Answer
$12H^+ + 4MnO_4^- + 5CH_3OH \rightarrow 4Mn^{2+} +11H_2O+5HCO_2H$
Work Step by Step
Half reaction 1: MnO4- —> Mn 2+
Add 4 H2O on right side, add 8 H+ on left side, add 5e- on left side
5e- +8H+ +MnO4- —>Mn2+ +4H2O
Half reaction 2: CH3OH —> HCO2H
Add 1 H2O on left side, add 4H+ on right side, add 4e- on right side
H2O + CH3OH —> HCO2H +4H+ + 4e-
Reaction 1 times 4, reaction 2 times 5, so that e- can be cancelled
