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.