10) Balance the following redox reaction in basic solution. Fe(OH)2(s)+MnO4^-(aq

Question

10) Balance the following redox reaction in basic solution. Fe(OH)2(s)+MnO4^-(aq)?MnO2(s)+Fe(OH)3(s)

Please show the steps so I can understand.

Explanation / Answer

reduction half is MnO4- ---> MnO2 to balance O's adding 4OH- to the right and 2H2O to the left MnO4 + 2H2O ---> MnO2 + 4OH- to balance the charge adding 4e-s on left MnO4 + 2H2O + 4e- ---> MnO2 + 4OH- ---(1) oxidation half Fe(OH)2 ----> Fe(OH)3 to balance O's adding OH- on left Fe(OH)2 + OH- ------> Fe(OH)3 to balance charge adding 1 e- on right Fe(OH)2 + OH- ---> Fe(OH)3 + e- multiplying above equation with 4 and adding to 1 you get 4Fe(OH)2 + MnO4 + 2H2O ----> MnO2 + 4Fe(OH)3

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