Question

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

Answer 1

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)`

Answer 2

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)`