pH implications of redox balancing half reaction method I have seen similar questions, but I’m asking a new question as far as I know. Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations? asks about using hydroxide instead of water and that is closest.

I have seen example problems where $\ce{H+}$ is produced or consumed, $\ce{OH-}$ is produced or consumed, and water is used to balance. My thought is that each of these changes affects the $\mathrm{pH}$ or $\mathrm{pOH}$ of each solution. 

If the solution is already an acid, there is an abundance of $\ce{H+}$ and it makes sense to me that a reaction may use excess $\ce{H+}$ to make the solution more neutral. One could make the same argument about using up excess $\ce{OH-}$ in a basic solution. What doesn’t make sense to me is adding to the basic or acidic nature. See example below which was specified to be in a basic solution. It appears to be making the solution more basic by producing more $\ce{OH-}$ instead of using it up. Shouldn’t that be less likely to occur? Why not? What am I missing?

$$\ce{6I- (aq) + 4H2O + 2MnO4- (aq) —&gt; 3I2 + 2MnO2 (aq) + 8OH- (aq) }$$
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
$$\ce{ KI  + 6 KOH + 6 KMnO4 -&gt;  KIO3 + 6 K2MnO4 + 3 H2O }$$