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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) . |
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Answer» Solution :(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)` |
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