A) \[{{H}_{2}}\]
B) \[SO_{4}^{2-}\]
C) \[\text{S}{{\text{O}}_{\text{2}}}\]
D) \[{{\text{O}}_{\text{2}}}\]
Correct Answer: D
Solution :
Pure water does not conduct electricity. But when small amount of acid say \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] is added to it, water ionises. On passing electricity, it decomposes. \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] being a strong electrolyte ionise completely whereas water is feebly ionised \[{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}2{{H}^{+}}+SO_{4}^{2-}\] \[{{H}_{2}}O{{H}^{+}}+O{{H}^{-}}\] During electrolysis, the hydrogen ions migrate towards the cathode and discharge here in the form of hydrogen gas \[2{{H}^{+}}+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}\uparrow \] At anode, the concentration of \[\text{O}{{\text{H}}^{-}}\] ions is too low to maintain a reaction and sulphate ions are not oxidised but remain in solution. Thus water molecules must be the species reacting at anode \[2{{H}_{2}}O\xrightarrow{{}}{{O}_{2}}+4{{H}^{+}}+4{{e}^{-}}\] The overall reactions are: At cathode \[2{{H}^{+}}+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}\] \[4{{H}^{+}}+4{{e}^{-}}\xrightarrow{{}}2{{H}_{2}}\] At anode \[2{{H}_{2}}O\xrightarrow{{}}{{O}_{2}}+4{{H}^{+}}+4{{e}^{-}}\] Overall cell reaction: \[4{{H}^{+}}+2{{H}_{2}}O2{{H}_{2}}+{{O}_{2}}+4{{H}^{+}}\] as we see that acid used is regenerated at the end. Therefore, the whole electrolysis reaction is the dissociation of water to give oxygen at anode and hydrogen at cathode catalysed by \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{.}\]
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