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Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:
(a) Fe3+(aq) and I-(aq) (b) Ag+   (aq) and Cu(s)
(c) Fe3+(aq) and Cu(s) (d)Ag(s) and Fe3+(aq)
(e) Br2 (aq) and Fe3+ (aq).

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Fe3+(aq) and i- (aq) Oxidation half- reaction is 2i- -> i2 + 2 e( electron) Excell= -0.54 Reduction half cell reaction is ×2 Ecell= +.077 And finally addition oxidation and reduction We get E.M.F = (-0.54)+.077= +0.23 . And emf is positive so reaction is possible .. same option b; c;...
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Fe3+(aq) and i- (aq) 

Oxidation half- reaction is 

2i-    -> i2 + 2 e( electron) 

Excell= -0.54 

Reduction half cell reaction is

[Fe3+  + e ( electron ) -> fe2+ ]×2

Ecell= +.077

And finally addition oxidation and reduction 

We get 

E.M.F = (-0.54)+.077= +0.23 .

And emf is positive so reaction is possible .. same option b; c; d ; option reaction possible..

But option is not feasible because it's ecell is negative..if you want more details . Email

 

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