# Question 065fa

Nov 21, 2017

Here's what I got.

#### Explanation:

In the redox reaction given to you, the dichromate anions are being reduced to chromium(III) cations and the iron(II) cations are being oxidized to iron(III) cations.

Chromium has a $\textcolor{b l u e}{+ 6}$ oxidation number in the dichromate anions, but I won't go into why that is the case because I assume that you're familiar with how oxidation numbers work.

The fact that you have hydroxide anions present on the products' side tells you that this redox reaction takes place in basic medium.

$\textcolor{w h i t e}{\frac{a}{a}}$
So, the oxidation half-reaction looks like this

stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + "e"^(-)

Notice that the half-reaction is balanced in terms of charge because you have

$\left(2 +\right) \to \left(3 +\right) + \left(1 -\right)$

$\textcolor{w h i t e}{\frac{a}{a}}$
The reduction half-reaction looks like this

stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+)

Here each atom of chromium gains $3$ electrons to go from an oxidation state of $\textcolor{b l u e}{+ 6}$ on the reactants' side to an oxidation state of $\textcolor{b l u e}{+ 3}$ on the products' side, so two atoms of chromium will take in $6$ electrons.

In order to balance out the atoms of oxygen, add water to the side that needs oxygen and hydrogen cations, ${\text{H}}^{+}$, to the other side.

You will have

$14 {\text{H"_ ((aq))^(+) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 7"H"_ 2"O}}_{\left(l\right)}$

Now, because you're in basic medium, you need to add hydroxide anions to both sides of the equation in order to neutralize the added protons.

This means that you have

overbrace(14"OH"_ ((aq))^(-) + 14"H"_ ((aq))^(+))^(color(red)(=14"H"_ 2"O")) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 7"H"_ 2"O"_ ((l)) + 14"OH"_ ((aq))^(-)

which is equivalent to

$\stackrel{\textcolor{red}{7}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{14}}}} {\text{H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + color(red)(cancel(color(black)(7"H"_ 2"O"_ ((l))))) + 14"OH}}_{\left(a q\right)}^{-}$

You can simplify the half-reaction to

$7 {\text{H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 14"OH}}_{\left(a q\right)}^{-}$

Notice that the half-reaction is balanced in terms of charge because you have

$\left(2 -\right) + 6 \times \left(1 -\right) = 2 \times \left(3 +\right) + 14 \times \left(1 -\right)$

$\textcolor{w h i t e}{\frac{a}{a}}$
So there you have it, the two balanced half-reactions are

 {(color(white)(aaaaaaaaaaaaaaaaaa)stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + "e"^(-)), (7"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)) :}

$\textcolor{w h i t e}{\frac{a}{a}}$
To get the balanced chemical equation, you need to use the fact that in every redox reactions, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

So you would multiply the oxidation half-reaction by $6$ and add the two half-reactions to get

 {(color(white)(aaaaaaaaaaaaaaaaa)6stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> 6stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + 6"e"^(-)), (7"H"_ 2"O"_ ((l)) + stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 6"e"^(-)-> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)) :}
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 7"H"_ 2"O"_ ((l)) + 6stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) + color(red)(cancel(color(black)(6"e"^(-)))) -> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + color(red)(cancel(color(black)(6"e"^(-)))) + 6stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)

which is exactly what you started with

stackrel(color(blue)(+6))("Cr")_ 2"O"_ (7(aq))^(2-) + 7"H"_ 2"O"_ ((l)) + 6stackrel(color(blue)(+2))("Fe") ""_ ((aq))^(2+) -> 2stackrel(color(blue)(+3))("Cr")""_ ((aq))^(3+) + 6stackrel(color(blue)(+3))("Fe") ""_ ((aq))^(3+) + 14"OH"_ ((aq))^(-)#