Question #ea374

1 Answer
Apr 25, 2015

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 theE^(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