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(i) MnO_(4)^(-) +Sn^(2+)to Mn^(2+)+Sn^(4+) |
Answer» Solution : EQUALISE the increase/ decrease in O.N by multiplying the oxidant and REDUCTANT by suitable numbers. `2MnO_(4)^(-) +5Sn^(2+) to 2Mn^(2+)+5Sn^(4+)` Balance all other atoms exceptO and H `2MnO_(4)^(-) +5Sn^(2+)to 2Mn^(2+)+5Sn^(4+)` Balance O atom by adding water on the SIDE falling short of oxygen atoms. `2MnO_(4)^(-)+5Sn^(2+) to 2Mn^(2+) +Sn^(4+)+8H_(2)O` Balance H atom by adding `H+` on the side falling short of hydrogen atoms. `2MnO_(4)^(-)+5Sn^(2+) 16 H^(+) to 2Mn^(2+) +5Sn^(4+)+8H_(2)O` |
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