Question ddcdf

Sep 25, 2016

Here's what I got.

Explanation:

For starters, ignore the hydroxide anions present on the reactants' side, they will come into play later.

The half-reaction given to you describes the oxidation of chromium from an oxidation state of $\textcolor{b l u e}{+ 3}$ on the reactants' side to an oxidation state of $\textcolor{b l u e}{+ 6}$ on the products' side.

stackrel(color(blue)(+3))("Cr")("OH")_ 3 -> stackrel(color(blue)(+6))("Cr")"O"_4^(2-) + 3"e"^(-)

Here one atom of chromium loses $3$ electrons to go from an oxidation satte of $\textcolor{b l u e}{+ 3}$ to an oxidation state of $\textcolor{b l u e}{+ 6}$.

Now, balance the oxygen atoms by adding water, $\text{H"_2"O}$, to the side that needs oxygen. In this case, you have $3$ oxygen atoms on the reactants' side and $4$ on the products' side, so add one molecule of water on the reactants' side

${\text{H"_ 2"O" + stackrel(color(blue)(+3))("Cr")("OH")_ 3 -> stackrel(color(blue)(+6))("Cr")"O"_4^(2-) + 3"e}}^{-}$

Now focus on the hydrogen atoms. You have a total of $5$ hydrogen atoms on the reactants' side and none on the products' side, so add $5$ protons, ${\text{H}}^{+}$, on the products' side

${\text{H"_ 2"O" + stackrel(color(blue)(+3))("Cr")("OH")_ 3 -> stackrel(color(blue)(+6))("Cr")"O"_4^(2-) + 3"e"^(-) + 5"H}}^{+}$

Because you're in basic medium, the protons will be neutralized by hydroxide anions, ${\text{OH}}^{-}$. To neutralize the five protons present on the products' side, add $5$ hydroxide anions.

To keep the equation balanced, add $5$ hydroxide anions to the reactants' side as well. You will have

5"OH"^(-) + "H"_ 2"O" + stackrel(color(blue)(+3))("Cr")("OH")_ 3 -> stackrel(color(blue)(+6))("Cr")"O"_4^(2-) + 3"e"^(-) + overbrace( 5"H"^(+) + 5"OH"^(-))^(color(darkgreen)(=5"H"_2"O"))#

The $5$ protons and the $5$ hydroxide anions will neutralize each other to form $5$ water molecules. Since you already have one water molecule on the reactants' side, you will be left with $4$ water molecules on the products' side

$5 \text{OH"^(-) + color(red)(cancel(color(black)("H"_ 2"O"))) + stackrel(color(blue)(+3))("Cr")("OH")_ 3 -> stackrel(color(blue)(+6))("Cr")"O"_4^(2-) + 3"e"^(-) + color(red)(cancel(color(black)(5)))^4"H"_2"O}$

Therefore, the balanced half-reaction will look like this

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5 {\text{OH"^(-) + "Cr"("OH")_ 3 -> "CrO"_ 4^(2-) + 4"H"_ 2"O" + 3"e}}^{-}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$