# Balancing a redox reaction equation with a mystery reactant

The answer to the following question from NSAA 2021 Chemistry Section 2 is C.

Consider the following chemical equation: $$\ce{vQ + wP4 + x H2O -> yPH4I + zH3PO4}$$ where Q is a binary compound. The molecules of Q are hexatomic and contain phosphorus in the +2 oxidation state. Using the lowest integer values for all the coefficients $$v, w, x, y$$ and $$z$$, what is the value of $$w$$ when the equation is balanced?

A. 1

B. 2

C. 13

D. 16

E. 24

F. 26

I assumed that the unknown compound Q would be composed of P and I, and more specifically to be $$\ce{P2I4}$$ as it is stated that the compound is hexatomic and the oxidation state of P is +2. However, I can not find a way to balance the equation. Does anyone have a solution for this?

Well, $$\ce{P2I4}$$ has to be reduced to form $$\ce{PH4I}$$, and $$\ce{P4}$$ must undergo disproportionation. Otherwise, you don't have enough phosphorus to balance the amount of iodine in forming $$\ce{PH4I}$$ where the phosphorus has an oxidation state of -3. So you can write three different equations:

$$\ce{P2I4 + 8H+ + 10e- -> 2PH4+ + 4I-}$$ (eq.1)

$$\ce{P4 + 16H2O -> 4H3PO4 + 20H+ + 20e-}$$ (eq.2)

$$\ce{P4 + 8I- -> 2P2I4 + 8e-}$$ (eq.3)

Let's combine eq.1 and eq.3 to cancel out the $$\ce{I-}$$ ions. Note: $$\ce{2PH4+ + 4I- \equiv 2PH4I + 2I-}$$:

$$\ce{4P2I4 + 32H+ + 40e- + P4 + 8I- -> 8PH4I + 8I- + 2P2I4 + 8e-}$$

$$\ce{4P2I4 + 32H+ + 32e- + P4 -> 8PH4I + 2P2I4}$$

$$\ce{2P2I4 + 32H+ + 32e- + P4 -> 8PH4I}$$ (eq.4)

Now we need to combine this eq.4 with the eq.2 to get rid of the $$\ce{H+}$$ in the equations, which should also cancel out the electron terms. You can simply multiply eq.4 by a factor of 5 and eq.2 by a factor of 8 and then combine. This would result in:

$$\ce{10P2I4 + 160H+ + 160e- + 5P4 + 8P4 + 128H2O -> 40PH3I + 32H3PO4 + 160H+ + 160e-}$$

Or finally the fully balanced equation:

$$\ce{10P2I4 + 13P4 + 128H2O -> 40PH4I + 32H3PO4}$$

So this is where you get the 13 from.

• I don't have 4 disproportionation reactions of P4. Only eq.1,2,3 are relevant. I just combine them in a systematic way to get the final balanced equation which can be verified.
– M.L
Sep 13, 2022 at 6:36

Once we accept that the mystery reactant is $$\ce{P2I4}$$, this is the equation we are asked to balance (this might or might not be a fantasy reaction - the question does not teach you much about chemistry):

$$\ce{vP2I4 + w P4 + x H2O -> y PH4I + z H3PO4}$$

This is a problem with 5 unknowns. You can come up with 5 constraints (the number of $$\ce{P, I, H, O}$$ atoms on either side has to be equal, and the coefficients should be the smallest non-zero integers). So unless any of these conditions are redundant, there should be a single solution.

I will show a simple way of balancing this without using half-reactions (even though this is a redox reaction). Then, I will discuss the complexity of trying to use half-reactions.

### Using the atom balance to find the coefficients

Notice that all species are neutral, so we have an automatic charge balance. I will start with an atom present in only two species, oxygen, to eliminate the variable $$x$$, then proceed to hydrogen, iodine, and finally phosphorus:

$$\ce{vP2I4 + wP4 + 4zH2O -> y PH4I + z H3PO4}\tag{O}$$

$$\ce{vP2I4 + wP4 + 4zH2O -> 5/4z PH4I + z H3PO4}\tag{H}$$

At this stage, I will multiply by 4 to remove the fraction:

$$\ce{4vP2I4 + 4w P4 + 16z H2O -> 5z PH4I + 4z H3PO4}$$

$$\ce{5/4z P2I4 + 4w P4 + 16z H2O -> 5z PH4I + 4z H3PO4}\tag{I}$$

At this stage, I will multiply by 4 to remove the fraction:

$$\ce{5z P2I4 + 16w P4 + 64z H2O -> 20z PH4I + 16z H3PO4}$$

$$\ce{5z P2I4 + 13/2z P4 + 64z H2O -> 20z PH4I + 16z H3PO4}\tag{P}$$

At this stage, I will set $$z$$ to 2 to get the lowest integers:

$$\ce{10 P2I4 + 13 P4 + 128 H2O -> 40 PH4I + 32 H3PO4}\tag{done}$$

### Less formal way of writing

If you solve this by inspection, you would leave out the unknowns and just write the coefficients in the correct ratios:

$$\ce{4 H2O -> H3PO4}\tag{O}$$

$$\ce{16 H2O -> 5 PH4I + 4 H3PO4}\tag{H}$$

$$\ce{5 P2I4 + 64 H2O -> 20 PH4I + 16 H3PO4}\tag{I}$$

$$\ce{10 P2I4 + 13 P4 + 128 H2O -> 40 PH4I + 32 H3PO4}\tag{P}$$

### Half-reaction method

The only atoms that have distinct oxidation states are the phosphorus atoms; there are four species, each one with a different oxidation state. The two reactants could either be oxidized to the +5 state (in $$\ce{H3PO4}$$) or reduced to the -3 state (in $$\ce{PH4I}$$). Knowing the coefficients of the balanced equation, we can make arbitrary decisions about how to write the half-reactions.

If we want to reduce all of the $$\ce{P2I4}$$, we get the following scheme:

The half-reactions would be: $$\ce{10 P2I4 +100 e- + 80 H+ -> 20 PH4I + 20 I- }\tag{red}$$ $$\ce{13 P4 + 20 I- + 128 H2O -> 20 PH4I + 32 H3PO4 + 80 H+ + 100 e-}\tag{ox}$$

If we want to oxidize all of the $$\ce{P2I4}$$, we get the following scheme:

I will leave it as an exercise to figure the half-reaction for this scenario. Or you can choose anything in between these two schemes, for example:

Now, both half-reactions contain both reactant species.

So you have non-unique half-reactions, but a unique solution to the combined reaction. To get from the half-reactions, you would have to combine them in a way that no ions are "left over". This makes the half-reaction scheme complicated and confusing in this case.

### Redundant conditions

This problem is different than being asked to balance a reaction like this: $$\ce{H2(g) + O2(g) + NaCl(s) -> H2O(l) + NaCl(aq)}$$

Like the original problem, we have four different types of atoms and five coefficients. However, the atom balance for sodium and for chlorine counts as only one constraint because they are redundant. In this case, the equation can be separated into:

$$\ce{H2(g) + O2(g) -> H2O(l) }$$ $$\ce{NaCl(s) -> NaCl(aq) }$$

and there is no unique way to balance the sum of two independent reactions. For the reaction given by the OP, balancing the reaction has a unique solution, but writing down the half-reactions doesn't. This is unusual and probably disturbing to some.

• What is unusual is using a chemical reaction to generate phosphine to react with the HI, phosphine could have been added independently. Sep 14, 2022 at 22:34
• @jimchmst I'm not sure that whoever wrote the problem knows about the chemistry behind it. I certainly don't; I took the chemical equation as correct (no intermediates that accumulate etc) and balanced it.
– Karsten
Sep 14, 2022 at 23:24
• @jimchmst You still have access to the post, and you are encouraged to edit your original answer, merging the two post to communicate your insight. Your second post wasn't an answer, it was a comment on the first post. At least that is what it looked like, so it was deleted.
– Karsten
Sep 15, 2022 at 21:15
• Thank you I am learning the protocols, slowly it seems! I rewrote my post Please read it. Sep 15, 2022 at 21:43

This involves two independent reactions that, if added properly, will result in a neutralization of the hydriodic acid formed in the first reaction with phosphine formed in the second by conversion [on paper at least] to phosphonium iodide. I think that this is a much simpler approach and that it makes more chemical sense. Whether it actually happens??

The first equation is the actual redox reaction: this is balanced by choosing one as the oxidant and the other as the reductant [it makes no difference which is which], writing the half-reactions and balancing. [it is easier to have phosphorus as P then convert to P4]

$$\ce{16 H2O + P4 + 2 P2I4}$$ = $$\ce{4 H3PO4 + 4 PH4I +4 HI}$$

The second equation is the disproportionation of P4: where P is the oxidant and the reductant; simply write a half-reaction for each and combine.

$$\ce{12 H2O + 2 P4}$$ = $$\ce{3 H3PO4 + 5 PH3}$$

To react the phosphine with the HI the two reactions must be kinetically similar or somehow be mechanistically combined. But to neutralize the HI with the phosphine the first reaction must be multiplied by 5 to give 20 HI and the second multiplied by 4 to give 20 PH3; they react to give 20 PH4I. Multiply, add, and collect:

$$\ce{128 H2O +13 P4 + 10 P2I4}$$ = $$\ce{32 H3PO4 + 40 P4I}$$

Mechanistic studies are necessary to determine what actually happens. I think that this is an improper use of a chemical equation because the two processes are not [or most likely not] mechanistically linked. The same result can be obtained by independently adding phosphine and if needed phosphoric acid.

As an added thought, these reactions, if actually possible in some form, destroy the possibility that the detection of phosphine in Venus's atmosphere MUST be linked to life.

• I must apologize the exact relationship furnishes the proper amount of Phosphine to react with the HI from the redox reaction. Sep 13, 2022 at 7:02
• You can always edit your answer. When you edit it, you might want to format the equations using mhchem.
– Karsten
Sep 13, 2022 at 13:48