# Question e437b

Apr 30, 2017

$\text{HOCH"_2"CH"_2"OH" + "10MnO"_4^"-" + "10OH"^"-" → "2CO"_2 + "10MnO"_4^"2-" + 8"H"_2"O}$

#### Explanation:

Let's rewrite the formula for ethylene glycol, $\text{HOCH"_2"CH"_2"OH}$, as ${\text{C"_2"H"_6"O}}_{2}$.

Then the skeleton equation is

$\text{C"_2"H"_6"O"_2 + "MnO"_4^"-" → "CO"_2 + "MnO"_4^"2-}$

One way to balance this equation is by the half-reaction method.

Step 1. Separate the equation into two half-reactions.

${\text{C"_2"H"_6"O"_2 → "CO}}_{2}$
$\text{MnO"_4^"-"→"MnO"_4^"2-}$

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

${\text{C"_2"H"_6"O"_2 → "2CO}}_{2}$
$\text{MnO"_4^"-"→"MnO"_4^"2-}$

Step 3. Balance $\text{O}$ by adding $\text{H"_2"O}$ molecules to the deficient side.

${\text{C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO}}_{2}$
$\text{MnO"_4^"-"→"MnO"_4^"2-}$

Step 4. Balance $\text{H}$ by adding $\text{H"^"+}$ ions to the deficient side.

$\text{C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2 + "10H"^"+}$
$\text{MnO"_4^"-"→"MnO"_4^"2-}$

Step 5. Balance charge by adding electrons to the deficient side.

$\text{C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2 + "10H"^"+" + "10e"^"-}$
$\text{MnO"_4^"-" + "e"^"-"→"MnO"_4^"2-}$

Step 6. Multiply each half-reaction by a number to equalize the electrons transferred.

1×["C"_2"H"_6"O"_2 + 2"H"_2"O" → "2CO"_2 + "10H"^"+" + "10e"^"-"]
10 × ["MnO"_4^"-" + "e"^"-"→"MnO"_4^"2-"]

Step 7. Add the two half-reactions.

$\text{C"_2"H"_6"O"_2 + "10MnO"_4^"-"+ 2"H"_2"O" → "2CO"_2 + "10MnO"_4^"2-" + "10H"^"+}$

Step 8. Convert to basic solution my adding the appropriate multiples of $\text{H"^"+" + "OH"^"-" → "H"_2"O}$.

"C"_2"H"_6"O"_2 + "10MnO"_4^"-"+ color(red)(cancel(color(black)("2H"_2"O"))) → "2CO"_2 + "10MnO"_4^"2-" + color(red)(cancel(color(black)("10H"^"+")))
color(red)(cancel(color(black)("10H"^"+"))) + 10"OH"^"-" → stackrelcolor(blue)(8)(color(red)(cancel(color(black)(10))))"H"_2"O"
stackrel(—————————————————————)("C"_2"H"_6"O"_2 + "10MnO"_4^"-" + "10OH"^"-" → "2CO"_2 + "10MnO"_4^"2-" + 8"H"_2"O")#

Step 9. Check that all atoms are balanced.

$\boldsymbol{\text{Atom"color(white)(m)"On the left"color(white)(m)"On the right}}$
$\textcolor{w h i t e}{m} \text{C} \textcolor{w h i t e}{m m m m m} 2 \textcolor{w h i t e}{m m m m m m m} 2$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m l} 16 \textcolor{w h i t e}{m m m m m m l l} 16$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m l} 52 \textcolor{w h i t e}{m m m m m m l l} 52$
$\textcolor{w h i t e}{m} \text{Mn} \textcolor{w h i t e}{m m m l l} 10 \textcolor{w h i t e}{m m m m m m l l} 10$

Step 10. Check that charge is balanced

$\boldsymbol{\text{On the left"color(white)(m)"On the right}}$
$\textcolor{w h i t e}{m m l} \text{20-"color(white)(mmmmmm)"20-}$

Everything checks! The equation is balanced.