Question #62a46

Your first unbalanced equation is

#"KClO"_3 → "KClO"_4 + "KCl"#

Step 1. Calculate the oxidation numbers of every atom:

#stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+5")("Cl")stackrelcolor(blue)("-2")("O")_3 → stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")("Cl")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("Cl")#

Step 2. Identify the atoms that change oxidation number.

The changes in oxidation number are:

#"Cl: +5 → +7; Change = +2"#
#"Cl: +5 → -1;" color(white)(ll)"Change ="color(white)(m) "-6"#

Step 3. Equalize the changes in oxidation number.

You need #"3 atoms of Cl"# that increase oxidation number for every #"1 atom of Cl"# that decreases oxidation number. . This gives us total changes of #"+6"# and #"-6"#.

Step 4. Insert coefficients to get these numbers.

#"KClO"_3 → color(red)(3)"KClO"_4 + color(blue)(1)"KCl"#

Step 5. Balance #"K"#.

#color(orange)(4)"KClO"_3 → color(red)(3)"KClO"_4 + color(blue)(1)"KCl"#

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

Step 6. Check that all atoms balance.

#"Atom"color(white)(m) "Left Hand Side"color(white)(m) "Right Hand Side"#
#color(white)(m)"K"color(white)(mmmmmm) 4 color(white)(mmmmmmml)4#
#color(white)(m)"Cl" color(white)(mmmmmll)4 color(white)(mmmmmmml)4#
#color(white)(m)"O"color(white)(mmmmml)12 color(white)(mmmmmmm)12#

The balanced equation is

#"4KClO"_3 → "3KClO"_4 + "KCl"#

Now, can you use the same technique to balance the second equation?

Here's a video that might help.

I'm going to assume it is in an acidic solution.

The first step would be to separate the overall equation into two half reactions, but you've already done that. So next would be to balance everything except for the H and O.

#color(navy)(2KClO_3 rarr KClO_4 + KCl)#

#color(maroon)(I_2 + HNO_3 rarr 2HIO_3 + NO + H_2O)#

Then balance the oxygen by adding H2O.

#color(navy)(2KClO_3 rarr KClO_4 + KCl + 2H_2O)#

#color(maroon)(I_2 + HNO_3 + 5H_2O rarr 2HIO_3 + NO + H_2O)#

Now balance the hydrogen by adding H+.

#color(navy)(2KClO_3 + 4H^+ rarr KClO_4 + KCl + 2H_2O)#

#color(maroon)(I_2 + HNO_3 + 5H_2O rarr 2HIO_3 + NO + H_2O + 7H^+)#

Add e- as needed to balance the charges on each side of the equation.

#color(navy)(2KClO_3 + 4H^+ + 4e^(-) rarr KClO_4 + KCl + 2H_2O)#

#color(maroon)(I_2 + HNO_3 + 5H_2O rarr 2HIO_3 + NO + H_2O + 7H^+ + 7e^-)#

Finally, multiply each half-reaction by an integer so that when they are added, the e- will cancel out. Then add them and simplify.

7(#color(navy)(2KClO_3 + 4H^+ + 4e^(-) rarr KClO_4 + KCl + 2H_2O)#)

+

4(#color(maroon)(I_2 + HNO_3 + 5H_2O rarr 2HIO_3 + NO + H_2O + 7H^+ + 7e^-)#)

=

#color(navy)(14KClO_3 + 28H^+ + 28e^(-) + color(maroon)(4I_2 + 4HNO_3 + 20H_2O) color(black)(rarr) 7KClO_4 + 7KCl + 14H_2O + color(maroon)(8HIO_3 + 4NO + 4H_2O + 28H^+ + 28e^-))#

#color(navy)(14KClO_3 + cancel(28H^+) + cancel(28e^(-)) + color(maroon)(4I_2 + 4HNO_3 + cancel(20)2H_2O) color(black)(rarr) 7KClO_4 + 7KCl + cancel(14H_2O) + color(maroon)(8HIO_3 + 4NO + cancel(4H_2O) + cancel(28H^+) + cancel(28e^-)))#

#14KClO_3 + 4I_2 +4HNO_3 + 2H_2O rarr 7KClO_4 + 7KCl + 8HIO_3 + 4NO#