# What does it mean to balance a redox reaction in acidic or basic mediums? [closed]

Until now it was clear to me what balancing a reaction was. Balancing redox reaction also make sense to me. What does it mean to balance a redox reaction in acidic or basic medium?

• Well, some reactions may turn out different ways depending on the medium. Nov 28, 2019 at 16:14
• I think you are confused by the word "medium." Most of the reactions are carried out in the liquid phase. When you are able to adjust the pH of a liquid phase, and redox reaction occurs, one can talk about an "acidic or basic" medium. Nov 28, 2019 at 16:55

In acidic medium the equation must mention somewhere that some $$\ce{H+}$$ ions appear somewhere in the equation. In basic medium, the equation must mention somewhere that some $$\ce{OH-}$$ ions appear somewhere in the equation,

For example, you may say that permanganate ion reacts in acidic conditions to produce $$\ce{Mn^2+}$$. In basic conditions, it could not produce $$\ce{Mn^2+}$$, because this ion will react with $$\ce{OH-}$$ ions to produce a precipitate of $$\ce{Mn(OH)2}$$.

As an application, the permanganate ion reacts with oxalic acid $$\ce{H2C2O4}$$ in acidic conditions according to :

$$\ce{2 MnO4- + 5 H2C2O4 + 6 H+ -> 2 Mn^2+ + 10 CO2 + 8 H2O}$$

Here, one sees that 6 ions $$\ce{H+}$$ are necessary for the reaction to happen. If no $$\ce{H+}$$ are present in the mixture permanganate + oxalic acid, this reaction cannot take place. Another reaction is produced without $$\ce{H+}$$. If you mix permanganate ion with oxalic acid in a basic solution, the first reaction that happen is the neutralization of the acid by the base namely :

$$\ce{H2C2O4 + 2 OH- -> C2O4^2- + 2 H2O}$$

So, here, it is not oxalic acid that reacts with permanganate, but the oxalate ion $$\ce{C2O4^2-}$$. The reaction will be :

$$\ce{2 MnO4- + 3 C2O4^2- + 4 OH- -> 2 MnO2 + 6 CO3^2- + 2 H2O}$$

Here you see that some $$\ce{OH-}$$ are needed for the reaction to proceed. It cannot happen in acidic medium.

• Protons or hydroxide ions must appear as reagents, if the solution is acidic or basic; not just somewhere. Nov 28, 2019 at 19:12

Say we have a given redox reaction: $$\ce{MnO4- + Fe^2+ -> Fe^3+ + Products}$$ This is an unbalanced redox reaction, just depicting how $$\ce{Fe^2+}$$ is oxidized to $$\ce{Fe^3+}$$ by $$\ce{KMnO4}$$. Now, depending on the medium, the following reactions may take place:

• In an acidic medium, $$\ce{MnO4-}$$ is reduced to $$\ce{Mn^2+}$$ by accepting 5 electrons $$\ce{MnO4- + 5Fe^2+ + 8H+ -> 5Fe^3+ + Mn^2+ + 4H2O}$$

• In a neutral/weakly basic medium, $$\ce{MnO4-}$$ is reduced to $$\ce{MnO2}$$ by accepting 3 electrons $$\ce{MnO4- + 3Fe^2+ + 2H2O -> 3Fe^3+ + MnO2 + 4OH-}$$

• In a strongly basic medium, $$\ce{MnO4-}$$ is reduced to $$\ce{MnO4^2-}$$ by accepting 1 electron $$\ce{MnO4- + Fe^2+ -> Fe^3+ + MnO4^2-}$$

This is just one example of how different mediums can affect a given redox process; other redox reactions may also be influenced, depending on whether the medium is acidic or basic.

• In basic medium, $\ce{Fe^3+}$ does not exist. It is transformed in Fe(OH)3 Nov 28, 2019 at 17:46