Question

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`).

Answer 1

(a) At a particular spot of an object made of iron , oxidation take 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 reactions are :
At Anode : ` " " 2Fe(s) to 2Fe^(2+) + 4e^(-)`
At Cathode : `O_(2)(g) + 4H^(+) (aq) + 4e^(-) to 2H_(2)O (l)`.
The overall reaction is
`2Fe(s) + O_(2)(g) + 4H^(+) (aq) to 2Fe^(2+) (aq) + 2H_(2)O(l)`
(b) `" " E_("cell")^(@) = E_("cathod")^(@) - E_("anode")^(@)`
= `-0.40 -(-0.44)` ltbr =`-0.40 + 0.44 = 0.04 V`
Since `" " E_("cell")^(@) = (0.059)/(n) "log" Kc`
`0.04 = (0.059)/(2) "log" Kc`
log Kc = `(2 xx 0.04)/(0.059) = 1.356`
`therefore " " ` Kc = anti-log (1.356)= 22.70