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Balance the following reaction by oxidation number method. |
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Answer» Solution :`MnO_(4)^(-1) +H_(2)S+H^(+) to MN^(2+)+S` (Acidic Medium). (i) Write oxidation number of elements `MnO_(4)^(-1) +H_2S to Mn^(2+)+S` `(+7)(-2) (+1)(-2) (+2) 0` (ii) Balance the number of atoms of the elements in which oxidation number changes `MnO_(4)^(-1) +H_2S to Mn^(2+)+S` `(+7) (-2) (+2) 0` (iii) Decide the oxidation and REDUCTION reaction on the basis of difference of oxidation number. Increase in oxidation number by 2 (Oxidation)  Decrease in oxidation number by 5( Reduction) (IV) On multiplying oxidation reaction by 5 and reduction reaction by 2 to balance the change in oxidation number. `2MnO_(4)^(-1) +5H_(2)S to 2Mn^(2+)+5S` (v) Balance the electric charge and atoms which do not change in oxidation number (spectators). `2MnO_(4)^(-1) +5H_(2)S+6H^(+) to 2Mn^(2+)+5S+8H_(2)O` `2(-1)5(0)+6(+1)=2(+2)+5(0)+8(0)` `-2+6=+4` `+4=+4` In the above reaction the reactants and products are balanced in terms of electric charge and mass EQUIVALENCE. |
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