Balance the following equation in basic solution:

Question

Balance the following equation in basic solution:

MnO₄^-(aq) +Fe(OH)₂(s) → MnO₂(s) + Fe(OH)₃(s)

What is the coefficient of water?

the answer is: 2

please show steps so I can follow, thank you!

Answer 1

MnO₄^- ==> MnO₂ + 2H₂O ... to balance oxygens

Fe(OH)₂ + H2O ==> Fe(OH)₃

MnO₄^- + 4H⁺ ==> MnO₂ + 2H₂O ... to balance hydrogens

Fe(OH)₂ + H₂O ==> Fe(OH)₃ + H⁺

MnO₄^- + 4H⁺ + 4OH^- ==> MnO₂ + 2H₂O + 4OH^- ... converting to basic solution

Fe(OH)₂ + H₂O + 4OH^- ==> Fe(OH)₃ + H⁺ + 4OH^-

MnO₄^- + 4H⁺ + 4OH^- + 3e- ==> MnO₂ + 2H₂O + 4OH^- ... to balance charge

Fe(OH)₂ + H₂O + 4OH^- ==> Fe(OH)₃ + H⁺ + 4OH^- + 1e-

MnO₄^- + 4H⁺ + 4OH^- + 3e- ==> MnO₂ + 2H₂O + 4OH^- ... to balance electrons

3Fe(OH)₂ + 3H₂O + 12OH^- ==> 3Fe(OH)₃ + 3H⁺ + 12OH^- + 3e-

Adding half reactions, combining like terms and canceling where possible, we end up with ...

MnO₄^- + 3Fe(OH)₂ + 2H₂O ==> MnO₂ + 3Fe(OH)₃ + OH^- ... balanced for mass and charge