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Using the standard electrode potentials, predict if the reaction between the following is feasible : (a) Fe^(3+) (aq) and I^(-)(aq), (b) Ag^(+) (aq) and Cu(s) (iii) Fe^(3+) (aq) and Br^(-) (aq), (d) Ag (s) and Fe^(3+) (aq) ( c) Br_(2)(aq) and Fe^(2+) (aq). Given standard electrode potentials: E_(1//2, I_(2))^(@) =0.541 V, E_(Cu^(2+),Cu)^(@) = 0.34 V E_(1//2,Br_(2),Br^(-))^(@) = 1.09 V, E_(Ag^(+)//Ag)^(@)= 0.80 V E_(Fe^(3+),Fe^(2+))^(@) = 0.77 V |
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Answer» SOLUTION :A REACTION FEASIBLE if emf of the cell constituted is +ve (a) `Fe^(3+)(aq) + I^(-)(aq) to Fe^(2+)(aq) +1/2 I_(2), i.E, Pt | It | I^(-) (aq) || Fe^(3+) (aq) | Fe^(2+) (aq) | Pt` `therefore E_("cell")^(@) = (E_(Fe)^(3+), Fe^(2+)) - (E_(1//2I_(2),I^(-))^(@)) = 0.77 - 0.54 = 0.23 V` (feasible) (b) `Ag^(+)(aq) + Cu to Ag(s) + Cu^(2+)(aq)`, i.e. `Cu | Cu^(2+)(aq) || Ag^(+) (aq) | Ag` `E_("cell")^(@) = 0.77 - 1.09 = -0.32 V` (not feasible) (d) `Ag(s) + Fe^(3+) (aq) to Ag^(+) + Fe^(2+)` (aq) `E_("cell")^(@) = 0.77 - 0.80 =-0.03 V` (not feasible) ( e) `1/2Br_(2)(aq) + Fe^(2+) (aq) to Br^(-) + Fe^(3+)`, `E_("cell")^(@) = 1.09 - 0.77 = 0.32 V` (feasible) |
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