How would you balance the following reactions using the oxidation number change method?

#KClO_3 -> KCl + O_2#
#SbCl_5 + KI -> SbCl_3 + KCl + I_2#

2 Answers
May 19, 2017

Answer:

#"KClO"_3"(s)" + Deltararr"KCl"+3/2"O"_2"(g)"uarr#

Explanation:

In the given equation, #stackrel(+V)"Cl"# is reduced to #stackrel(-I)"Cl"#; oxide is oxidized to dioxygen gas.

And for #SbCl_5# we use the method of half equations:

#"SbCl"_5+"2e"^(-) rarr "SbCl"_3 + "2Cl"^(-)# #(ii)#

#"I"^(-) rarr 1/2"I"_2 + e^(-)# #(iii)#

And we takes #(ii)+2xx(iii)# to get............

#"2I"^(-) +"SbCl"_5rarr "I"_2 + "SbCl"_3 + "2Cl"^(-)#

Charge is balanced and mass is balanced, so the equation is reasonable.

May 19, 2017

Answer:

Here's how to balance the second equation by the oxidation number method.

Explanation:

We start with the unbalanced equation:

#"Sb""Cl"_5 + "K""I" → "Sb""Cl"_3 + "K""Cl" + "I"_2#

Step 1. Identify the atoms that change oxidation number

#stackrelcolor(blue)("+5")("Sb")stackrelcolor(blue)("-1")("Cl")_5 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("I") → stackrelcolor(blue)("+3")("Sb")stackrelcolor(blue)("-1")("Cl")_3 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("-1")("Cl") + stackrelcolor(blue)(0)("I")_2#

The atoms that change oxidation number are:

#"Sb: +5 → +3; Change ="color(white)(ll) "-2 (reduction)"#
#"I:"color(white)(mll) "-1" → color(white)(ll)"0; Change = +1 (oxidation)"#

Step 2. Equalize the changes in oxidation number

We need 2 atoms of #"I"# for every 1 atom of #"Sb"#.

This gives us total changes of +2 and -2.

Step 3. Insert coefficients to get these numbers

#color(red)(1)"Sb""Cl"_5 + color(red)(2)"K""I" → color(red)(1)"Sb""Cl"_3 + "K""Cl" + color(red)(1)"I"_2#

Step 4. Balance #"Cl"#

We have fixed 5 #"Cl"# atoms on the left and 3 on the right. We need 2 more#"Cl"# atoms on the right. Put a 2 before #"KCl"#.

#color(red)(1)"Sb""Cl"_5 + color(red)(2)"K""I" → color(red)(1)"Sb""Cl"_3 + color(blue)(2)"K""Cl" + color(red)(1)"I"_2#

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

Step 5. Check that all atoms are balanced.

#bb"On the left"color(white)(l)bb "On the right"#
#color(white)(mm)"1 Sb"color(white)(mmmmll) "1 Sb"#
#color(white)(mm)"5 Cl"color(white)(mmmmm) "5 Cl"#
#color(white)(mm)"2 K"color(white)(mmmmml) "2 K"#
#color(white)(mm)"2 I"color(white)(mmmmmll) "2 I"#

The balanced equation is

#color(blue)("Sb""Cl"_5 + "2K""I" → "Sb""Cl"_3 + "2K""Cl" + "I"_2)#