# How do you write a balanced chemical equation for the fermentation of sucrose (C_12H_22O_11) by yeasts in which the aqueous sugar reacts with water to form aqueous ethyl alcohol (C_2H_5OH) and carbon dioxide gas?

Feb 19, 2017

WARNING! Long answer! The balanced equation is

${\text{C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO}}_{2}$

#### Explanation:

This is a reduction-oxidation reaction.

You can find the general technique for balancing redox equations here.

We can use the method of oxidation numbers to balance this equation.

${\text{C"_12"H"_22"O"_11 + "H"_2"O" → "C"_2"H"_5"OH" + "CO}}_{2}$

Step 1. Identify the atoms that change oxidation number

Determine the oxidation numbers of every atom in the equation.

${\stackrel{\textcolor{b l u e}{0}}{\text{C")_12stackrelcolor(blue)("+1")("H")_22stackrelcolor(blue)("-2")("O")_11 + color(white)(l)stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-2")("C")_2stackrelcolor(blue)("+1")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-2")("O}}}_{2}$
$\textcolor{w h i t e}{{\stackrel{\textcolor{b l u e}{0}}{\text{C")_12stackrelcolor(blue)("+22")("H")_22stackrelcolor(blue)("-22")("O")_11 + stackrelcolor(blue)("+2")("H")_2stackrelcolor(blue)("-2")("O") → stackrelcolor(blue)("-4")("C")_2stackrelcolor(blue)("+5")("H")_5stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H") + stackrelcolor(blue)("+4")("C")stackrelcolor(blue)("-4")("O}}}_{2}}$

We see that the oxidation number of $\text{C}$ in sucrose is reduced to -2 in $\text{C"_2"H"_5"OH}$ and increased to +4 in ${\text{CO}}_{2}$.

This is a disproportionation reaction.

The changes in oxidation number are:

$\text{C: 0 → -2";color(white)(l) "Change ="color(white)(m) "-2 (oxidation)}$
$\text{C: 0 → +4"; "Change ="color(white)(l) "+4 (reduction)}$

Step 2. Equalize the changes in oxidation number

We need 2 atoms of $\text{C}$ that become ethanol for every 1 atom of $\text{C}$ that becomes ${\text{CO}}_{2}$.

We must also have a total of 12 $\text{C}$ atoms from the sucrose.

That means we need 1 molecule of sucrose, 4 of ethanol, and 4 of ${\text{CO}}_{2}$.

Step 3. Insert coefficients to get these numbers

$\textcolor{red}{1} {\text{C"_12"H"_22"O"_11 + "H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO}}_{2}$

Step 4. Balance $\text{O}$

We have fixed 12 $\text{O}$ atoms on the right and 11 $\text{O}$ atoms on the right, so we need 1 more $\text{O}$ atoms on the left. Put a 1 before $\text{H"_2"O}$.

$\textcolor{red}{1} {\text{C"_12"H"_22"O"_11 + color(blue)(1)"H"_2"O" → color(red)(4)"C"_2"H"_5"OH" + color(red)(4)"CO}}_{2}$

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

Step 5. Check that all atoms are balanced.

$\boldsymbol{\text{On the left"color(white)(l) "On the right}}$
$\textcolor{w h i t e}{m m} \text{12 C"color(white)(mmmml) "12 C}$
$\textcolor{w h i t e}{m m} \text{24 H"color(white)(mmmml) "24 H}$
$\textcolor{w h i t e}{m m} \text{12 O"color(white)(mmmml) "12 O}$

The balanced equation is

$\textcolor{b l u e}{{\text{C"_12"H"_22"O"_11 + "H"_2"O" → "4C"_2"H"_5"OH" + "4CO}}_{2}}$