Chapter 5: Unit 3. Redox Reaction

Redox Reactions

Oxidation-Reduction or redox reactions are electron transfer reactions. Reactions in batteries and rusting are very common examples of redox reaction. Even in our body, digestion process involves redox reaction.

Oxidation results in the loss of electrons. Metals and anions tend to undergo loss of electrons. In some reactions oxidation involves addition of oxygen and removal of hydrogen. Reduction results in the gain of electrons. Nonmetals and cations tend to undergo reduction reaction. Reduction results in the loss of O atom and gain of H atoms.

A redox is a type of chemical reaction that involves a transfer of electrons between two species. Electron transfer occurs simultaneously, i.e., number of electrons gain is equal to number of electrons lost.

The reducing agent is the reactant that is being oxidized (and thus causing something else to be reduced), so in our case C is the reducing agent.

The oxidizing agent is reactant that is being reduced (and thus causing something else to be oxidized), so in our case Fe2O3, is the oxidizing agent.

Let’s see the following reaction:

In this reaction Al is oxidized from 0 to 3+, loss of electrons.Copper is reduced from 2+ to , gaining electrons. Al is reducing agent, copper is oxidizing agent. In general, higher the positive charge of  a substance means loss of electron and higher the negative charge of a substance means gain of electrons.

Let’s look at the reaction more in detail:

Reduction half-reaction:

The reduction half-reaction shows the reactants and products participating in the reduction step. Since Cu²⁺ is being reduced to Cu(s) we can write the following:

Cu²⁺(aq)→Cu(s)

Which is not charge-balanced. There is a net charge of +2, plus on the reactant side and 0 on the product side. We can balance the charges by including the electrons (which have negative charge) being transferred, and then we will get our reduction half-reaction right and balanced:

Cu²⁺(aq)+2e−→Cu(s)

now, if we add up the charges, including the electrons that have been transferred, we end up with an overall charge of 0 at both sides of the reductions half-reaction. The balanced half-reaction tells us that Cu²⁺  is gaining 2 electrons per copper atom to form solid Cu.

Let’s do the same with the other half of the reactions:

Oxidation half-reaction:  This reaction will include the oxidation of Al(s) to Al³⁺  . let’s add to the half reaction the electrons that have been transferred from Al(s):

Al(s)→Al³⁺(aq)+3e−

Now, we will combine the two balanced half-reactions to get the balanced overall equation. In doing so, we need to multiply the reduction half-reaction by 3 and multiply the oxidation half-reaction by 2 so both reactions involve the transfer of 6 electrons:

3×[Cu²⁺(aq)+2e−→Cu(s)]    3×reduction half-reaction

2×[Al(s)→Al³⁺(aq)+3e−]    2×oxidation half-reaction

The last step of this method involves adding the half reactions together to get our overall balanced equation,  and check to see if any reactants and products appear on both sides.

6e−+3Cu²⁺(aq) →   3Cu(s)                                      3×reduction half-reaction

2Al(s)        →2Al3+(aq)+6e−                                     2×oxidation half-reaction

2Al(s)+3Cu²⁺(aq)→2Al³⁺(aq)+3Cu(s)                    Overall balanced reaction

Another example:

For more example watch the following video:

Other example:

Mg(s) + 2H⁺(aq) – > ; Mg²⁺(aq) + H_2(g)

Zn(s) + MnO₄(aq) —>; ZnO(aq) + Mn₂;O₃(aq)( unbalanced)

Another approach to understand redox Reaction:

Oxidation state

The oxidation state is an important concept in chemistry, essential in order to understand the REDOX reactions. The oxidation state of an element corresponds to the number of electrons, that an atom loses or gains when joining with other atoms in compounds.

We can see the oxidation state also as the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present form or state.

Oxidation involves an increase in oxidation state

Reduction involves a decrease in oxidation state

To determine the oxidation state of an element, we can use the following rules:

• The oxidation state of an individual atom is 0. This is because it hasn’t been either oxidised or reduced yet!
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. The sum of the oxidation states of all the atoms in an ion is equal to the charge on the
• The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. (Fluorine is the most electronegative element followed by oxygen)
• Group 1 metals have an oxidation state of +1 and Group 2 an oxidation state of +2
• The oxidation state of fluorine is -1 in compounds
• Hydrogen generally has an oxidation state of +1 in compounds
• Oxygen generally has an oxidation state of -2 in compounds

In binary metal compounds, Group 17 elements have an oxidation state of -1, Group 16 elements of -2, and Group 15 elements of -3.

How redox work

Let’s get an example and understand better how redox reactions work:

How do we know if the above reaction is a redox? In order to be a redox, a transfer in electrons needs to happen, therefore we need to see if there is an electron transfer occurring, and we can do that by checking if any oxidation numbers change from the reactants to the products.

From the above figure, we can see that the oxidation numbers for carbon and iron are changing during the reaction from a transfer of electrons.

Carbon is being oxidized because it is losing electrons as the oxidation number increases from 0 to +4.
Iron is being reduced because it is gaining electrons as the oxidation number decreases from +3 to 0.

Questions:

1. Which substance is oxidized and which substance is reduced?
2. Ag + Cl → AgCl
3. 2SO₃ → 2SO₂ + O₂

Ans: a) Ag is oxidized and Cl is reduced

b) S is reduced and O is oxidized