A) \[Fe\] is oxidised to \[F{{e}^{2+}}\] and dissolved oxygen in water is reduced to \[O{{H}^{-}}\]
B) \[Fe\] is oxidised to \[F{{e}^{3+}}\] and \[{{H}_{2}}O\] is reduced to \[O_{2}^{2-}\]
C) \[Fe\] is oxidised to \[F{{e}^{2+}}\] and \[{{H}_{2}}O\] is reduced to \[O_{2}^{-}\]
D) \[Fe\] is oxidised to \[F{{e}^{2+}}\] and \[{{H}_{2}}O\] is reduced to \[{{O}_{2}}\]
Correct Answer: A
Solution :
\[~Fe\xrightarrow{{}}F{{e}^{2+}}+2{{e}^{-}}\left( anode\text{ }reaction \right)\] \[{{O}_{2}}+2{{H}_{2}}O+4{{e}^{-}}\xrightarrow{{}}4O{{H}^{-}}(cathode\,\,reaction)\] The overall reaction is \[2Fe+{{O}_{2}}+2{{H}_{2}}O\xrightarrow{{}}2Fe{{(OH)}_{2}}\] \[Fe{{(OH)}_{2}}\] may be dehydrated to iron oxide \[FeO\], or further oxidised to \[Fe{{(OH)}_{3}}\]and then dehydrated to iron rust, \[F{{e}_{2}}{{O}_{3}}.\]
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