# How to balance the following redox problems using both methods?

## K2Cr2O7 + HI => CrI3 + KI + I2 + HOH K2Cr2O7 + H2C2O4 + HCl => CrCl3 + CO2 + KCl + HOH Sb2(SO4)3 + KMnO4 + HOH => H3SbO4 + K2SO4 + MnSO4 + H2SO4 Mn(NO3)2 + NaBiO3 + HNO3 => Bi(NO3)2 + HMnO4 + NaNO3 + HOH

Jan 21, 2017

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

#### Explanation:

${\text{Sb"_2("SO"_4)_3 + "KMnO"_4 + "HOH" → "H"_3"SbO"_4 + "K"_2"SO"_4 + "MnSO"_4 + "H"_2"SO}}_{4}$

Step 1. Identify the atoms that change oxidation number

$\stackrel{\textcolor{b l u e}{\text{+3")("Sb")_2(stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4)_3 + stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")("Mn")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") → stackrelcolor(blue)("+1")("H")_3stackrelcolor(blue)("+5")("Sb")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("K")_2stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+2")("Mn")stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O")_4 + stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("+6")("S")stackrelcolor(blue)("-2")("O}}}{_} 4$

The changes in oxidation number are:

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

Step 2. Equalize the changes in oxidation number

We need 5 atoms of $\text{Sb}$ for every 2 atoms of $\text{Mn}$ or 10 atoms of $\text{Sb}$ for every 4 atoms of $\text{Mn}$. This gives us total changes of +20 and -20.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{5} {\text{Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + "HOH" → color(red)(10)"H"_3"SbO"_4 + "K"_2"SO"_4 + color(red)(4)"MnSO"_4 + "H"_2"SO}}_{4}$

Step 4. Balance $\text{K}$

We have fixed 4 $\text{K}$ on the left, so we need 4 $\text{K}$ on the right. Put a 2 in front of ${\text{K"_2"SO}}_{4.}$

$\textcolor{red}{5} {\text{Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + "HOH" → color(red)(10)"H"_3"SbO"_4 + color(blue)(2)"K"_2"SO"_4 + color(red)(4)"MnSO"_4 + "H"_2"SO}}_{4}$

Step 5. Balance $\text{S}$

We have fixed 15 $\text{S}$ atoms on the left and 6 $\text{S}$ atoms on the right, so we need 9 more $\text{S}$ atoms on the right. Put a 9 before ${\text{H"_2"SO}}_{4}$.

$\textcolor{red}{5} {\text{Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + "HOH" → color(red)(10)"H"_3"SbO"_4 + color(blue)(2)"K"_2"SO"_4 + color(red)(4)"MnSO"_4 + color(purple)(9)"H"_2"SO}}_{4}$

Step 6. Balance $\text{O}$

We have fixed 76 $\text{O}$ atoms on the left and 100 $\text{O}$ atoms on the right, so we need 24 more $\text{O}$ atoms on the left. Put a 24 before $\text{HOH}$.

$\textcolor{red}{5} {\text{Sb"_2("SO"_4)_3 + color(red)(4)"KMnO"_4 + color(brown)(24)"HOH" → color(red)(10)"H"_3"SbO"_4 + color(blue)(2)"K"_2"SO"_4 + color(red)(4)"MnSO"_4 + color(purple)(9)"H"_2"SO}}_{4}$

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

Step 7. Check that all atoms are balanced.

$m a t h b f \left(\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}\right)$
$\textcolor{w h i t e}{m l} \text{Sb} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m m m} 10$
$\textcolor{w h i t e}{m l} \text{S} \textcolor{w h i t e}{m m m m m} 15 \textcolor{w h i t e}{m m m m m m m} 15$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{m m m m l} 100 \textcolor{w h i t e}{m m m m m m l} 100$
$\textcolor{w h i t e}{m l} \text{K} \textcolor{w h i t e}{m m m m m l} 4 \textcolor{w h i t e}{m m m m m m m l l} 4$
$\textcolor{w h i t e}{m l} \text{Mn} \textcolor{w h i t e}{m m m m l l} 4 \textcolor{w h i t e}{m m m m m m m l l} 4$
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{m m m m m} 48 \textcolor{w h i t e}{m m m m m m m} 48$

The balanced equation is

$\textcolor{red}{5 {\text{Sb"_2("SO"_4)_3 + 4"KMnO"_4 + 24"HOH" →10"H"_3"SbO"_4 + 2"K"_2"SO"_4 + 4"MnSO"_4 +9"H"_2"SO}}_{4}}$

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

${\text{Sb"_2("SO"_4)_3 + "KMnO"_4 + "HOH" → "H"_3"SbO"_4 + "K"_2"SO"_4 + "MnSO"_4 + "H"_2"SO}}_{4}$

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.

$\text{Sb"^"3+" + "SO"_4^"2-" + "K"^"+" + "MnO"_4^"-" + "HOH" → "H"^"+" + "SbO"_4^"3-" + "K"^"+" + "SO"_4^"2-" + "Mn"^"2+" + "SO"_4^"2-" + "H"^"+" + "SO"_4^"2-}$

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

We also eliminate $\text{H"^"+}$ and $\text{H"_2"O}$, 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

$\text{Sb"^"3+" + "MnO"_4^"-" → "SbO"_4^"3-" + "Mn"^"2+}$

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

Now we are ready to balance the equation.

Step 1: Write the two half-reactions.

$\text{Sb"^"3+" → "SbO"_4^"3-}$
$\text{MnO"_4^"-" → "Mn"^"2+}$

Step 2: Balance all atoms other than $\text{H}$ and $\text{O}$.

Done.

Step 3: Balance $\text{O}$.

$\text{Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-}$
$\text{MnO"_4^"-" → "Mn"^"2+" + "4H"_2"O}$

Step 4: Balance $\text{H}$.

$\text{Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-" + "8H"^"+}$
$\text{MnO"_4^"-" + "8H"^"+" → "Mn"^"2+" + "4H"_2"O}$

Step 5: Balance charge.

$\text{Sb"^"3+" + "4H"_2"O" → "SbO"_4^"3-" + "8H"^"+" + "2e"^"-}$
$\text{MnO"_4^"-" + "8H"^"+" + "5e"^"-" → "Mn"^"2+" + "4H"_2"O}$

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")))
$5 \text{Sb"^"3+" + 2"MnO"_4^"-" + 12"H"_2"O" → 5"SbO"_4^"3-" + "2Mn"^"2+" + "24H"^"+}$

Step 8: Check mass balance.

$m a t h b f \left(\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}\right)$
$\textcolor{w h i t e}{m l} \text{Sb} \textcolor{w h i t e}{m m m m l} 5 \textcolor{w h i t e}{m m m m m m m} 5$
$\textcolor{w h i t e}{m l} \text{Mn} \textcolor{w h i t e}{m m m m} 2 \textcolor{w h i t e}{m m m m m m m} 2$
$\textcolor{w h i t e}{m l} \text{O} \textcolor{w h i t e}{m m m m l} 20 \textcolor{w h i t e}{m m m m m m l} 20$
$\textcolor{w h i t e}{m l} \text{H} \textcolor{w h i t e}{m m m m l} 24 \textcolor{w h i t e}{m m m m m m l} 24$

Step 9: Check charge balance.

$m a t h b f \left(\text{On the left"color(white)(mml)"On the right}\right)$
$\text{15 - 2 = +13"color(white)(m)"-15 + 4 + 24 = +13}$

The ionic equation is balanced.

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

$5 \text{Sb"^"3+" + 2"MnO"_4^"-" + 12"H"_2"O" → 5"SbO"_4^"3-" + "2Mn"^"2+" + "24H"^"+}$

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

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

$10 \text{Sb"^"3+" + 4"MnO"_4^"-" + 24"H"_2"O" → 10"SbO"_4^"3-" + "4Mn"^"2+" + "48H"^"+}$

Let's rewrite this as

$\textcolor{w h i t e}{m} 10 \text{Sb"^"3+" + 4"MnO"_4^"-" + 24"H"_2"O" → 10"H"_3"SbO"_4 + "4Mn"^"2+" + "18H"^"+}$
$\text{+ 15SO"_4^"2-" + "4K"^"+"color(white)(mmmmmmmmmmmmm) + "4SO"_4^"2-" + color(white)(ll)"9SO"_4^"2-}$

The balanced molecular equation is

$\textcolor{red}{5 {\text{Sb"_2("SO"_4)_3 + "4KMnO"_4 + "24H"_2"O" → "10H"_3"SbO"_4 + "2K"_2"SO"_4 + "4MnSO"_4 + "9H"_2"SO}}_{4}}$