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I have seen similar questions, but I’m asking a new question as far as I know. Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations? asks about using hydroxide instead of water and that is closest.

I have seen example problems where $\ce{H+}$ is produced or consumed, $\ce{OH-}$ is produced or consumed, and water is used to balance. My thought is that each of these changes affects the $\mathrm{pH}$ or $\mathrm{pOH}$ of each solution.

If the solution is already an acid, there is an abundance of $\ce{H+}$ and it makes sense to me that a reaction may use excess $\ce{H+}$ to make the solution more neutral. One could make the same argument about using up excess $\ce{OH-}$ in a basic solution. What doesn’t make sense to me is adding to the basic or acidic nature. See example below which was specified to be in a basic solution. It appears to be making the solution more basic by producing more $\ce{OH-}$ instead of using it up. Shouldn’t that be less likely to occur? Why not? What am I missing?

$$\ce{6I- (aq) + 4H2O + 2MnO4- (aq) —> 3I2 + 2MnO2 (aq) + 8OH- (aq) }$$

enter image description here

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  • $\begingroup$ Just because it is catalyzed by a low pOH doesn't mean that the reaction makes the solution more acidic. Often, the pH or pOH of the solution only catalyze the reaction because they favor the reaction mechanism, not the products. $\endgroup$ Apr 5 '20 at 18:02
  • $\begingroup$ This reaction cannot occur in solution, because it would produce both $I_2$ and $OH^-$. Unfortunately $I_2$ and $OH^-$ cannot exist simultaneously in solution : they would react and produce a mixture of iodide $I^-$ and iodate $IO_3^-$ ions, according to the equation : $$\ce{3 I_2 + 6 OH^- -> 5 I^- + IO_3^- + 3 H_2O}$$ $\endgroup$
    – Maurice
    Apr 5 '20 at 20:01
  • $\begingroup$ @Maurice that came directly from the textbook. I believe you, but that’s why I’m frustrated. Makes no sense to me half the time. $\endgroup$
    – BSD
    Apr 5 '20 at 21:16
  • $\begingroup$ It may happen that textbooks are wrong. They just propose exercices to see whether the student has understood the theory. It is not the first time that it happens. I have once seen a problem where the student had to calculate the pH of a solution 20 M in H2SO4. $\endgroup$
    – Maurice
    Apr 6 '20 at 9:54
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What you are missing is understanding of general nature of redox reactions. Redox reactions need two half reactions to complete. You can't arbitrarily select these two half reactions. They has to be chosen as instructions given in the problem. For example, for your given problem, it should be noted the medium of the reaction, whether it is acidic or basic or neutral.

Suppose the question asked is: Balance the following redox equation in acidic medium. $$\ce{I- (aq) + MnO4- (aq) -> MnO2 (s) + I2 (aq)}$$

To solve this, you can directly get two half reactions from a textbook or balance it old-fashioned. The oxidation half reaction is: $$\ce{2I- <=> I2 }$$ It is already balanced by mass, so add electrons (no mass) to balance the negative charge: $$\ce{2I- <=> I2 + 2e-}$$ Now oxidation half reaction is completed. The reduction half reaction is: $$\ce{MnO4- <=> MnO2 }$$ However, it was stated that reaction is in acid medium. Therefore we should add acid to reactants' side: $$\ce{MnO4- + H+ <=> MnO2 }$$ Now, we have to balance mass first. Since the reaction is in acidic aqueous medium we have plenty of $\ce{H2O}$ molecules around to balance oxigen as well as hydrogen molecules (by $\ce{H+}$ ions as needed): $$\ce{MnO4- + 4H+ <=> MnO2 + 2H2O }$$ Now equation is balanced by mass. Doing so, we increase net positive charges in left hand side by 3, so we should neutralize it by adding 3 electrons to left hand side because right hand side is neutral: $$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O }$$ Now reduction half reaction is also completed. To achieve balanced redox reaction, simply add balanced oxidation and reduction half reactions in order to cancel unwanted electrons: $$\ce{2MnO4- + 8H+ + 6I- -> 2MnO2 + 3I2 + 4H2O }$$

This redox reation is forward reaction because it has a net positive potential (refer reduction potentials of two half reactions). The major point is the reaction starts in acidic medium and stay acidic at the end. Keep in mind that if you have used very strong acidic conditions, $\ce{MnO2}$ would be further reduced to $\ce{Mn^2+}$ according to the half reaction: $\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O }$. But oxidation half reaction would stay unaffected, yet final redox reaction would be: $$\ce{2MnO4- + 16H+ + 10I- -> 2Mn^2+ + 5I2 + 8H2O }$$

Main point is changing the conditions most certainly change the reaction. We cannot arbitrarily choose it. For example, to balance the given equation, we have to use what's available for us. We cannot use what is not available or would be available in near future as a product (e.g., as $\ce{OH-}$ in the example in oxidation reaction in neutral medium you are considering).

For the shake of the argument, let's consider that reaction:

Suppose the question asked is: Balance the following redox equation in neutral medium. $$\ce{I- (aq) + MnO4- (aq) -> MnO2 (s) + I2 (aq)}$$

Again, to solve this problem, you can directly get two half reactions from a textbook or balance it old-fashioned as described above. The oxidation half reaction does not changed by the conditions and following is the completed equation: $$\ce{2I- <=> I2 + 2e-}$$ The reduction half reaction is: $$\ce{MnO4- <=> MnO2 }$$ However, it was stated that reaction is in a neutral medium (the product $\ce{MnO2}$ is possible both in acidic or neutral medium). Therefore we cannot add acid or base to reactants' side but water in either side. Therefore, we can balance loss of oxygen in product side by using water because it is in aqueous medium and we have plenty of $\ce{H2O}$ molecules around: $$\ce{MnO4- <=> MnO2 + 2H2O}$$ Now we have extra hydrogen in product side, so we don't have a choice but balance that by $\ce{H+}$ in reactant side for now: $$\ce{MnO4- + 4 H+ <=> MnO2 + 2H2O}$$ It is fact that acid-base neutralization reaction produces $\ce{H2O}$ predominantly. Thus, we can add $\ce{OH-}$ to both sides of the reaction so that it can keep the mass balance of the reaction. Since the reaction is in aqueous medium this won't affect the reaction: $$\ce{MnO4- + 4H+ + 4OH- <=> MnO2 + 2H2O + 4OH-}$$ $$\ce{MnO4- + 2H2O <=> MnO2 + 4OH-}$$ Now the equation is balanced by mass. However, we have increased net negative charges in right hand side by 3, so we should neutralize it by adding 3 electrons to left hand side to cancel the charges: $$\ce{MnO4- + 2H2O + 3e- <=> MnO2 + 4OH-} \tag{1}$$ Now reduction half reaction is also completed. To achieve balanced redox reaction, simply add balanced oxidation and reduction half reactions in order to cancel unwanted electrons: $$\ce{2MnO4- + 4H2O + 6I- -> 2MnO2 + 3I2 + 8OH- }$$

The major point is this reaction starts in neutral medium, but become basic at the end. Again, we didn't have choice to keep it neutral because that't the way this happens in nature. Keep in mind that we did not add anything unavailable during balancing the equation. If you check literature for reduction half reaction of $\ce{MnO4-}$ in neutral medium, you would get nothing but equation $(1)$ as we derived.

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  • $\begingroup$ Okay. I’m just making notes. You said “The major point is the reaction starts in acidic medium and stay acidic at the end.” Maybe but if we reduce the pH by 8 moles of H+, aren’t we at least changing it? It seems that by adding these ions I would be changing the pH. $\endgroup$
    – BSD
    Apr 5 '20 at 21:29
  • $\begingroup$ “Now we have extra hydrogen in product side, so we don't have a choice but balance that by H+ in reactant side“ okay, that makes sense. Then you add OH on the reactant side because the solution is neutral. I can see that. The 4H+ had to come from somewhere leaving the OH- behind. $\endgroup$
    – BSD
    Apr 5 '20 at 21:32
  • $\begingroup$ Alright I’ve read it all closely, but I still don’t understand why the pH should change. “The major point is this reaction starts in neutral medium, but become basic at the end.” If I add these ions, I wouldn’t expect any of them to alter the pH. I is not a base. I suppose MnO4- is a conjugate base, but I would expect to see the extra H+ in that case. It does not make sense to me for anything but a base to make the solution more basic. Please explain. $\endgroup$
    – BSD
    Apr 5 '20 at 21:39
  • $\begingroup$ Comment 1: Remember, adding $\ce{H+}$ to equation doesn't mean you are adding to solution. Solution is acidic already by the using conditions, we are just using the exsisting condition to balance the equation. $\endgroup$ Apr 5 '20 at 22:59
  • $\begingroup$ This is true for the remaining comments. For last one, $\mathrm{pH}$ increases as reaction progress. $\endgroup$ Apr 5 '20 at 23:02
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If the reaction happens at neutral or mildly alkalic $\mathrm{pH}$:

$$\ce{6I- + 4 H2O + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 8 OH-}$$

If the reaction happens at mildly aciding $\mathrm{pH}$, it could be as well:

$$\ce{6I- + 8 H+ + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 4 H2O }$$

In any case, either spending $\ce{H+}$ either production $\ce{OH-}$ increases $\mathrm{pH}$. Both iodide and permanganate carry a negative charge while their spent forms are neutral. This negative charge must be passed to other anions like $\ce{OH-}$ or neutralized by $\ce{H+}$.

If one considers further disproportionation of iodine to iodide and iodate in alkalic solutions, then the total reaction is:

$$\ce{ I- + H2O + 2 MnO4- -> IO3- + 2 MnO2(s) + 2 OH-}$$

In strongly alkalic it could be also $$\ce{ I- + 6 OH- + 6 MnO4- -> IO3- + 6 MnO4^2- + 3 H2O }$$

It may be obvious if the equation is written together with spectator ions:

Mildly acidic: $$\ce{6KI + 8 HA + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 4 H2O + 8 KA}$$ resp neutral/mildly alkalic $$\ce{6KI + 4 H2O + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 8 KOH}$$ resp alkalic $$\ce{ KI + H2O + 2 KMnO4 -> KIO3 + 2 MnO2(s) + 2 KOH}$$ resp. strongly alkalic $$\ce{ KI + 6 KOH + 6 KMnO4 -> KIO3 + 6 K2MnO4 + 3 H2O }$$

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  • $\begingroup$ Okay. That does help, but KOH is still a base and would still affect the pH. In other words, I add KI and KMnO4 which are both neutral and I suddenly arrive at a base. Why? $\endgroup$
    – BSD
    Apr 5 '20 at 21:43
  • $\begingroup$ Thanks for editing the notation. I’m not familiar with it but I’ll try to figure it out. $\endgroup$
    – BSD
    Apr 5 '20 at 21:44
  • $\begingroup$ Because all anions of acids are removed and nothing but OH- can carry negative charge. Do you see other option ? If suddenly OH- is created, pH suddenly increases. $\endgroup$
    – Poutnik
    Apr 6 '20 at 5:53
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For the reaction:

$$\ce{6I- (aq) + 4 H2O + 2 MnO4- (aq) ->[excess I-] 3 I3- + 2 MnO2 (s) + 8OH-}$$

what you are missing is the Nernst Equation. The two reduction half cell reactions are:

$$\ce{I3− + 2e- <=> 3I− \quad\quad EMF = +0.53}$$

$$\ce{MnO4− + 2H2O + 3e− <=> MnO2(s) + 4 OH− \quad\quad EMF =+0.595}$$

The gist of the Nernst Equation is that the voltages change as the concentration of the ionic species change. The standard EMF is for 1 molar concentration of all ionic species.

For the $\ce{I3-/I-}$ half cell there is no pH consideration in that HI is a strong acid.

For the permanganate reduction:

  • As the permanganate is reduced and the hydroxide builds up, the Nernst equation shows that the EMF of the half cell drops towards zero.

  • If the initial permanganate was 1 molar and the hydroxide was 0.1 molar, then the EMF would be even more positive than 0.595 V.


As a bit of a side note $\ce{I2}$ has a very limited solubility in solution, hence an excess of $\ce{I-}$ is typically used so that the soluble $\ce{I3-}$ complex is formed.

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