(aq) + 6H+
(aq) + 3H2
(aq) + 2e-
(aq)The question asks me to write a balanced equation for the spontaneous reaction that occurs when this pair of half-cells is combined.
From a table of E° (298 K)
values, +1.09 V is reported for the top reaction and +0.54 V for the one below. This tells me that, following the counterclockwise rule, the spontaneous reaction would progress from H+
(an oxidized species) to I-
(a reduced species) but iodide's presence in both half-cells confuses me. Do I need to combine its two appearances and end up with 3I-
in the final balanced equation? I know E°cell
is independent of the amount of material present, but I see stoichiometry in half-cells all the time.
I determined oxidation numbers for each: in [IO3]-, +5 for I and -2 for each O. +1 for each H+. -1 for each I-. +1 for each hydrogen and -2 for oxygen in H2O(l).
(I used colors to segment each species, for ease in vision.) To my mind, however, this doesn't help me.
In short, I'm lost. Can anyone offer any clarity on steps to take next or steps for me to re-examine if I was wrong in any of my assumptions?