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Perdisulphuric acid (H_2S_2O_8) or Marshall's acid can be prepared by the electrolytic oxidation of H_2SO_4. AT anode O_2 and H_2 are obtained as side products. After passing a current of 0.5 A for a certain time, the volume of H_2 and O_2 collected was found to be 10.08 and 2.24 L, respectively, at STP. What is the weight of H_2S_2O_8 produced during the same time? Also find the duration of electrolysis (in seconds) assuming 75% efficiency of electronysis. Give all the electronde reaction. |
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Answer» Solution :(a). Ectrode reactions: `{:(2H_2SO_4toH_2S_2O_8+2e^(-)),(2overset(ɵ)(O)Hto(1)/(2)O_2+H_2O+2e^(-)):}]` At anode SINCE two reactions are occuring at anode and one reaction at cathode, number of equivalent of `H_2S_2O_8` `+` number of equivalent of `O_2=` number of equivalent of `H_2`. LET the weight of `H_2S_2O_8` be x g " Eq of "`H_2S_2O_8=(x)/((194)/(2))` 1 " mol of "`H_2=2Eq` of `H_2=22.4L` at STP " Eq of "`H_2=(10.08)/((22.4)/(2))` 1 " mol of "`O_2=4 Eq` of `O_2=22.4L` " Eq of "`H_2S_2O_8+ Eq` of `O_2=Eq` of `H_2` `(x)/((194)/(2))+(2.24)/((22.4)/(4))=(10.08)/((22.4)/(2))` `(x)/(97)+0.4=0.9becausex=48.5g` Weight of `H_2S_2O_8=48.5g` (b). At cathode, only one reaction takes PLACE, so the number of faradays for electrolysis can be calculated from cathode reaction. `1Eq` of `H_2 gas =1F` `(10.08)/(11.2)Eq=0.9F` For `75%` efficiency, Number of faradays delivered`=0.9xx(100)/(75)` `=(90)/(75)xx96500` coloumb `Ixxt=(90xx96500)/(75)` `5xxt=(90xx06500)/(75)` `t=23160s=6.43hr` |
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