Question #ddcdf

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)|)))#