(a) Calculate DeltaG^(@)and log K_( c)for the following reaction at 298 K : 2Cr(s) + 3Fe^(2+) (aq) to 2Cr^(3+) (aq) + 3Fe (s) (b) (b) Using the E° values of A and B, predict which is better for coating the surface of iron [E^(@)(Fe^(2+)//Fe) = -0.44 V] to prevent corrosion and why? Given : E_(A^(2+)//A)^(@) = -2.37 V: E_(B^(2+)//B)^(@) = -0.14 V

(a) Calculate DeltaG^(@)and log K_( c)for the following reaction at 29

1.	(a) Calculate DeltaG^(@)and log K_( c)for the following reaction at 298 K : 2Cr(s) + 3Fe^(2+) (aq) to 2Cr^(3+) (aq) + 3Fe (s) (b) (b) Using the E° values of A and B, predict which is better for coating the surface of iron [E^(@)(Fe^(2+)//Fe) = -0.44 V] to prevent corrosion and why? Given : E_(A^(2+)//A)^(@) = -2.37 V: E_(B^(2+)//B)^(@) = -0.14 V
Answer» Solution :(a) Given that `E_("cell")^(@) = +0.30 V, F = 96500 C mol^(-1)` `2Cr (s) to 2Cr^(3+) + 6E^(-)` `3Fe^(2+)(AQ)+ 6e^(-) to 3Fe (s)` Thus, n=6 Apply the relation: `Delta_(r)G^(@) =-n xx F xx E_("cell")^(@)` or `Delta_(r)G^(@)=-6 xx 96500 C mol^(-1) xx 0.30 V = -173700 J mol^(-1)` or `-173.7 kJ mol^(-1)` `log K_( c) = (nE_("cell")^(@))/0.059` or `log K_( c) = (6 xx 0.30)/0.059` or `log K_( c) = 30.5` (b) A is better for coating the surface of iron. This is because E° value for A is more NEGATIVE than that of Fe. Therefore A has greater TENDENCY for oxidation than Fe. Thus A will oxidise in preference to Fe and protect it from corrosion.