Electrochemistry?

Question

Using tabulated standard reduction potentials from your text, calculate the standard cell potential, E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:

IO3^- (aq) + Fe^2+(aq) <--> Fe^3+(aq) + I2(s)


Can someone please help me with this because I've been retrying it for soo long and I just can't get the right answer. Please anyone help me. Thanks soo much.

Answer 1

I am not sure if the Standard Reduction Potential at 25C I listed here is exactly the same as on your book or not, but they should be close enough:
IO3-(aq) + 6H+(aq) +5e- => (1/2)I2(s) +3H2O(l), Eº=1.195V..(1)
Fe(3+)(aq) +e- ==> Fe(2+)(aq), Eº = 0.771V...(2)
Since Fe(2+) is to be oxidized to Fe(3+), you need to reverse half reaction (2) and therefore change the sign of Eº:
Fe(2+)(aq) ==> Fe(3+)(aq) + e-, Eº = -0.771V...(3)
Now adding (1) and (3) you get the result:
IO3-(aq) + Fe(2+)(aq) + H+(aq) => I2(s) + Fe(3+)(aq) + H2O(l), ∆E =1.195V - 0.771V = 0.424V