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

Answer:

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

Answer:

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)#