Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere.
(b) Calculate the equilibrium constant for the equilibrium reaction
`Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd_((s))`
(Given : `E_(Cd^(2+)|Cd)^(@) = -0.40 V , E_(Fe^(2+)|Fe)^(@) = -0.44V`).

Explanation

(a). At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode. Electrons released at anodic spot move through the metal and go to another spot of metal which behaves as cathode.
The half reaction are
At anode: `2Fe(s)to2Fe^(2+)+4e^(-)`
At cathode: `O_(2)(g)+4H^(+)(aq)+4e^(-)to2H_(2)O(l)`
The overally reaction is
`2Fe(s)+O_(2)(g)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l)`
(b). `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)`
`=-0.40-(-0.44)=0.04V`
Since `E_(cell)^(@)=(0.059)/(n)logK_(c)`
`0.04=(0.059)/(2)logK_(C)`
`logK_(c)=(2xx0.4)/(0.059)=1.356`
`K_(C)="anti-log"(1.356)=22.70`