# Can we balance a redox reaction in basic medium just like as acidic medium and add OH- ions on both sides to cancel out H+ Ions? [closed]

My teacher told me that the method (which I mentioned above) is wrong but I find many references of saying the same method correct.

• Provide an example please. – M. Farooq Mar 6 at 6:13
• Probably the teacher would prefer introducing hydroxide ions directly rather than first using solvated hydrogen ions and then canceling them out. – Oscar Lanzi Mar 6 at 10:29
• Yes my teacher uses the same method as you mentioned here. I want to ask which is more correct or which is more accurate? – Bunny pink Mar 6 at 13:02
• – user55119 Mar 6 at 19:18

There are of course two ways of balancing redox half-equations in basic solution : writing it directly with $$\ce{OH-}$$ ions or starting to do it in acidic conditions and then destroying the $$\ce{H+}$$ ions by adding $$\ce{OH-}$$ ions. Let's compare these two approaches with chromate ions being reduced to chromium(III).
First method (acidic solution). The redox half-equation in acidic conditions is quickly obtained : $$\ce{CrO4^{2-} + 8 H+ + 3 e- -> Cr^{3+} + 4 H2O} \tag{1}$$ In basic solution, $$\ce{8 OH-}$$ have to be added on both sides to destroy the $$\ce{8 H+}$$ and get :$$\ce{CrO4^{2-} + 8 H2O + 3 e- -> Cr^{3+} + 4 H2O + 8 OH-} \tag{2}$$ which can be simplified thus : $$\ce{CrO4^{2-} + 4 H2O + 3 e- -> Cr^{3+} + 8 OH-} \tag{3}$$
Second method (basic solution). Balancing the redox half-equation from $$\ce{CrO4^{2-}}$$ to $$\ce{Cr^{3+}}$$ without $$\ce{H+}$$ is not so easy. Because the $$4$$ Oxygen atoms of the chromate ion are of course transformed into $$\ce{4OH-}$$ ions, provided enough $$\ce{H}$$ are available. So this requires enough $$\ce{H2O}$$ on the left-hand-side to compensate for the $$\ce{4 H}$$ atoms included in these $$\ce{4 OH-}$$ : $$\ce{2 H2O}$$. But these $$\ce{2 H2O}$$ molecules bring new oxygen atoms. It is not obvious to see that, at the end, $$\ce{4 H2O}$$ (and not $$2$$) have to be added on the left-hand side. The final half-equation is :$$\ce{CrO4^{2-} + 4 H2O + 3 e- -> Cr^{3+} + 8 OH-} \tag{4}$$ This is equal to $$(3)$$. But it not so easy to obtain.
Final remarks. 1. Whatever the method used, it may be useful to state that the $$\ce{Cr^{3+}}$$ ion does not exist and makes a precipitate in basic solution so that the final equation should be written $$\ce{CrO4^{2-} + 4 H2O + 3 e- -> Cr(OH)3 + 5 OH-} \tag{5}$$ 2. The same reasoning could have been done starting from the ion $$\ce{Cr2O7^{2-}}$$ in acidic conditions. With the same conclusion.