InterviewSolution
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InterviewSolution
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Write the electrochemical mechanism of corrosion. |
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Answer» SOLUTION :(i) The formation of rust requires both oxygen and water. Since it is an electrochemical redox process, it requires an anode and cathode in different places on the iron. (ii) The iron surface and a droplet of water on the surface form a tiny galvanic cell. (iii) The REGION enclosed by water is exposed to low amount of oxygen and it acts as the anode. (iv) The remaining area has high amount of oxygen and it act a cathode. So an electro chemical cell is formed. Corrosion occurs at the anode i,e., in the region enclosed by the water.  Anode (oxidation) Iron dissolves in the anode region : `2Fe_((s)) rarr 2Fe_((aq))^(2+)+4^(-)""E^(@)=0.44V` The electrons move through the iron metal from the anode to the cathode area where the oxygen dissolved in water, is reduced to water. Cathode (reduction) : The REACTION of atmospheric carbon dioxide with water gives carbonic acid which furnishes the `H^(+)` ions for reduction. `O_(2(g))+4H_((aq))^(+)+4e^(-) rarr 2H_(2)O(l)""E^(@)=1.23V` The electrical circuit is completed by the migration of ions through water droplet. The overall redox reactions is, `2Fe_((s))+O_(2(g))+4H_((aq))^(+) rarr 2Fe^(2+)""_((aq))+2H_(2)O(l)` `""E^(@)=0.444+1.23=1.67V` The positive emf value indicates that the reaction is spontaneous. `Fe^(2+)` ions are further oxidised to `Fe^(3+)` which on further reaction with oxygen to form rust. `4Fe_((aq))^(2+)+O_(2(g))+4H_((aq))^(+) rarr 4Fe_((aq))^(3+)+2H_(2)O(l)` `2Fe_((aq))^(3+)+4H_(2)O(l) rarr Fe_(2)O_(3).H_(2)O_((s))+6H_((aq))^(+)` |
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