):

$\ce{FNO3 + OH^- -> F^- + NO3^- + O2 + H2O}$.

The reduction half-reaction is difficult to render via atomic oxidation states because if we try to work out oxidation states in $\ce{FNO3}$, we find different oxygen atoms in different oxidation states. Use the entire $\ce{FNO3}$ molecule as the oxidizing agent:

$\ce{FNO3 + 2 e^- -> F^- + NO3^-}$.

The oxidation half-reaction involves just the water and hydroxide ions giving the familiar result

$\ce{4 OH ^- -> 2 H2O +O2 + 4 e^-}$.

The combination of these half-reactions to give the balanced full reaction is as it would be for any redox reaction.

Answer 3

I agreed with EdV about OP giving an errotic equation. Since $\ce{NO3-}$ remained unchanged in RHS of the equation, we need a compound to reduce $\ce{F+}$ to $\ce{F-}$ in this redox reaction. Since $\ce{O2}$ is being a product, we can choose one of the following half reactions to do the trick based on the medium (acidic or basic):

$$\ce{O2 + 4H+ + 4e- <=> 2H2O } \tag1$$ $$\ce{O2 + 2H2O + 4e- <=> 4OH- } \tag2$$

Thus, if the reaction is in acidic medium (equation $(1)$), we can correct the OP's equation as:

$$\ce{ FNO3 + H2O -> F- + O2 + H+ + NO3- } \tag3$$

Or, if it is in basic medium (equation $(2)$), we can correct the OP's equation as:

$$\ce{ FNO3 + OH- -> F- + O2 + H2O + NO3- } \tag4$$

If it's according to the equation $(3)$, we can write two corresponding half reactions:

$$\ce{ F+ <=> F- } \tag5$$ $$\ce{ H2O <=> O2 + H+ } \tag6$$

You can balance these two equations as follows:

• Balance center atom first.
• Then, balance any oxygen by $\ce{H2O}$ molecules (we have plenty of water in the medium).
• Adding water gives extra $\ce{H}$ atoms, which can be balanced by $\ce{H+}$ ions, because you are doing the reaction in acidic medium.
• Now, balance the all chargers by $\ce{e-}$s, which is relatively massless.
• If the reaction is in basic medium, neutralize the $\ce{H+}$ ions by available $\ce{OH-}$ ions and add same amount to other side of the equation to balance the mass because you already have balanced the equation.
• Finally, add oxidation and reduction half reactions in order to cancel the electrons.

Lets follow this order in equations $(5)$ and $(6)$:

$$\ce{ F+ + 2e+ <=> F- } \tag7$$ $$\ce{ 2H2O <=> O2 + 4H+ + 4e-} \tag8$$

Add the equations $(8)$ and $(5) \times 2$ to cancel electrons:

$$\ce{2F+ + 2H2O -> 2F- + O2 + 4H+} \tag9$$

Since $\ce{NO3-}$ is a spectator ion in this reaction, you can add that to both side as follows:

$$\ce{2F+ + 2NO3- + 2H2O -> 2F- + O2 + 4H+ + 2NO3-} $$

Or:

$$\ce{2FNO3 + 2H2O -> 2HF + O2 + 2HNO3} \tag{10}$$

The equation $(10)$ is the complete balanced redox reaction in acidic medium. EdV has given it in basic medium.

Answer 4

I have more succinctly arrived at Eq(9) as proposed by Ed V above, by noting observed reactions as reported in the literature.

For the record, I do not recommend working with FNO3, a shock-sensitive explosive. My expected action of FNO3, with say dry KOH, is better described simply by the term energetic. Working with dilute aqueous alkaline solutions is likely more feasible in practice.

Some supporting background, starting with Wikipedia on Flourine nitrate, to quote:

Fluorine nitrate is an unstable derivative of nitric acid with the formula $\ce{FNO3}$. It is shock-sensitive.[1] Due to its instability, it is often produced from chlorine nitrate as needed...It decomposes in water to form oxygen gas, oxygen difluoride, hydrofluoric acid, and nitric acid.[1]

So, an example of a possible water reaction could be given by:

$\ce{6 FNO3 + 4 H2O -> O2 + 2 OF2 + 2 HF + 6 HNO3}$

Upon further adding 8 OH- to both sides of the above suggests an alkaline reaction (but with no cancelling of water as of yet) :

$\ce{6 FNO3 + 4 H2O + 8 OH- -> O2 + 2 OF2 + 2 F- + 6 NO3- + 8 H2O}$

Further, any formed $\ce{OF2}$ only slowly dissolves in water per Wikipedia on oxygen difluoride. Also, per another source on the alkaline hydrolysis of OF2, it apparently proceeds, again not rapidly, as follows:

$\ce{OF2 (aq) + 2 OH- (aq) -> O2 (g) + 2 F- (aq) + H2O}$

So, upon adding the above intermediate reaction times 2 to my second cited equation above implies a net reaction (assuming all the oxygen difluoride is eventually removed from the system) of:

$\ce{6 FNO3 + 4 H2O + 12 OH- -> 3 O2 + 6 F- + 6 NO3- + 10 H2O}$

Or, upon canceling out like terms and dividing by 6:

$\ce{FNO3 + 2 OH- -> 1/2 O2 + F- + NO3- + H2O}$

Those anticipating only a friendly oxygen gas, should be warned, again per the Wikipedia source, to quote:

Oxygen difluoride is considered an unsafe gas due to its oxidizing properties.

So, with sufficient time elapse, the cited REDOX per Ed V is likely observed.