You can find the detailed procedure here.
The skeleton equation is
#"I"^"-" + "NiO"_2 → "I"_2 +"Ni(OH)"_2#
1. Half-reactions
#"I"^"-" → "I"_2#
#"NIO"_2 → "Ni(OH)"_2#
2. Balance all atoms other than #"H"# and #"O"#
#color(red)(2)"I"^"-" → color(red)(1)"I"_2#
#color(blue)(1)"NiO"_2 → color(blue)(1)"Ni(OH)"_2#
3. Balance O
Done.
4. Balance #"H"#
#color(red)(2)"I"^"-" → color(red)(1)"I"_2#
#color(blue)(1)"NiO"_2 +color(olive)(2)"H"^"+" → color(blue)(1)"Ni(OH)"_2#
5. Balance electrons
#color(red)(2)"I"^"-" → color(red)(1)"I"_2 + color(orange)(2)"e"^"-"#
#color(blue)(1)"NiO"_2 +color(olive)(2)"H"^"+" + color(orange)(2)"e"^"-" → color(blue)(1)"Ni(OH)"_2#
6. Equalize electrons transferred
#1 × [color(red)(2)"I"^"-" → color(red)(1)"I"_2 + color(orange)(2)"e"^"-"]#
#1 × [color(blue)(1)"NiO"_2 +color(olive)(2)"H"^"+" + color(orange)(2)"e"^"-" → color(blue)(1)"Ni(OH)"_2]#
7. Add the two half-reactions
#1 × [color(red)(2)"I"^"-" → color(red)(1)"I"_2 + color(red)(cancel(color(black)(color(orange)(2)"e"^"-")))]#
#1 × [color(blue)(1)"NiO"_2 +color(olive)(2)"H"^"+" + color(red)(cancel(color(black)(color(orange)(2)"e"^"-"))) → color(blue)(1)"Ni(OH)"_2]#
#stackrel(——————————————————)(color(red)(2)"I"^"-" + color(blue)(1)"NiO"_2 + color(olive)(2)"H"^"+" → color(red)(1)"I"_2 + color(blue)(1)"Ni(OH)"_2)#
8. Convert to base
#color(red)(2)"I"^"-" + color(blue)(1)"NiO"_2 + color(red)(cancel(color(black)(color(olive)(2)"H"^"+"))) → color(red)(1)"I"_2 + color(blue)(1)"Ni(OH)"_2#
#2"H"_2"O" →color(red)(cancel(color(black)(2"H"^"+"))) + 2"OH"^"-"#
#stackrel(————————————————————)(color(red)(2)"I"^"-" + color(blue)(1)"NiO"_2 + 2"H"_2"O"→ color(red)(1)"I"_2 + color(blue)(1)"Ni(OH)"_2 + 2"OH"^"-")#
The balanced equation is
#"2I"^"-" + "NiO"_2 + 2"H"_2"O"→ "I"_2 + "Ni(OH)"_2 + 2"OH"^"-"#
Now, can you use the same method to balance the second equation?
