# Question #e70fa

Feb 15, 2017

You got it right, then. REDOX usually only affects two elements.

#### Explanation:

NOT every element in the overall reaction will necessarily change its oxidation state. In this case, the carbon is shifting its bonds between oxygen and hydrogen, but its oxidation state does not change.

Feb 15, 2017

Here's how you do it.

#### Explanation:

You use the oxidation numbers of $\text{O}$ and $\text{H}$ to figure out the oxidation numbers of $\text{C}$.

$\stackrel{\textcolor{b l u e}{0}}{\text{C")_2 stackrelcolor(blue)("+1")("H")_4 stackrelcolor(blue)("-2")("O")_2 + stackrelcolor(blue)(0)("H")_2 → stackrelcolor(blue)("-2")("C")_2 stackrelcolor(blue)("+1")("H")_6 stackrelcolor(blue)("-2")("O")+ stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O}}$
$\stackrel{\textcolor{b l u e}{0}}{\textcolor{w h i t e}{\text{C"))color(white)(l)stackrelcolor(blue)("+4")(color(white)("H"))color(white)(ll)stackrelcolor(blue)("-4")(color(white)("O"))color(white)(mmmmm)stackrelcolor(blue)("-4")(color(white)("C"))color(white)(l)stackrelcolor(blue)("+6")(color(white)("H"))color(white)(ll)stackrelcolor(blue)("-2")(color(white)("O"))color(white)(ml)stackrelcolor(blue)("+2")(color(white)("H"))color(white)(ll)stackrelcolor(blue)("-2")(color(white)("O}}}$

You write the oxidation numbers of $\text{O}$ and $\text{H}$ above the atoms and their total oxidation number beneath them.

Thus, for ${\text{C"_2"H"_4"O}}_{2}$, the sum of their oxidation numbers is $+ 4 - 4 = 0$.

Hence, the sum of the oxidation numbers of the two $\text{C}$ atoms is zero, so their average oxidation number is zero.

Similarly, in $\text{C"_2"H"_6"O}$, the sum of the oxidation numbers of $\text{O}$ and $\text{H}$ is $+ 6 - 2 = + 4$, so the two $\text{C}$ atoms must add up to -4, or -2 for each $\text{C}$ atom.

We see that the oxidation number of $\text{C}$ decreases from 0 to -2, so the $\text{C}$ atoms are reduced.

Similarly, the oxidation number of $\text{H}$ increases from 0 in ${\text{H}}_{2}$ to +1 in water, so the hydrogen is oxidized.