A) \[M,{{X}_{2}}\]
B) \[{{H}_{2}},{{X}_{2}}\]
C) \[{{H}_{2}},{{O}_{2}}\]
D) \[M,{{O}_{2}}\]
Correct Answer: B
Solution :
Given, \[E_{{{M}^{+}}/M}^{o}=-1.2\,V\] \[E_{{{x}_{2}}/{{x}^{-}}}^{o}=-1.1\,V\] \[E_{{{O}_{2}}/{{H}_{2}}O}^{o}=1.23\,V\] On electrolysis of aqueous solution of salt MX, the products obtained depends on the given\[E{}^\circ \]values i.e. cathodic reaction will be one with higher\[E_{red}^{o}\]value. Anodic reaction will be one with higher\[E_{oxi}^{o}\]. value or lower\[E_{red}^{o}\]value. Hence, the products are\[{{H}_{2}}\]and\[{{X}_{2}}\].
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