Question #53cf1

1 Answer
Aug 16, 2017

Answer:

WARNING! Long answer! The balanced equation is

#"I"_2 + "10HNO"_3 → "2HIO"_3 + "10NO"_2 + 4"H"_2"O"#

Explanation:

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"#