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(vi) MnO_(4)^(-)+Fe^(2+)to Mn^(2+)+Fe^(3+). |
Answer» Solution : Equalise the increase/ decrease in Oxidation number by multiplying Mn species by 1 and Fe species by 5. `MnO_(4)^(-)+5Fe^(2+)to Mn^(2+)+5Fe^(3+)` Balance all other ATOMS EXCEPT O and H `MnO_(4)^(-) +5Fe^(2+)to Mn^(2+)+5Fe^(3+)`. Balance O atom by adding `H_2O` on the side falling short of hydrogen and equal number `OH^(-)` on the opposite side. `MnO_(4)^(-)+5Fe^(2+)+8H_2Oto Mn^(2+)+5Fe^(3+)+4H_(2)O+8OH^(-)` `MnO_(4)^(-)+5Fe^(2+)+8H_(2)O to Mn^(2+)+5Fe^(3+)+4H_(2)O+8OH^(-)`. |
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