# Question #ea374

Apr 25, 2015

Only (3) is feasible.

We prove it by using standard electrode potentials.

List them most -ve to most +ve:

${I}_{2 \left(s\right)} + 2 e r i g h t \le f t h a r p \infty n s 2 {I}_{\left(a q\right)}^{-}$ ${E}^{0} = + 0.54 \text{V}$

$F {e}_{\left(a q\right)}^{3 +} + e r i g h t \le f t h a r p \infty n s F {e}_{\left(a q\right)}^{2 +}$ ${E}^{0} = + 0.77 \text{V}$

$2 B {r}_{2 \left(l\right)} + 2 e r i g h t \le f t h a r p \infty n s B {r}_{\left(a q\right)}^{-}$ ${E}^{0} = + 1.066 \text{V}$

$C {l}_{2 \left(g\right)} + 2 e r i g h t \le f t h a r p \infty n s 2 C {l}_{\left(a q\right)}^{-}$ ${E}^{0} = + 1.36 \text{V}$

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