(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`

Explanation

(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.