InterviewSolution
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InterviewSolution
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(a). What is the equivalent weight of (i) `Fe(HC_2O_4)_2`, (b). Equivalent weight of `Cl_2` is 42.6 in the following disproportionation reaction: `Cl_2+overset(ɵ)(O)HtoCl^(ɵ)+H_2O+?` (oxidised product) identify the oxidised product. (c). What is the equivalent weight of `K_2S_2O_3` in the reaction with `I_2` in acidic medium? |
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Answer» (a). (i). Ew of `Fe(HC_2O_4)_2^(-1xx2)` as reducing agent `Fe^(2+)` is oxidised to `Fe^(3+)` and `2C_2O_4^(2-)` to `4CO_2` `{:(Fe^(2+)toFe^(3+)+e^(-)),(2C_2O_4^(2-)to4CO_2+4e^(-)):}]n=5` Mw of `Fe(HC_2O_4)_2` as an acid. There are `2H^(o+)` `Ew=(234)/(2)=117g` (ii). Ew of `Fe(HC_2O_4)_3^(-1xx3)` as reducing agent. `Fe^(3+)` does not oxidise only `3C_2O_4^(2-)` is oxidised to `6CO_2` `3C_2O_4^(2-)to^CO_2+6e^(-)` Mw of `Fe(HC_2O_4)_3` `=56+3(1+2xx12+4xx16)=323g` `Ew=(234)/(3)=78.0g` (b). `2e^(-)+Cl_2to2Cl^(ɵ)` (reduction) `Cl_2` can be oxidised eithr to `ClO^(ɵ)` (in cold) or `ClO_3^(ɵ)` (in hot). Since condition is not given, so check which is formed. `{:(Cl_2to2ClO^(ɵ)+2e^(-)):}` `Ew=((Mw)/(2)+(Mw)/(2))=(71)/(2)+(71)/(2)=71g` `{:(Cl_2to2ClO_3^(ɵ)+10e^(-)):}` `Ew=((Mw)/(2)+(Mw)/(10))=(71)/(2)+(71)/(10)` `=35.5+7.1=42.6` Given `Ew=42.6, so ClO_3^(ɵ)` is formed. (c). Mw of `K_2S_2O_3=2xx39+2xx32+3xx16=190` In acidic medium `2S_2O_3^(2-)toS_4O_6^(2-)+2e^(-)(n=(2)/(2)=1)` `Ew=(190)/(1)=190.0g` |
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