Reduction of Organic compounds I've encountered in chemistry quite a lot of reducing agents. However one question strikes me quite often and  any research done on the same hasn't been fruitful. This is in context to organic chemistry. Now for example, in reductive ozonolysis, we often use ozone in combination with zinc and water. Zinc's presence ensures reductive results and not oxidative ones. Now, some sources(not books) which I have seen, say that some sulphur compounds would do the same like $\ce{Me2S}$. Also, in case of 1,3- dinitrobenzene,  $\ce{Sn + HCl}$ converts both the $\ce{-NO2}$ groups to $\ce{-NH2}$ groups. Now if we use $\ce{NH4HS}$ or some other nitrogen and sulphur containing ammonia like compounds, we get reduction again, but only for one group out of the two. So my question is, how do we :

Decide that there is going to be reduction
Know the extent of reduction (partial/full)

Can some insight into the inorganic properties of the elements provide an answer like d-orbital availability, lone pair, electronegativity, size/charge ratio? I've tried reasoning but haven't gotten anywhere. Or is it that memorization of the reagents is the only way to know it's function.
Note: This question is only in context to some less commonly used reagents, so as to recognize it's function if we ever see one in problems present in question books.
I will try to explain the difficulty of this problem by using a simpler reaction, the reduction of nitrobenzene, which is already difficult enough. The reduction of dinitrobenzene should produce more different possibilities.
Nitrobenzene can be reduced into at least five different compounds. It depends on the choice of the reducing agent. I am afraid nobody can justify the particularly mechanisms used by a given reducing agent to get to each final result.
With ($\ce{Sn + HCl}$) or with ($\ce{Fe + HCl}$), it gives aniline, $\ce{C6H5NH2}$.
With sodium methoxide ($\ce{NaOCH3}$), it gives azoxybenzene, $\ce{C6H5N(O)NC6H5}$
With zinc, it can produce three different products:

With zinc + $\ce{NH4Cl}$ at elevated temperature, it gives
phenylhydroxylamine, $\ce{C6H5NHOH}$.

With zinc and an alcoholic solution of $\ce{NaOH}$, it gives azobenzene
$\ce{C6H5N=NC6H5}$.

With zinc and an aqueous solution of $\ce{NaOH}$ at $\pu{60 ^\circ C}$, it give hydrazobenzene $\ce{C6H5NH-NHC6H5}$.