(a) Calcualte E_("cell")^(@) for the following reaction at 298 K 2Al(s) + 3Cu^(2+) (0.01 M) to 2Al^(3+) (0.01 M) + 3Cu(s) Given: E_("cell") = 1.98 V (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) Calcualte E_("cell")^(@) for the following reaction at 298 K 2Al(

1.	(a) Calcualte E_("cell")^(@) for the following reaction at 298 K 2Al(s) + 3Cu^(2+) (0.01 M) to 2Al^(3+) (0.01 M) + 3Cu(s) Given: E_("cell") = 1.98 V (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) n=6 in this case `2Al (s) + 3Cu^(2+) (0.01 M) to 2Al^(3+) (0.01 M) + 3Cu(s)` `E_("cell")^(@) = E_("cell") + (0.0591 V)/n log ([AL^(3+)]^(2))/([CU^(2+)])^(3)` `E_("cell")^(@) =E_("cell")^(2)+ (0.0591 V)/6 log ([0.01])^(2)/([0.01])^(2)` `E_("cell")^(@) = 1.98 V + 0.0098 V xx log [0.01]^(-1)` `=1.98 V + 0.0098 V xx 2` `=1.9996 V` = 2V (Rounded) (b) A Is better for coating the surface of iron. Tliis is because `E^(@)`value for A is more negative than that of Fe. THUS A has greater tendency for OXIDATION than Fe. Therefore A will get oxidised in preference to B and PROTECT it from corrosion.