Reducing character of group 15 hydrides
When your book says that reducing character increases down the group, it PRIMARILY meant in gaseous state. What you have described above are all ionic reactions, i.e., in some solution. 

NOW in gaseous state, as we go down the group, the bond energy of hydrides reduces (due to increasing atomic size as we go down the group). So they can be more easily broken to release nascent hydrogen (see my note at the bottom), which is the cause of reducing nature. So reducing ability increases down the group in gaseous state.

NOTE : Nascent hydrogen is referred to the hydrogen which has just evolved post some reaction, and has not yet formed its molecule. This is not the same as hydrogen atom so keep that in mind.

Reducing property in a solution

This is the catch. In aqueous solution, reduction takes place due to the group 15 ion, and not hydrogen ion. So it depends on how easily the group 15 ion can lose an electron (reduction is gain of electrons)
Down the group, electronegativity decreases and hence it becomes easier to remove electrons for reduction. So reduction ability decreases down the group in solutions too.

Now to answer your question, Sb(3+) will be oxidised to Sb(5+) as it is what is causing the reduction. You should be able to answer your following question now.

Btw, the E0 value for (H-/H2)is 0.00. Note that all oxidation reduction reactions of hydrogen gas have this value as 0.00