pH implications of redox balancing half reaction method
If the reaction happens at neutral or mildly alkalic $\mathrm{pH}$:

$$\ce{6I- + 4 H2O + 2 MnO4- -&gt; 3 I2 + 2 MnO2 (s) + 8 OH-}$$

If the reaction happens at mildly aciding $\mathrm{pH}$, it could be as well:

$$\ce{6I- + 8 H+ + 2 MnO4- -&gt; 3 I2 + 2 MnO2 (s) + 4 H2O }$$

In any case, either spending  $\ce{H+}$ either production $\ce{OH-}$ increases $\mathrm{pH}$. Both iodide and permanganate carry a negative charge while their spent forms are neutral. This negative charge must be passed to other anions like $\ce{OH-}$ or neutralized by $\ce{H+}$. 

If one considers further disproportionation of iodine to iodide and iodate in alkalic solutions, then the total reaction is:

$$\ce{ I-  +  H2O + 2 MnO4- -&gt;  IO3- + 2 MnO2(s) + 2 OH-}$$

In strongly alkalic it could be also
$$\ce{ I-  + 6 OH- + 6 MnO4- -&gt;  IO3- + 6 MnO4^2- + 3 H2O }$$

It may be obvious if the equation is written together with spectator ions:

Mildly acidic:
$$\ce{6KI + 8 HA + 2 KMnO4 -&gt; 3 I2 + 2 MnO2 (s) + 4 H2O + 8 KA}$$
resp neutral/mildly alkalic
$$\ce{6KI + 4 H2O + 2 KMnO4 -&gt; 3 I2 + 2 MnO2 (s) + 8 KOH}$$
resp alkalic
$$\ce{ KI  +  H2O + 2 KMnO4 -&gt;  KIO3 + 2 MnO2(s) + 2 KOH}$$
resp. strongly alkalic
$$\ce{ KI  + 6 KOH + 6 KMnO4 -&gt;  KIO3 + 6 K2MnO4 + 3 H2O }$$