Write a balanced half-reaction for the oxidation of manganese ion mn+2 to solid manganese dioxide mno2 in basic aqueous solution. be sure to add physical state symbols where appropriate.

The oxidation state of of Mn in Mn²⁺ = +2
the oxidation state of Mn in MnO₂ - +4
from +2 to +4 the oxidation state has increased, hence this is an oxidation reaction 
eventually on both sides of the equation, both charges and masses must be balanced 
lets construct this equation stepwise
Mn²⁺ > MnO₂
Mn atoms are balanced, however right side has O atoms, to balance the O atoms, add H₂O molecules with the coefficient to suit the number of O atoms on the right 
Mn²⁺ + 2H₂O > MnO₂ 
Since there are  2 O in MnO₂, 2 H₂O molecules are added 
Now to balance the H atoms, add H⁺ ions to the right side, since there are 2 H₂O molecules, add 4 protons
Mn²⁺ + 2H₂O > MnO₂ + 4H⁺
Next we have to balance the charges, as of now, left side has +2 net charge and right side has +4 charge. since charges of both sides should be equal, add 2 electrons to the right side, to reduce charge from +4 to +2
Mn²⁺ + 2H₂O > MnO₂ + 4H⁺ + 2e
Since this is a basic medium add 4 OH⁻ ions to both sides of the equation to neutralise the 4 H⁺ ions
Mn²⁺ + 2H₂O + 4OH⁻ > MnO₂ + 4H⁺ + 4OH⁻+  2e
4H⁺ and 4OH⁻ neutralise to form water molecules
Mn²⁺ + 2H₂O + 4OH⁻ > MnO₂ + 4H₂O +  2e
since there water molecules on both sides, theres a net amount of 2 H₂O molecules on the right side, hence water is cancelled from the left side 

the balanced half reaction is 
Mn²⁺ + 4OH⁻ > MnO₂ + 2H₂O +  2e

Afraidsugar lol it’s the third one