aBetterChemText: Redox Reactions

Let’s balance the simple redox reaction here:

C r_{2} O_{7}^{- 2} + C l^{- 1} \to C r (O H)_{3} + C l_{2}

Once again, here are the steps for balancing redox reactions:

Break the reaction into half-reactions
Balance everything except H and O
Balance O by adding $H_{2} O$
Balance H by adding $H^{+ 1}$ ions
IF the reaction is in basic solution, add enough $O H^{- 1}$ ions to both sides to combine with the $H^{+ 1}$ ions
Balance the charge by adding electrons
Match the number of electrons between the two half-reactions
Add the half-reactions and cancel (if possible)

Now let’s apply those steps to our reaction.

Step 1 - Break the reaction into half-reactions

This is simple. We choose one thing on the left side of the reaction and match it with the obvious pair on the right side. Then repeat.

C r_{2} O_{7}^{- 2} \to C r (O H)_{3}

C l^{- 1} \to C l_{2}

Step 2 - Balance everything except H and O

For this reaction, that means making sure that the Cr and Cl are balanced. In this case, neither is. So we balance those elements, giving us:

C r_{2} O_{7}^{- 2} \to 2 C r (O H)_{3}

2 C l^{- 1} \to C l_{2}

Step 3 - Balance O by adding $H_{2} O$

Here, the chlorine reaction has no oxygen atoms, so there is nothing to do. The chromium reaction however has oxygen atoms. Since the

C r_{2} O_{7}^{- 2}

has 7 oxygens and the 2

C r (O H)_{3}

molecules have 6 oxygens between them, we will add 1

H_{2} O

molecule to the right side of the reaction. That will give us:

C r_{2} O_{7}^{- 2} \to 2 C r (O H)_{3} + H_{2} O

2 C l^{- 1} \to C l_{2}

Step 4 - Balance H by adding $H^{+ 1}$ ions

Again, the chlorine reaction has no hydrogen atoms, so we can ignore that reaction. There are 8 H atoms on the right side of the chromium reaction (thanks to the 2

C r (O H)_{3}

molecules and the water molecule we just added). So, we’ll add 8

H^{+ 1}

ions to the left side of that reaction. That will give us:

C r_{2} O_{7}^{- 2} + 8 H^{+ 1} \to 2 C r (O H)_{3} + H_{2} O

2 C l^{- 1} \to C l_{2}

Step 5 - IF the reaction is in basic solution, add enough $O H^{- 1}$ ions to both sides to combine with the $H^{+ 1}$ ions.

In this case our reaction is basic. There are two ways that we could know this. This first is that you could be told “this reaction is in basic solution.” The second way to know that a reaction is basic is to see hydroxide ions (

O H^{- 1}

) in the reaction. Since this reaction contains

C r (O H)_{3}

, we know that this is a basic reaction.

The chlorine reaction has no

H^{+ 1}

ions, so we don’t need to do anything to it. The chromium reaction has 8

H^{+ 1}

ions, so we will add 8

O H^{- 1}

ions to BOTH sides. We add to both sides to maintain the balance of O and H atoms that we had already achieved. This will give us:

C r_{2} O_{7}^{- 2} + 8 H^{+ 1} + 8 O H^{- 1} \to 2 C r (O H)_{3} + H_{2} O + 8 O H^{- 1}

2 C l^{- 1} \to C l_{2}

We know that

H^{+ 1} + O H^{- 1} \to H_{2} O

, so we can rewrite the chromium reaction to be:

C r_{2} O_{7}^{- 2} + 8 H_{2} O \to 2 C r (O H)_{3} + H_{2} O + 8 O H^{- 1}

We have

H_{2} O

on both sides of the reaction, so we can simplify by subtracting 1

H_{2} O

from each side, giving:

C r_{2} O_{7}^{- 2} + 7 H_{2} O \to 2 C r (O H)_{3} + 8 O H^{- 1}

Our two half-reactions now look like this:

C r_{2} O_{7}^{- 2} + 7 H_{2} O \to 2 C r (O H)_{3} + 8 O H^{- 1}

2 C l^{- 1} \to C l_{2}

Step 6 - Balance the charge by adding electrons

The left side of the chromium reaction has a charge of -2, while the right side has a charge of -8. To balance those, we’ll need to add 6 electrons to the left side.

C r_{2} O_{7}^{- 2} + 7 H_{2} O + 6 e^{- 1} \to 2 C r (O H)_{3} + 8 O H^{- 1}

The charges for the chlorine reaction are -2 and 0, so we’ll add two electrons to the right.

2 C l^{- 1} \to C l_{2} + 2 e^{- 1}

Step 7 - Match the number of electrons between the two half-reactions

The chromium reaction requires 6 electrons, but the chlorine reaction only gives 2 electrons. To make that work, we’ll multiply the chlorine reaction by 3. That gives us:

C r_{2} O_{7}^{- 2} + 7 H_{2} O + 6 e^{- 1} \to 2 C r (O H)_{3} + 8 O H^{- 1}

6 C l^{- 1} \to 3 C l_{2} + 6 e^{- 1}

Step 8 - Add the half-reactions and cancel (if possible)

C r_{2} O_{7}^{- 2} + 7 H_{2} O + 6 e^{- 1} + 6 C l^{- 1} \to 2 C r (O H)_{3} + 8 O H^{- 1} + 3 C l_{2} + 6 e^{- 1}

Then cancelling the electrons that appear on both sides, we get the final, balanced reaction:

C r_{2} O_{7}^{- 2} + 7 H_{2} O + 6 C l^{- 1} \to 2 C r (O H)_{3} + 8 O H^{- 1} + 3 C l_{2}

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Tuesday, July 13, 2021

Redox Reactions - Example 2

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