Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent? The question is the same as the title states. In my textbook,[NCERT Chemistry I for Class 12 , pg no 217.] the following is written

$\ce{Cr^{2+}}$ gets converted to $\ce{Cr^{3+}}$ as the +3 oxidation state has half-filled $\mathrm{t_{2g}}$ orbitals; thus it is a good reducing agent.
On the other hand $\ce{Mn^{3+}}$ gets converted to $\ce{Mn^{2+}}$ as its +2 oxidation state has a half-filled d-subshell.

Now my questions are:

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$?

Why can't the half-filled $\mathrm{d}$-orbital argument be used for $\ce{Cr}$?


$\ce{Mn^3+}$ could be oxidized to $\ce{Mn^4+}$, which would have a half-filled $\mathrm{t_{2g}}$ set. As oxidation and reduction both can take place in aqueous medium, why wouldn't $\mathrm{t_{2g}^3}$ or $\mathrm{d^3}$ configuration be more stable than $\mathrm{d^5}$ configuration in aqueous medium?
A similar observation is made in case of $\ce{Fe^2+}$ and $\ce{Cr^2+}$, where it observed that $\ce{Cr^2+}$ is more powerful reducing agent than $\ce{Fe^2+}$ (due to reasons explained above).
If possible, I would like to see the crystal field splitting calculations that show this to be the case.
Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$ ?


Because the number of $\mathrm{t_{2g}}$ electrons is not changing upon going from $\ce{Mn^3+}$ to $\ce{Mn^2+}$.  In both $\ce{Mn^2+}$ and $\ce{Mn^3+}$ there are 3 $\mathrm{t_{2g}}$ electrons.  I think the point they are trying to make with $\ce{Cr^3+}$ being half filled is that there are 3 $\mathrm{t_{2g}}$ electrons and no $\mathrm{e_g}$ electrons.  $\ce{Mn^3+}$ and $\ce{Mn^2+}$ both have $\mathrm{e_g}$ electrons. 

You could make a similar argument for $\ce{Mn^3+}$ being oxidized to $\ce{Mn^4+}$, which would have half filled $\mathrm{t_{2g}}$ and empty $\mathrm{e_g}$.


  Why can't the half-filled d-orbital argument be used for Cr?


Because neither $\ce{Cr^3+}$ nor $\ce{Cr^2+}$ has a half filled d subshell.  $\ce{Cr^3+}$ has 3 electrons while $\ce{Cr^2+}$ has 4 electrons.