A) \[Cs{{O}_{2}}+{{H}_{2}}O\]
B) \[Ba{{O}_{2}}+dil.\,{{H}_{2}}S{{O}_{4}}\] (or) Conc. \[{{H}_{2}}S{{O}_{4}}\]
C) \[Mn{{O}_{2}}+Cone.\text{ }{{H}_{2}}S{{O}_{4}}\]
D) Electrolysis of 50% \[{{H}_{2}}S{{O}_{4}}\] followed by complete hydrolysis of the electrolysis product.
Correct Answer: C
Solution :
[c] Statement (3) is INCORRECT Superoxides contains \[{{O}_{2}}^{\Theta }\]ion \[(K{{O}_{2}},Cs{{O}_{2}},Rb{{O}_{2}})\] react with \[{{H}_{2}}O\] to give \[{{H}_{2}}{{O}_{2}}\]and \[{{O}_{2}}\] \[Cs{{O}_{2}}+2{{H}_{2}}O\xrightarrow{{}}2KOH+{{H}_{2}}{{O}_{2}}+{{O}_{2}}\] (2) Peroxides contains \[{{O}_{2}}^{2-}\] ion \[(N{{a}_{2}}{{O}_{2}},Ba{{O}_{2}})\] react with \[{{H}_{2}}O\] to give \[{{H}_{2}}{{O}_{2}}\] with dil. acids and give \[({{H}_{2}}{{O}_{2}}+{{O}_{2}})\]with cone. acids. (i) \[Ba{{O}_{2}}+{{H}_{2}}S{{O}_{4}}(dil.)\xrightarrow{{}}BaS{{O}_{4}}+2{{H}_{2}}{{O}_{2}}\] (ii) \[Ba{{O}_{2}}+2{{H}_{2}}S{{O}_{4}}(conc.)\xrightarrow{{}}2BaS{{O}_{4}}+2{{H}_{2}}O+{{O}_{2}}\] (3) INCORRECT Dioxides contains \[{{o}^{2-}}\] ion \[(Mn{{O}_{2}},Pb{{O}_{2}})\] react with cone. acids to give \[{{O}_{2}}\] \[2Mn{{O}_{2}}+2{{H}_{2}}S{{O}_{4}}(conc)\xrightarrow{{}}\] \[2MnS{{O}_{4}}+2{{H}_{2}}O+{{O}_{2}}\] (4) Anode \[2S{{O}_{4}}^{2-}\xrightarrow{{}}{{S}_{2}}{{O}_{8}}^{2-}+2\bar{e}\] ?.(i) \[2{{H}_{2}}O\xrightarrow{{}}{{O}_{2}}+4{{H}^{\oplus }}+4\bar{e}\] ?.(ii) Cathode: \[2{{H}_{2}}O+2\bar{e}\xrightarrow{{}}{{H}_{2}}+2\overset{\Theta }{\mathop{O}}\,H\] When \[{{H}_{2}}S{{O}_{4}}\] is \[50%\] both reactions (i) and (ii) occurs simultaneously at anode. So at anode \[{{S}_{2}}{{O}_{8}}^{2-}\] and \[{{O}_{2}}\] are obtained and at cathode \[{{H}_{2}}(g)\] is product. (ii) For dilute \[{{H}_{2}}S{{O}_{4}},\] reaction (ii) is preferred So at anode: \[{{O}_{2}}(g)\]; At cathode: \[{{H}_{2}}(g)\] The product obtained from reaction (i) on complete hydrolysis gives \[{{H}_{2}}{{O}_{2}}\] \[{{S}_{2}}{{O}_{8}}^{2-}+2{{H}^{\oplus }}\xrightarrow{{}}\underset{\begin{smallmatrix} Peroxodisulphuric\text{ }acid \\ (Marshall's\,\,acid) \end{smallmatrix}}{\mathop{{{H}_{2}}{{S}_{2}}{{O}_{8}}}}\,\] \[{{H}_{2}}{{S}_{2}}{{O}_{8}}+2{{H}_{2}}O\xrightarrow{{}}2{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\] On partial hydrolysis, \[{{H}_{2}}{{S}_{2}}{{O}_{8}}\] gives \[{{H}_{2}}S{{O}_{4}}\]and \[{{H}_{2}}S{{O}_{5}}\] (peroxo sulphuric acid or caro's acid) \[{{H}_{2}}{{S}_{2}}{{O}_{8}}+{{H}_{2}}O\xrightarrow{{}}{{H}_{2}}S{{O}_{4}}+{{H}_{2}}S{{O}_{5}}\]
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