# Determining the redox half-reactions of single displacement reactions

For the following double displacement half reaction, the half reactions can be determined easily using the oxidation numbers. $$\ce{MnO4- {(aq)} + IO3- {(aq)} -> MnO2 {(s)} + IO4- {(aq)}}$$ We know this is the reduction half reaction because the oxidation number of $\ce{Mn}$ changes from +7 to +2:

$$\ce{MnO4- {(aq)} -> MnO2 {(s)}}$$ and we know this is the oxidation half reaction because the oxidation number of $\ce{I}$ changes from +5 to +7: $$\ce{IO3- {(aq)} -> IO4- {(aq)}}$$

However, for the following reactions, this strategy cannot be used. So how do we determine the half reaction equations for the following reactions:

1. $\ce{Mn(OH)2 {(s)} + H2O2 {(aq)} ->T[Base] Mn(OH)3 {(s)}}$
2. $\phantom{iiiiiiiiiiiiiiiii}\ce{MnO4^2- {(aq)} ->[Acid] MnO2 {(s)} + MnO4- {(aq)}}$
3. $\phantom{iiiiiiiii}\ce{IO3- {(aq)} + I- {(aq)} ->T[Acid] I3 {(aq)}}$
4. $\phantom{iiiiiiiiii}\ce{P {(s)} + PO3^3- {(aq)} ->T[Base] HPO3^2- {(aq)}}$

Let's take the first equation in the question by example: $$\ce{Mn(OH)2 (s) + H2O2 (aq) -> Mn(OH)3- (s)}$$ In this equation ion $\ce{Mn(II)}$ is oxidised into $\ce{Mn(III)}$, and oxygen in $\ce{H2O2}$ (oxidation number -1) is reduced into ion hydroxide (oxidation number -2): \begin{align} \ce{Mn^{2+} &-> Mn^{3+} +e^- \tag 1}\\ \ce{H_2O2 (aq) + 2e^- &-> 2OH^- \tag 2} \end{align} If we multiply the first equation by 2 and add the equations $(1)$ and $(2)$: $$\phantom{iiii}\ce{2Mn^{2+} + H_2O2 (aq) -> 2 Mn^{3+} + 2OH^- \tag 3}$$ By adding four ions hydroxide to the two sides of equation $(3)$: $$\ce{2Mn^{2+} + H_2O2 (aq) +4OH^- -> Mn^{3+} + 6OH^- }\phantom{iiiiiiiiii}$$ By rearranging the equation $(3)$, we find: $$\ce{2Mn(OH)2 (s) + H2O2 (aq) -> 2 Mn(OH)3 (s) \tag 4}\phantom{iiiiiii}$$ We do the same for the other equations.
• For the first one, I see the question about $\ce{H_2O_2}$ acting weird in redox-reactions here and on other forums quite often. It can do both, reduce and oxidize. And there is a disproportionation of $\ce{H_2O_2}$ reacting with itself which is catalyzed by things like $\ce{MnO_2}$ or $\ce{KI}$. So in these cases, it's all about the redox-partner or catalyst and the pH-environment.
• For the next one, this is a classic disproportionation reaction. In your case, this depends on the situation. Of course, if permanganate was given and the note that acid is added your teacher may also ask you for the formation of $\ce{Mn_2O_7}$. If the topic of the lecture or exam is redox-reactions however and no other partner is given, and the acid is written on the arrow instead of a redox-partner, then you could think about a disproportionation. To predict these at home, so if you have access to literature or the internet, the so-called 'Frost-Ebsworth'-diagram is of help. You can check them on the net and learn how to draw them. I use them all of the time when I think of the possible site reactions my redox-reactions could do. It helps you to decide whether a species in solution is unstable and will do disproportionation or not. If you draw one for the correct pH-environment here you can predict it and it will also come up with the answer given in your example.
• The same should be true for the last equation, also I didn't try that on my own yet. I know that $\ce{P_4}$ will react with water and does disproportionate under basic conditions to phosphane and some oxidized species.
• For the third one, you'd have the oxidized iodine and the reduced iodine synproportionate to form elemental iodine, which, in the presence of iodide and in water form the brown solution of triiodide. At least this would seem logical if there was a negative charge in your equation. Unless there is another molecular $\ce{I_3}$ species I haven't heard about yet (if someone has info on that I'd be interested).