French Authors: Ryan Zanganeh

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1. Redox Reactions:

A redox reaction is a reaction that involves both oxidation and reduction. All three of these reactions can be described in terms of electron transfer and respective change in oxidation number.

In terms of electron transfer:

- Oxidation is the loss of electrons

- Reduction is the gain of electrons

In order to remember this, use the acronym “LEO the lion goes GER”.

- Lose Electrons; Oxidation

- Gain Electrons; Reduction

Consider the following reaction:

Cu(s) + ½ O2 → CuO

In order to determine whether or not a reaction is indeed a redox reaction, the oxidation number of each element in the reaction need to be determined.

Rules for assigning oxidation numbers:

- The oxidation number of an atom in its pure, elemental form is zero.

- The oxidation number of a monatomic ion is equal to the charge on that ion (Li+, F-, Na+).

- The algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero (NaCl).

- The algebraic sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion (CO32-).

- The most common oxidation number of hydrogen is +1; in hydrides, it’s -1.

- The most common oxidation number of oxygen is -2. (Exceptions: In peroxides, it’s -1. In combination with fluorine, it’s +2).

- The alkali metals have oxidation numbers of +1.

- The alkali earth metals have oxidation numbers of +2

- Aluminum has an oxidation number of +3

Using the rules in the following reaction, let’s assign the oxidation numbers:

Cu(s) + ½ O2 → CuO

- Cu(s) is an atom in its pure, elemental form (Hint: It’s in a solid state). Therefore, Cu(s) has an oxidation number of 0.

- Similarly, O2 is also an atom in its pure, elemental form. Therefore, O2 has an oxidation number of 0.

- In CuO, the oxygen can be assigned an oxidation number of -2 since it’s the most common oxidation number.

- If oxygen is -2, and the algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero, then the oxidation number of Cu is +2. (x – 2 = 0, x = +2)

Now that all oxidation numbers have been assigned, let’s look at the reaction again:

Cu(s) + ½ O2 → CuO

0 + 0  →  (+2)(-2)

Again, a redox reaction is a reaction that involves both oxidation and reduction. Let’s determine what’s being oxidized and what’s being reduced.

Notice in the reaction:

- Cu starts with an oxidation number of 0, and finishes with an oxidation number of +2. This indicates that Cu has lost 2 electrons; hence the +2. This also tells us that Cu is the element being oxidized in the reaction (LEO).

- Similarly, O2 starts with an oxidation number of 0, and finishes with an oxidation number of +2. This indicates that Oxygen has gained 2 electrons; hence the -2. This also tells us that O2 is the element being reduced in the reaction (GER).

Therefore, Cu is the element being oxidized and Oxygen is the element being reduced. Hence, the reaction is indeed a redox reaction.

2. Balancing Chemical Equations:

Redox reactions can be split into two half reactions. In one half reaction, oxidation occurs; electrons are produced. In the other half reaction, reduction occurs; electrons are consumed.

Consider the following redox reaction:

Mg(s) + Cl2(g) → MgCl2(s)

There are two half reactions occurring (it can also help to assign oxidation numbers to all elements):

Mg → Mg2+ + 2e-  which is the oxidation reaction (0 → +2, therefore LEO)

Cl2 + 2e- → 2Cl- which is the reduction reaction (0 → -2, therefore GER)

Since the sum of the oxidation numbers in a neutral polyatomic compound must equal zero, the redox reaction is balanced (+2 – 2 = 0).

Some reactions can also involve acidic solutions. There is a specific order to follow when balancing half reactions in acidic solutions:

1. Write the skeleton equation.

2. Balance the species other than oxygen and hydrogen.

3. Balance for oxygen using one water molecule for each oxygen you require.

4. Balance for hydrogen using a hydrogen ion for each hydrogen you require.

5. Balance for charge by adding electrons to either the product or reactant side.

Consider the following example: Balance the half reaction of ethanol (C2H5OH) going to ethanoic acid (CH3COOH) in acid solution.

Step 1: Write the skeleton equation.

C2H5OH → CH3COOH

Step 2: Balance the species other than oxygen and hydrogen. (No change)

C2H5OH → CH3COOH

Step 3: Balance for oxygen using one water molecule for each oxygen you require.

As shown in the skeleton reaction, there is one oxygen molecule in the reactants, and there are two oxygen molecules in the products. Therefore, one water molecule is added to the reactants side of the skeleton equation to balance the total number of oxygens.

C2H5OH + H2O  CH3COOH

As a result, both the reactants and products side have two oxygen atoms (equal number on both sides).

Step 4: Balance for hydrogen using a hydrogen ion for each hydrogen you require

If you count the total number of hydrogen atoms on each side of the reactant, the reactants have 8 hydrogen atoms, whereas the products have 4 hydrogen atoms. Therefore, 4 hydrogen ions are added to the products.

C2H5OH + H2O → CH3COOH + 4H+

As a result, both the reactants and products side have 8 hydrogen atoms (equal number on both sides).

Step 5: Balance for charge by adding electrons to either the product or reactant side.

Looking at the reactants side of the reaction, there is no charged molecule, therefore the charge of the reactants side remains zero. However, the products side has four H+, therefore the products side holds a charge of +4 (4 x +1). Hence, we add 4 electrons to balance.

C2H5OH + H2O → CH3COOH + 4H+ + 4e-

As a result of adding the 4 electrons to the products, the charge of the products side now also becomes zero; since electrons are negatively charged. (+4 – 4 = 0). The reactants and products have the same charge of zero, and therefore are balanced.

The final result is:

C2H5OH + H2O → CH3COOH + 4H+ + 4e-

Also, notice how this is also an oxidation reaction since 4e- are being lost (LEO).

Consider another example: Balance the half reaction of dichromate ion (Cr2O72-) going to chromium (III) ion (Cr3+).

Step 1: Write the skeleton equation.

Cr2O72- → Cr3+

Step 2: Balance the species other than oxygen and hydrogen.

Notice how in this reaction, there are two chromium atoms in the dichromate ion on the reactants side, whereas only one chromium ion on products side of the reaction. Therefore, we add the number 2 to the products side.

Cr2O72- → 2Cr3+

Now, both sides are balanced with the same number of chromium. (2 on each side.

Step 3: Balance for oxygen using one water molecule for each oxygen you require

There are 7 oxygens in the reactants side, and there are no oxygens in the product side of the reaction. Therefore, we add 7 water molecules to the products side.

Cr2O72- → 2Cr3+ + 7H2O

Now, both sides have equal number of oxygen. (7 on each side)

Step 4: Balance for hydrogen using a hydrogen ion for each hydrogen you require

On the reactants side, there are no hydrogens. On the products side, there are 7 water molecules with 2 hydrogen atoms per water molecule. Therefore, a total of 14 hydrogens on the products side. (7 x 2 = 14). There, we add 14 hydrogen ions to the reactants side of the reaction.

Cr2O72- + 14H+  2Cr3+ + 7H2O

Hence, now both sides have an equal number, 14, of hydrogens.

Step 5: Balance for charge by adding electrons to either the product or reactant side.

On the reactants side, there is one dichromate ion with a -2 charge, and 14 hydrogen ions with a charge of +1. Therefore, the charge of the reactants equal (1 x -2) + (14 x +1) = -2 + 14 = +12. On the products side, there are two chromium (III) ions with a +3 charge, and 7 water molecules with no charge; 0. Therefore the charge of the products equal (2 x +3) + (7 x 0) = +6

As shown mathematically, the reactants have an overall charge of +6 more than the products side. Therefore, we add 6 electrons to the reactants to balance. (+12 – 6 = +6)

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

As a result, both sides of the reaction have the same charge of +6. Therefore, the reaction is balanced.

The final result is:

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

Also, notice how this is also a reduction reaction since 6e- are being gained (GER).

To go even further, we have one oxidation half reaction, and one reduction half reaction. Let’s put the two together and make a redox reaction!

C2H5OH + H2O → CH3COOH + 4H+ + 4e- (Oxidation)

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O (Reduction)

In order to add the two reactions together, we must get rid of all the electrons. In order to do so, we must find the lowest common multiple (LCM) of the electrons on both the reactants and products side. In this case, we have 4 and 6. Therefore, the LCM is 12. As a result, we multiple the oxidation reaction by 2 and the reduction reaction by 3:

3 x [C2H5OH + H2O → CH3COOH + 4H+ + 4e-] (Oxidation)

2 x [Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O] (Reduction)

As a result, we get:

3C2H5OH + 3H2O → 3CH3COOH + 12H+ + 12e-

2Cr2O72- + 28H+ + 12e- → 4Cr3+ + 14H2O

Now, we can cancel the electrons since we have 12 electrons on both side. (12 – 12 = 0)

3C2H5OH + 3H2O → 3CH3COOH + 12H+ + 12e-

2Cr2O72- + 28H+ + 12e- → 4Cr3+ + 14H2O

We can continue to simplify for both the water molecules and the hydrogen ions: (14 – 3 = 11) & (28 – 12 = 16)

3C2H5OH + 3H2O → 3CH3COOH + 12H+

2Cr2O72- + 28H+ → Cr3+ + 14H2O

Therefore, the result is:

3C2H5OH → 3CH3COOH

2Cr2O72- + 16H+ → 4Cr3+ + 11H2O

Now, we can put the two reactions together; since nothing else can be simplified:

3C2H5OH + 2Cr2O72- + 16H+ → 3CH3COOH + 4Cr3+ + 11H2O

The complete redox reaction.

Now, just in case you’re worried you made a mistake. A trick you can always use to make sure you balanced correctly is to check the charge on each side of the reaction.

Reactants side: (3 x 0) + (2 x -2) + (16 x +1) = -4 + 16 = +12

Products side: (3 x 0) + (4 x +3) + (11 x 0) = +12

Therefore, both the reactants and products side have the same charge; +12. Hence, the redox reaction is correctly balanced.

3. Redox Reaction Rates:

We will be updating this section shortly!

4. Galvanic Cell:

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