You can find the general technique for balancing redox equations in acid solution here.
We see that "I"_2 is oxidized to "HIO"_3 and "HNO"_3 is reduced to "NO"_2.
Step 1: Write the two half-reactions.
"I"_2 → "HIO"_3
"HNO"_3 → "NO"_2
Step 2: Balance all atoms other than "H" and "O".
"I"_2 → "2HIO"_3
"HNO"_3 → "NO"_2
Step 3: Balance "O".
"I"_2 + 6"H"_2"O" → "2HIO"_3
"HNO"_3 → "NO"_2 + "H"_2"O"
Step 4: Balance "H".
"I"_2 + 6"H"_2"O" → "2HIO"_3 + 10"H"^"+"
"HNO"_3 + "H"^"+"→ "NO"_2 + "H"_2"O"
Step 5: Balance charge.
"I"_2 + 6"H"_2"O" → "2HIO"_3 + 10"H"^"+" + 10"e"^"-"
"HNO"_3 + "H"^"+" + "e"^"-"→ "NO"_2 + "H"_2"O"
Step 6: Equalize electrons transferred.
1 × ["I"_2 + 6"H"_2"O" → "2HIO"_3 + 10"H"^"+" + 10"e"^"-"]
10 × ["HNO"_3 + "H"^"+" + "e"^"-"→ "NO"_2 + "H"_2"O"]
Step 7: Add the two half-reactions.
"I"_2 + color(red)(cancel(color(black)(6"H"_2"O"))) → "2HIO"_3 + color(red)(cancel(color(black)(10"H"^"+"))) + color(red)(cancel(color(black)(10"e"^"-")))
ul(10"HNO"_3 + color(red)(cancel(color(black)(10"H"^"+"))) + color(red)(cancel(color(black)(10"e"^"-")))→ 10"NO"_2 + stackrelcolor(blue)(4)(color(red)(cancel(color(black)(10))))"H"_2"O")
"I"_2 + "10HNO"_3 → "2HIO"_3 + "10NO"_2 + 4"H"_2"O"
Step 8: Check mass balance.
ul(mathbf("Atom"color(white)(m)"On the left"color(white)(m)"On the right"))
color(white)(ml)"I"color(white)(mmmmml)2color(white)(mmmmmmm)2
color(white)(ml)"H"color(white)(mmmml)10color(white)(mmmmmml)10
color(white)(ml)"N"color(white)(mmmml)10color(white)(mmmmmml)10
color(white)(ml)"O"color(white)(mmmml)30color(white)(mmmmmml)30
Step 9: Check charge balance.
ul(mathbf("On the left"color(white)(m)"On the right"))
color(white)(mmm)0color(white)(mmmmmmm)0
∴ The balanced equation is
"I"_2 + "10HNO"_3 → "2HIO"_3 + "10NO"_2 + 4"H"_2"O"
