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(iv) Sb^(3+) + MnO_(4)^(-) to Sb^(5+)+Mn^(2+). |
Answer» Solution : Eqalise the INCREASE / decrease in Oxidation NUMBER by multiplying with SUITABLE numbers. `5Sb^(3+)+2MnO_(4)^(-)to Sb^(5+)+Mn^(2+)` Balance all other atoms except O and H `5Sb^(3+)+2MnO_(4)^(-)to 5Sb^(5+)+2MN^(2+)` Balance Oxygen atom by adding `H_2O` on the side falling short of oxygen. `5Sb^(3+)+2MnO_(4)^(-)to 5Sb^(5+)+2Mn^(2+)+8H_(2)O` Balance hydrogen atom by adding `H^(+)` on the side falling short of hydrogen atoms. `5Sb^(3+)+2MnO_(4)^(-) to 5Sb^(5+)+2Mn^(2+)+8H_(2)O+16H^(+)`. |
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