A) \[{{O}_{2}}\]
B) \[{{H}_{2}}\]
C) \[S{{O}_{2}}\]
D) \[S{{O}_{3}}\]
Correct Answer: A
Solution :
During electrolysis of dilute\[{{H}_{2}}S{{O}_{4}}\]using platinum electrodes, the oxygen gas is evolved at the anode and hydrogen is liberated at cathode. At cathode: \[{{H}^{+}}+{{e}^{-}}\xrightarrow{{}}\frac{1}{2}{{H}_{2}}\] At anode: \[2{{H}_{2}}O(l)\xrightarrow{{}}{{O}_{2}}(g)+4{{e}^{-}};\] \[\Delta {{E}^{o}}=+\text{ }1.23V\text{ }(I)\] \[2SO_{4}^{2-}(aq)\xrightarrow{{}}{{S}_{2}}O_{8}^{2-}+2{{e}^{-}};\] \[E=+1.96\,V\,(II)\] For dilute solution reaction (I) is preferred and \[{{O}_{2}}\]liberates at anode but at higher concentration reaction (II) is preferred.
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