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Hint: Corrosion coats the surfaces of metallic objects with oxides or other salts of the metal slowly. The rusting of iron, green coating on copper and bronze and tarnishing of silver are the examples of corrosion. Rusting or corrosion of iron occurs in the presence of water and air.
Complete step by step answer:
In corrosion, a metal is oxidised by loss of electrons to oxygen which leads to formation of oxides. It is considered as an electrochemical phenomenon. Let us look at the mechanism of rusting of iron:
- At any spot of a substance made of iron that spot behaves as anode because oxidation occurs at that place; the reaction involved is
Anode: $\text{Fe}\left( \text{s} \right)\to \text{F}{{\text{e}}^{2+}}\left( \text{aq}\text{.} \right)+\text{2}{{\text{e}}^{-}}\text{ E}_{\left( \text{Fe/F}{{\text{e}}^{2+}} \right)}^{-}=0.44\text{ volt}$
- Electrons released at that spot move to another spot on the metal and reduce oxygen from air in presence of ${{\text{H}}^{+}}$ which is obtained from ${{\text{H}}_{2}}\text{C}{{\text{O}}_{3}}$; formed by the dissolution of carbon dioxide from air into water. ${{\text{H}}^{+}}$ in water can also be present due to dissolution of other acidic oxides from the atmosphere). Now, this spot behaves as cathode; the reaction
Cathode: ${{\text{O}}_{2}}\left( \text{g} \right)+4{{\text{H}}^{+}}\left( \text{aq}\text{.} \right)+4{{\text{e}}^{-}}\to 2{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\text{ E}_{\left( {{\text{H}}^{+}}/{{\text{O}}_{2}}/{{\text{H}}_{2}}\text{O} \right)}^{-}=1.23\text{ volts}$
The overall reaction of rusting is $\text{2Fe}\left( \text{s} \right)+{{\text{O}}_{2}}\left( \text{g} \right)+4{{\text{H}}^{+}}\left( \text{aq}\text{.} \right)\to 2\text{F}{{\text{e}}^{2+}}\left( \text{aq}\text{.} \right)+2{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\text{ E}_{\text{cell}}^{\text{o}}=\text{1}\text{.67 volt}$
Ferrous ions $\left( \text{F}{{\text{e}}^{2+}} \right)$ formed are further oxidised by oxygen to ferric ions $\left( \text{F}{{\text{e}}^{3+}} \right)$ which come out as ‘Rust’ in the form of hydrated ferric oxide ($\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}\text{.x}{{\text{H}}_{2}}\text{O}$) and hydrogen ions are formed.
The answer to this question is option ‘A’ as they have asked for the reaction involved in rusting of iron on the surface of iron, the iron serves as anode here, the reaction involved is $\text{Fe}\left( \text{s} \right)\to \text{F}{{\text{e}}^{2+}}\left( \text{aq}\text{.} \right)+2{{\text{e}}^{-}}$.
Hence the correct option is A
Note:
Prevention of corrosion is very much needed. It not only saves money but also helps in avoiding accidents such as a bridge collapse. One method of preventing corrosion is to prevent the surface of the metallic surface from coming in contact with the atmosphere which can be done by covering that surface with paint or by some chemicals.
Complete step by step answer:
In corrosion, a metal is oxidised by loss of electrons to oxygen which leads to formation of oxides. It is considered as an electrochemical phenomenon. Let us look at the mechanism of rusting of iron:
- At any spot of a substance made of iron that spot behaves as anode because oxidation occurs at that place; the reaction involved is
Anode: $\text{Fe}\left( \text{s} \right)\to \text{F}{{\text{e}}^{2+}}\left( \text{aq}\text{.} \right)+\text{2}{{\text{e}}^{-}}\text{ E}_{\left( \text{Fe/F}{{\text{e}}^{2+}} \right)}^{-}=0.44\text{ volt}$
- Electrons released at that spot move to another spot on the metal and reduce oxygen from air in presence of ${{\text{H}}^{+}}$ which is obtained from ${{\text{H}}_{2}}\text{C}{{\text{O}}_{3}}$; formed by the dissolution of carbon dioxide from air into water. ${{\text{H}}^{+}}$ in water can also be present due to dissolution of other acidic oxides from the atmosphere). Now, this spot behaves as cathode; the reaction
Cathode: ${{\text{O}}_{2}}\left( \text{g} \right)+4{{\text{H}}^{+}}\left( \text{aq}\text{.} \right)+4{{\text{e}}^{-}}\to 2{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\text{ E}_{\left( {{\text{H}}^{+}}/{{\text{O}}_{2}}/{{\text{H}}_{2}}\text{O} \right)}^{-}=1.23\text{ volts}$
The overall reaction of rusting is $\text{2Fe}\left( \text{s} \right)+{{\text{O}}_{2}}\left( \text{g} \right)+4{{\text{H}}^{+}}\left( \text{aq}\text{.} \right)\to 2\text{F}{{\text{e}}^{2+}}\left( \text{aq}\text{.} \right)+2{{\text{H}}_{2}}\text{O}\left( \text{l} \right)\text{ E}_{\text{cell}}^{\text{o}}=\text{1}\text{.67 volt}$
Ferrous ions $\left( \text{F}{{\text{e}}^{2+}} \right)$ formed are further oxidised by oxygen to ferric ions $\left( \text{F}{{\text{e}}^{3+}} \right)$ which come out as ‘Rust’ in the form of hydrated ferric oxide ($\text{F}{{\text{e}}_{2}}{{\text{O}}_{3}}\text{.x}{{\text{H}}_{2}}\text{O}$) and hydrogen ions are formed.
The answer to this question is option ‘A’ as they have asked for the reaction involved in rusting of iron on the surface of iron, the iron serves as anode here, the reaction involved is $\text{Fe}\left( \text{s} \right)\to \text{F}{{\text{e}}^{2+}}\left( \text{aq}\text{.} \right)+2{{\text{e}}^{-}}$.
Hence the correct option is A
Note:
Prevention of corrosion is very much needed. It not only saves money but also helps in avoiding accidents such as a bridge collapse. One method of preventing corrosion is to prevent the surface of the metallic surface from coming in contact with the atmosphere which can be done by covering that surface with paint or by some chemicals.
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