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Marshall's acid (H_(2)S_(2)O_(8)) or peroxodisulphuric acid is prepared by the electrolytic oxidation of mmolH_(2)SO_(4) as : 2H_(2)SO_(4) rarr H_(2)S_(2)O_(8)+2H^(o+)+2e^(-) O_(2)(g) and H_(2)(g) are obtained as byproducts. In such electrolysis 4.48L of H_(2)(g) and 1.12L or O_(2)(g) were produced at STP. The weight of H_(2)S_(2)O_(8) formed is a. 9.7g""b.19.4g""c.14.5g""d.29.1g |
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Answer» Solution :`b.` Refer Section `(point h)`. Aqueous `UNDERSET((dil or conc))(H_(2)SO_(4)) overset(El ectrolysis)rarr2H^(o+)+SO_(4)^(2-)` At anode, two reactions are COMPETING `:` `i. 2SO_(4)^(2-) rarr S_(2)O_(8)^(2-)+2e^(-)` `ii. H_(2)O rarr (1)/(2)O_(2)+2H^(o+)+2e^(-)` At cathode, only one reaction occurs `:` `2H^(o+) (` from `H_(2)SO_(4)) +2e^(-) rarr H_(2)(g)` SINCE onereaction at cathode and two reactions at anode are taking place, therefore, the EQUIVALENT of `H_(2)(g)` produced at cathode at cathode should be equal to the equibalent of `O_(2)(g)` produced `+` equivalent of `H_(2)S_(2)O_(8)` formed. `:.` Equivalent of `H_(2)=`Equivalent of `O_(2)+` Equivalent of `H_(2)S_(2)O_(8)` `(4.48)/(22.4//2)=(1.12)/(22.4//4)+(W_(H_(2)S_(2)O_(8)))/(194//2)` `0.4=0.2+(W)/(97)` `:. W_(H_(2)S_(2)O_(8)=19.4g` |
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