How to balance this reaction which is taking place in a basic medium? $$\ce{Al + MnO4- -> MnO2 + Al(OH)4-}$$

My Attempt

  • Reduction half reaction:

$$ \begin{align} \ce{4 H+ + MnO4- + 3 e- &-> MnO2 + 2 H2O}\\ \ce{28 H+ + 7 MnO4- + 21 e- &-> 7 MnO2 + 14 H2O} \end{align} $$

  • Oxidation half reaction:

$$ \begin{align} \ce{Al + 4 H2O &-> Al(OH)4- + 8 H+ +7 e-}\\ \ce{3 Al + 12 H2O &-> 3 Al(OH)4- + 24 H+ + 21 e-} \end{align} $$

  • Adding half reactions:

$$ \ce{3 Al + 7 MnO4- + 28 H+ + 12 H2O + 21 e- -> 7 MnO2 + 3 Al(OH)4- + 14 H2O + 24 H+ + 21 e-} $$

Adding $\ce{OH-}$ and eliminating $\ce{e-}$ and formed $\ce{H2O}$.

  • Final equation:

$$ \ce{3 Al + 7 MnO4- + 2 H2O -> 7 MnO2 + 3 Al(OH)4- + 4(OH)-} $$

Please tell if I am wrong.

  • $\begingroup$ The hydrogens don't seem to be balanced on both the sides of the equation $\endgroup$ – YUSUF HASAN Dec 27 '18 at 4:36
  • $\begingroup$ Just add 2 H2O to the reactants, maybe? $\endgroup$ – Oscar Lanzi Dec 27 '18 at 13:43
  • 1
    $\begingroup$ Thanks a lot to all of you those who have helped $\endgroup$ – Nick Dec 27 '18 at 14:11

Writing an answer since I can't comment yet.

You are using the ion-electron method for basic medium, but are balancing using $\ce{H+}$ and later adding $\ce{OH-}$? Sounds suspicious to me, but the outcome must be the same.

Generally, I add $\ce{OH-}$ ions on the oxygen deficient side and then add water to compensate the hydrogen deficiency on the other side.

Note: your first half-reaction balancing the conversion of permanaganate to $\ce{MnO2}$ must be regarded incorrect, as $\ce{MnO2}$, being soluble in acidic conditions (presence of $\ce{H+}$, as the equation indicates!) would yield to further reduction, and end up as $\ce{Mn^2+}$.

Here's how I balanced it:

  • Oxidation half

$$ \begin{align} \ce{Al &-> Al(OH)4-}\\ \ce{Al + 4OH- &-> Al(OH)4- + 3e-} \end{align} $$

  • Reduction half:

$$ \begin{align} \ce{MnO4- &-> MnO2}\\ \ce{MnO4- &-> MnO2 + 2OH-}\\ \ce{MnO4- + 2 H2O &-> MnO2 + 4 OH- + 3 e-} \end{align} $$

  • Adding:

$$ \ce{Al + MnO4- + 2 H2O -> Al(OH)4- + MnO2} $$

Your method is fine as well. The only mistake is the wrong $\ce{H+}$ coefficient in $\ce{Al}$ oxidation half-reaction, which will be 4 instead of 8.

P. S. (only concern): note the differences in half reactions in basic and acidic media. Conclusion: when products differ in acidic/basic conditions, it is best not to balance in the asker's method, as it cannot be considered entirely correct. Nothing more, nothing less.

  • $\begingroup$ It is quite acceptable to first balance the equation in acidic medium and then add hydroxide on both sides to neutralise the protons. This is because it is the same idea as first providing OH- on oxygen deficient side and then adding hydrogen on the opposite side, or doing it vice-versa as the OP (tried) to do. $\endgroup$ – YUSUF HASAN Dec 27 '18 at 6:58
  • $\begingroup$ Nonetheless the Al oxidation half reaction is likely to be technically incorrect, as the product differs in relation to pH. $\endgroup$ – William R. Ebenezer Dec 27 '18 at 7:38
  • $\begingroup$ @YUSUFHASAN thanks for pointing out, i did the necessary edits. $\endgroup$ – William R. Ebenezer Dec 27 '18 at 7:45
  • $\begingroup$ I converted your handwriting to text, however I wasn't able to recognize one line. Please feel free to add missing points; also check out How can I format math/chemistry expressions here? $\endgroup$ – andselisk Dec 27 '18 at 11:32

First off, for basic medium there should be no protons in any parts of the half-reactions. Use water and hydroxide-ions if you need to, like it's been done in another answer.

In basic aqueous solution permanganate $\ce{MnO4-}$ is going to be reduced to manganate $\ce{MnO4^2-}$, and not to manganese(IV) oxide $\ce{MnO2}$ ($\ce{MnO2}$ forms in neutral medium). This can be explained by applying to the standard electrode potentials (for example, see this answer to "Stability of MnO4- in basic vs acidic conditions") or simply deduced by applying the Le Chatelier's principle to the following half reaction you'd normally use for neutral medium:

$$ \ce{MnO4- + 2 H2O -> MnO2 + 4 \color{red}{OH-} + 3 e-} $$

Theoretically, $\ce{MnO4^2-}$ can be hydrolyzed to $\ce{MnO2}$, but for that an excess of a strong reducing agent and only mildly basic solution would be required. None of these conditions are fulfilled here as the strong alkali is requied to remove constantly forming passivating layer of $\ce{Al2O3}$/$\ce{Al(OH)3}$.

That's said, I'd rather propose the following redox scheme (numbers above the elements are the oxidation numbers):

\begin{align} \ce{3 \overset{+7}{Mn}O4- + 3e- &-> 3 \overset{+6}{Mn}O4^2-} \tag{red}\\ \ce{\overset{0}{Al} + 4 OH- &-> \overset{+3}{Al}(OH)4- + 3e-} \tag{ox}\\ \hline \ce{3 \overset{+7}{Mn}O4- + \overset{0}{Al} + 4 OH- &-> 3 \overset{+6}{Mn}O4^2- + \overset{+3}{Al}(OH)4-} \tag{redox} \end{align}


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