How to balance the following redox problems using both methods?

  1. K2Cr2O7 + HI => CrI3 + KI + I2 + HOH

  2. K2Cr2O7 + H2C2O4 + HCl => CrCl3 + CO2 + KCl + HOH

  3. Sb2(SO4)3 + KMnO4 + HOH => H3SbO4 + K2SO4 + MnSO4 + H2SO4

  4. Mn(NO3)2 + NaBiO3 + HNO3 => Bi(NO3)2 + HMnO4 + NaNO3 + HOH

2 Answers
Jan 21, 2017

WARNING! Long answer! Here's how to balance Part 3. by the oxidation number method.

Explanation:

We start with the unbalanced equation:

"Sb"_2("SO"_4)_3 + "KMnO"_4 + "HOH" → "H"_3"SbO"_4 + "K"_2"SO"_4 + "MnSO"_4 + "H"_2"SO"_4

Step 1. Identify the atoms that change oxidation number

stackrelcolor(blue)("+3")("Sb")_2(stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4)_3 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")("Mn")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") → stackrelcolor(blue)("+1")("H")_3stackrelcolor(blue)("+5")("Sb")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("K")_2stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+2")("Mn")stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4

The changes in oxidation number are:

"Sb:" color(white)(l)"+3" → "+5"; "Change = +2 (oxidation)"
"Mn:" "+7" → "+2"; "Change =" color(white)(ll) "-5 (reduction)"

Step 2. Equalize the changes in oxidation number

We need 5 atoms of "Sb" for every 2 atoms of "Mn" or 10 atoms of "Sb" for every 4 atoms of "Mn". This gives us total changes of +20 and -20.

Step 3. Insert coefficients to get these numbers

color(red)(5)"Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + "HOH" → color(red)(10)"H"_3"SbO"_4 + "K"_2"SO"_4 + color(red)(4)"MnSO"_4 + "H"_2"SO"_4

Step 4. Balance "K"

We have fixed 4 "K" on the left, so we need 4 "K" on the right. Put a 2 in front of "K"_2"SO"_4.

color(red)(5)"Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + "HOH" → color(red)(10)"H"_3"SbO"_4 + color(blue)(2)"K"_2"SO"_4 + color(red)(4)"MnSO"_4 + "H"_2"SO"_4

Step 5. Balance "S"

We have fixed 15 "S" atoms on the left and 6 "S" atoms on the right, so we need 9 more "S" atoms on the right. Put a 9 before "H"_2"SO"_4.

color(red)(5)"Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + "HOH" → color(red)(10)"H"_3"SbO"_4 + color(blue)(2)"K"_2"SO"_4 + color(red)(4)"MnSO"_4 + color(purple)(9)"H"_2"SO"_4

Step 6. Balance "O"

We have fixed 76 "O" atoms on the left and 100 "O" atoms on the right, so we need 24 more "O" atoms on the left. Put a 24 before "HOH".

color(red)(5)"Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + color(brown)(24)"HOH" → color(red)(10)"H"_3"SbO"_4 + color(blue)(2)"K"_2"SO"_4 + color(red)(4)"MnSO"_4 + color(purple)(9)"H"_2"SO"_4

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

Step 7. Check that all atoms are balanced.

mathbf("Atom"color(white)(m)"On the left"color(white)(m)"On the right")
color(white)(ml)"Sb"color(white)(mmmml)10color(white)(mmmmmmm)10
color(white)(ml)"S"color(white)(mmmmm)15color(white)(mmmmmmm)15
color(white)(ml)"O"color(white)(mmmml)100color(white)(mmmmmml)100
color(white)(ml)"K"color(white)(mmmmml)4color(white)(mmmmmmmll)4
color(white)(ml)"Mn"color(white)(mmmmll)4color(white)(mmmmmmmll)4
color(white)(ml)"H"color(white)(mmmmm)48color(white)(mmmmmmm)48

The balanced equation is

color(red)(5"Sb"_2("SO"_4)_3 + 4"KMnO"_4 + 24"HOH" →10"H"_3"SbO"_4 + 2"K"_2"SO"_4 + 4"MnSO"_4 +9"H"_2"SO"_4)

Jan 21, 2017

WARNING! Long answer! Here's how to balance Part 3. by the ion-electron method.

Explanation:

We must first convert the "molecular" equation to a skeletont ionic equation.

The molecular equation is

"Sb"_2("SO"_4)_3 + "KMnO"_4 + "HOH" → "H"_3"SbO"_4 + "K"_2"SO"_4 + "MnSO"_4 + "H"_2"SO"_4

To get the ionic equation, we write every ionic substance as separate ions.

We ignore the coefficients, because they will come back in during the balancing process.

"Sb"^"3+" + "SO"_4^"2-" + "K"^"+" + "MnO"_4^"-" + "HOH" → "H"^"+" + "SbO"_4^"3-" + "K"^"+" + "SO"_4^"2-" + "Mn"^"2+" + "SO"_4^"2-" + "H"^"+" + "SO"_4^"2-"

To get the skeleton ionic equation, we eliminate every ion that occurs on both sides of the equation.

We also eliminate "H"^"+" and "H"_2"O", because they come back through the balancing process.

"Sb"^"3+" +color(red)(cancel(color(black)( "SO"_4^"2-"))) + color(red)(cancel(color(black)("K"^"+"))) + "MnO"_4^"-" + color(red)(cancel(color(black)("HOH"))) → color(red)(cancel(color(black)("H"^"+"))) + "SbO"_4^"3-" + color(red)(cancel(color(black)("K"^"+"))) + color(red)(cancel(color(black)("SO"_4^"2-"))) + "Mn"^"2+" + color(red)(cancel(color(black)("SO"_4^"2-"))) + color(red)(cancel(color(black)("H"^"+"))) + color(red)(cancel(color(black)("SO"_4^"2-")))

The skeleton ionic equation is

"Sb"^"3+" + "MnO"_4^"-" → "SbO"_4^"3-" + "Mn"^"2+"

We see that "Sb"^"3+" is oxidized to "SbO"_4^"3-" and "MnO"_4^"-" is reduced to "Mn"^"2+".

Now we are ready to balance the equation.

Step 1: Write the two half-reactions.

"Sb"^"3+" → "SbO"_4^"3-"
"MnO"_4^"-" → "Mn"^"2+"

Step 2: Balance all atoms other than "H" and "O".

Done.

Step 3: Balance "O".

"Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-"
"MnO"_4^"-" → "Mn"^"2+" + "4H"_2"O"

Step 4: Balance "H".

"Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-" + "8H"^"+"
"MnO"_4^"-" + "8H"^"+" → "Mn"^"2+" + "4H"_2"O"

Step 5: Balance charge.

"Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-" + "8H"^"+" + "2e"^"-"
"MnO"_4^"-" + "8H"^"+" + "5e"^"-" → "Mn"^"2+" + "4H"_2"O"

Step 6: Equalize electrons transferred.

5×["Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-" + "8H"^"+" + "2e"^"-"]
2×["MnO"_4^"-" + "8H"^"+" + "5e"^"-" → "Mn"^"2+" + "4H"_2"O"]

Step 7: Add the two half-reactions.

5"Sb"^"3+" + stackrelcolor(blue)("+12")color(red)(cancel(color(black)(20)))"H"_2"O" → "5SbO"_4^"3-" + stackrelcolor(blue)(24)(color(red)(cancel(color(black)(40))))"H"^"+" + color(red)(cancel(color(black)("10e"^"-")))
2"MnO"_4^"-" + color(red)(cancel(color(black)("16H"^"+"))) + color(red)(cancel(color(black)("10e"^"-"))) → "2Mn"^"2+" + color(red)(cancel(color(black)("8H"_2"O")))
5"Sb"^"3+" + 2"MnO"_4^"-" + 12"H"_2"O" → 5"SbO"_4^"3-" + "2Mn"^"2+" + "24H"^"+"

Step 8: Check mass balance.

mathbf("Atom"color(white)(m)"On the left"color(white)(m)"On the right")
color(white)(ml)"Sb"color(white)(mmmml)5color(white)(mmmmmmm)5
color(white)(ml)"Mn"color(white)(mmmm)2color(white)(mmmmmmm)2
color(white)(ml)"O"color(white)(mmmml)20color(white)(mmmmmml)20
color(white)(ml)"H"color(white)(mmmml)24color(white)(mmmmmml)24

Step 9: Check charge balance.

mathbf("On the left"color(white)(mml)"On the right")
"15 - 2 = +13"color(white)(m)"-15 + 4 + 24 = +13"

The ionic equation is balanced.

Now we reinsert the spectator ions to get the molecular equation.

5"Sb"^"3+" + 2"MnO"_4^"-" + 12"H"_2"O" → 5"SbO"_4^"3-" + "2Mn"^"2+" + "24H"^"+"

We would have to add 2.5 "SO"_4^"2-" to match the charge on the 5"Sb"^"3+".

We multiply the ionic equation ×2 and then add the spectator ions.

10"Sb"^"3+" + 4"MnO"_4^"-" + 24"H"_2"O" → 10"SbO"_4^"3-" + "4Mn"^"2+" + "48H"^"+"

Let's rewrite this as

color(white)(m)10"Sb"^"3+" + 4"MnO"_4^"-" + 24"H"_2"O" → 10"H"_3"SbO"_4 + "4Mn"^"2+" + "18H"^"+"
"+ 15SO"_4^"2-" + "4K"^"+"color(white)(mmmmmmmmmmmmm) + "4SO"_4^"2-" + color(white)(ll)"9SO"_4^"2-"

The balanced molecular equation is

color(red)(5"Sb"_2("SO"_4)_3 + "4KMnO"_4 + "24H"_2"O" → "10H"_3"SbO"_4 + "2K"_2"SO"_4 + "4MnSO"_4 + "9H"_2"SO"_4)