Question

Question #ea374

Answer 1

Only (3) is feasible.

We prove it by using standard electrode potentials.

List them most -ve to most +ve:

`I_(2(s))+2erightleftharpoons2I_((aq))^(-)` `E^(0)=+0.54"V"`

`Fe_((aq))^(3+)+erightleftharpoonsFe_((aq))^(2+)` `E^(0)=+0.77"V"`

`2Br_(2(l))+2erightleftharpoonsBr_((aq))^(-)` `E^(0)=+1.066"V"`

`Cl_(2(g))+2erightleftharpoons2Cl_((aq))^(-)` `E^(0)=+1.36"V"`

Use the rule "bottom left oxidise top right".

1. Is not feasible because the`E^(0)` value for Fe3+/Fe2+ is not great enough. It would need to be > +1.36 V

2. Is not feasible for the same reason. The `E^(0)` value would need to be > + 1.066V

3. Is feasible because the `E^(0)` value for Fe3+/Fe2+ is greater than the `E^(0)` value for I2/I-. So bottom left will oxidise top right