Question

Question #ddcdf

Answer 1

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 `color(blue)(+3)` on the reactants' side to an oxidation state of `color(blue)(+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 `color(blue)(+3)` to an oxidation state of `color(blue)(+6)`.

Now, balance the oxygen atoms by adding water, `"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

`"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, `"H"^(+)`, on the products' side

`"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, `"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"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

`color(green)(bar(ul(|color(white)(a/a)color(black)(5"OH"^(-) + "Cr"("OH")_ 3 -> "CrO"_ 4^(2-) + 4"H"_ 2"O" + 3"e"^(-))color(white)(a/a)|)))`