My teacher told me that the method (which I mentioned above) is wrong but I find many references of saying the same method correct.
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7$\begingroup$ Provide an example please. $\endgroup$– AChemMar 6, 2021 at 6:13
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1$\begingroup$ Probably the teacher would prefer introducing hydroxide ions directly rather than first using solvated hydrogen ions and then canceling them out. $\endgroup$– Oscar LanziMar 6, 2021 at 10:29
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$\begingroup$ Yes my teacher uses the same method as you mentioned here. I want to ask which is more correct or which is more accurate? $\endgroup$– Bunny pinkMar 6, 2021 at 13:02
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$\begingroup$ Related: chemistry.stackexchange.com/questions/140462/… $\endgroup$– user55119Mar 6, 2021 at 19:18
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1 Answer
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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.
