Using the standard electrode potentials given is the Table 1 predict if the reaction between the following is feasible :
(a) `Fe^(3+)(aq) and I^(-) (aq)`
(b) `Ag^(+) (aq) and Cu(s)`
(c) `Fe^(3+) (aq) and Cu(s)`
(d) `Ag(s) and Fe^(3+)(aq)`
(e) `Br_2(aq) and Fe^(2+) (aq)` .

Explanation

(a) It is clear from the table that electrode potential `Fe^(3+)|Fe` (0.77V) is more than that of `I_2|I^(-) (0.54V)`, therefore , `Fe^(3+)` will be readily reduced and the following reaction is feasible.
`Fe^(3+)(aq) +I^(-) (aq) rarr Fe^(2+) (aq) +1/2I_2(s)`
(b) The electrod potential of `Ag^(+) |Ag` (0.80V) is more than that of `Cu^(2+)|Cu` (0.34V ) and therefore , `Ag^(+)` will be reduced by copper . The following reaction is feasible.
`2Ag^(+) (aq) +Cu(s) rarr 2Ag(s) +Cu^(2+) (aq)`
(c) The electrode potential of `Fe^(3+)|Fe^(2+)` (0.77V) is more than that of `Cu^(2+)|Cu` (0.34V), therefore , `Fe^(3+)` can be reduced . The following reaction is feasible .
`2Fe^(3+)(aq) +Cu(s) rarr 2Fe^(2+)(aq) +Cu^(2+) (aq)`
(d) The electrode potential of `Ag|Ag^(+)` (0.80V) is more than that of `Fe^(3+)|Fe` (0.77V) and therefore `Ag^+` will not be reduced by `Fe^(3+)` . Therefore , the reaction will not be feasible.
`Ag(s) +Fe^(3+)(aq) rarr Ag^(+) (aq) +Fe^(2+) (aq)`
(e) The electrode potential of `Br | Br^(-)` (1.09V) is more than that of `Fe^(3+)|Fe^(2+)` (0.77V) and therefore , Br will be able to reduce by `Fe^(3+)` . Therefore , the following reaction is feasible.
`1/2 Br_2(aq) +Fe^(2+)(aq)rarrBr^(-) (aq) +Fe^(3+)(aq)`