2 answers

How does #"iodic acid"#, #HIO_3#, react with iodide anion, #I^-#, to give elemental iodine?

Question:

How does #"iodic acid"#, #HIO_3#, react with iodide anion, #I^-#, to give elemental iodine?

Answers

This is a so-called #"comproportionation reaction"#.

Explanation:

#"Iodic acid"# is reduced to elemental iodine:

#HI^(+V)O_3 + 5H^(+)+5e^(-) rarr 1/2I_2 +3H_2O# #(i)#

#"Iodide"# is oxidized to elemental iodine:

#I^(-) rarr 1/2I_2+e^(-)# #(ii)#

So #(i)+5xx(ii)=#

#HI^(+V)O_3 + 5H^(+)+5I^(-) rarr 3I_2 +3H_2O#

This is AFAIK balanced with respect to mass and charge, and therefore a reasonable representation of chemical reality.

.

WARNING! Long answer! The balanced equation is

#"HIO"_3 + "5HI" → "3I"_2 + "3H"_2"O"#

Explanation:

We start with the unbalanced equation:

#"HIO"_3 + "HI" → "I"_2 + "H"_2"O"#

Step 1. Identify the atoms that change oxidation number

We start by determining the oxidation numbers of every atom in the equation.

#stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("I")stackrelcolor(blue)("-2")("O")_3 + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)(0)("I")_2 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O")#
#color(white)(stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("I")stackrelcolor(blue)("-6")("O")_3 + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)(0)("I")_2 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O"))#

We see that the oxidation number of #"I"# in #"HIO"_3# is reduced to 0 in #"I"_2# and the oxidation number if #"I"# in #"HI"# is increased to 0 in #"I"_2#.

This is a comproportionation reaction, a reaction in which an element in a higher oxidation state reacts with the same element in a lower oxidation state to give the element in an intermediate oxidation state.

The changes in oxidation number are:

#"I: +5 → 0";color(white)(l) "Change" =color(white)(l) "-5 (reduction)"#
#"I: -1 → 0"; color(white)(ll)"Change ="color(white)(l) "+1 (oxidation)"#

Step 2. Equalize the changes in oxidation number

We need 1 atom of #"I"# in #"HIO"_3# for every 5 atoms of #"I"# in #"HI"#.

That also means that we need a total of 6 #"I"# atoms.

Step 3. Insert coefficients to get these numbers

#color(red)(1)"HIO"_3 + color(red)(5)"HI" → color(red)(3)"I"_2 + "H"_2"O"#

Step 4. Balance #"O"#

We have fixed 3 #"O"# atoms on the left, so we need 3 #"O"# atoms on the right.

Put a 3 before #"H"_2"O"#.

#color(red)(1)"HIO"_3 + color(red)(5)"HI" → color(red)(3)"I"_2 + color(blue)(3)"H"_2"O"#

Every formula now has a coefficient. The equation should be balanced.

Step 7. Check that all atoms are balanced.

#bb("On the left"color(white)(l) "On the right")#
#color(white)(mm)"6 H"color(white)(mmmml) "6 H"#
#color(white)(mm)"6 I"color(white)(mmmmll) 6 color(white)(l)"I"#
#color(white)(mm)"3 O"color(white)(mmmml) "3 O"#

The balanced equation is

#color(red)("HIO"_3 + "5HI" → "3I"_2 + "3H"_2"O")#

.

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