Chemistry (9620) | Oxidation, reduction and redox equations

Welcome to Oxidation and Reduction!

Hello! This chapter, Oxidation, reduction, and redox equations, is one of the most fundamental and important topics in chemistry. Why? Because redox reactions are everywhere—they power your phone battery, they cause iron to rust, and they are essential for biological processes like respiration.

Don't worry if the names sound intimidating! At its heart, a redox reaction is simply a chemical process involving the transfer of electrons. We will break down how to track these electrons and write the equations that describe this transfer. Let's get started!

3.1.5 Oxidation, Reduction, and Redox Equations

1. Defining Oxidation and Reduction

A redox reaction is a reaction where both oxidation and reduction occur simultaneously. You cannot have one without the other—it's like a transaction where one party must lose something (electrons) and another must gain it.

The Key Definitions (The "OIL RIG" Trick!)

To help you remember the definitions, use this classic mnemonic:

• O I L: Oxidation Is Loss of electrons.
• R I G: Reduction Is Gain of electrons.

Oxidation is the process where a species loses electrons, resulting in an increase in its oxidation state.

Reduction is the process where a species gains electrons, resulting in a decrease in its oxidation state.

Oxidising Agents and Reducing Agents

In a redox reaction, we also talk about the agents involved:

• Oxidising Agent (or Oxidant): This is the chemical species that accepts electrons. Because it gains electrons, the oxidising agent itself is reduced.
• Reducing Agent (or Reductant): This is the chemical species that donates electrons. Because it loses electrons, the reducing agent itself is oxidised.

Analogy: Think of a bank robbery! The electron is the money. The person losing the money (the one being oxidised) is the store clerk, and the one taking the money (the oxidising agent) is the robber, who is "reduced" to taking the money.

Key Takeaway: Oxidation is electron loss (increase in state); Reduction is electron gain (decrease in state). The oxidising agent gets reduced, and the reducing agent gets oxidised.

2. Working Out Oxidation States

The most reliable way to track whether an element has been oxidised or reduced is by calculating its oxidation state (sometimes called oxidation number). The oxidation state is the hypothetical charge an atom would have if all its bonds were 100% ionic.

Rules for Assigning Oxidation States

Follow these rules in order of priority. (Don't worry, these rules are mostly common sense!)

1. Pure Elements: The oxidation state of an uncombined element is always 0.
   Example: \(Zn\), \(O_2\), \(S_8\).
2. Ions: The oxidation state of a simple ion is equal to its charge.
   Example: \(Na^+\) is +1, \(Cl^-\) is -1, \(Cu^{2+}\) is +2.
3. Group 1 and 2:
   • Group 1 metals (like Na, K) are always +1.
   • Group 2 metals (like Mg, Ca) are always +2.
4. Fluorine: Fluorine (F) is the most electronegative element and is always -1.
5. Hydrogen: Hydrogen (H) is usually +1, except when it is bonded to a metal (a metal hydride, e.g., \(NaH\)), where it is -1.
6. Oxygen: Oxygen (O) is usually -2. The exceptions are:
   • In peroxides (like \(H_2O_2\)), O is -1.
   • When bonded to fluorine (\(OF_2\)), O is +2.
7. Neutral Compounds: The sum of the oxidation states in a neutral compound must equal zero.
8. Polyatomic Ions: The sum of the oxidation states in a polyatomic ion must equal the charge of the ion.

Step-by-Step Example: Finding the Oxidation State of Manganese in Permanganate (\(MnO_4^-\))

1. We know the total charge is -1 (Rule 8).
2. We know Oxygen is usually -2 (Rule 6). There are four Oxygens.
3. Let the oxidation state of Manganese (Mn) be \(x\).
4. Set up the equation: \(x + 4 \times (-2) = -1\)
5. Solve for \(x\): \(x - 8 = -1\). Therefore, \(x = +7\).

The oxidation state of Manganese in the permanganate ion, \(MnO_4^-\), is +7.

3. Writing Half-Equations

Redox reactions are complex, so we often separate them into two hypothetical steps called half-equations: one for oxidation and one for reduction. These half-equations show the movement of electrons explicitly.

Step-by-Step Guide: Writing Half-Equations in Acidic Solution

This process is crucial for combining reactions later on. We usually assume the reaction happens in an acidic solution, meaning we can use \(H^+\) and \(H_2O\) to balance the equation.

1. Balance the main atom: Balance all atoms *except* oxygen (O) and hydrogen (H).
2. Balance Oxygen (O): Add \(H_2O\) molecules to the side that needs oxygen.
3. Balance Hydrogen (H): Add \(H^+\) ions to the side that needs hydrogen.
4. Balance Charge (Electrons, \(e^-\)): Add electrons (\(e^-\)) to the side with the more positive total charge to make the charges equal on both sides.

Example: Reduction of Dichromate ion (\(Cr_2O_7^{2-} \rightarrow Cr^{3+}\))

1. Balance Cr: \(\boldsymbol{Cr_2O_7^{2-}} \rightarrow \boldsymbol{2Cr^{3+}}\)
2. Balance O (add \(H_2O\)): \(Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + \boldsymbol{7H_2O}\)
3. Balance H (add \(H^+\)): \(\boldsymbol{14H^+} + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O\)
4. Balance Charge:
   • Left side charge: \(14(+1) + (-2) = +12\)
   • Right side charge: \(2(+3) + 0 = +6\)
   • Need to add 6 electrons to the left side (the reduction side):
   \[6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O\]

Since electrons are gained, this confirms it is a reduction half-equation.

4. Combining Half-Equations into a Full Redox Equation

The final step in describing a redox reaction is combining the separate oxidation and reduction half-equations into one overall equation.

The crucial principle here is that electrons lost must equal electrons gained. The electrons must cancel out completely in the final equation.

Step-by-Step Guide: Combining Half-Equations

1. Write the balanced half-equations for the oxidation and the reduction processes.
2. Multiply one or both half-equations by an integer so that the number of electrons lost equals the number of electrons gained.
3. Add the two equations together and cancel out any species (like electrons, \(H^+\), or \(H_2O\)) that appear on both sides of the arrow.

Example: Reaction between Zinc metal (Zn) and Copper ions (\(Cu^{2+}\))

A. Oxidation (Loss): Zinc metal loses two electrons to form a zinc ion.

\[Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\]

B. Reduction (Gain): Copper ions gain two electrons to form copper metal.

\[Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\]

C. Combine: Since both equations involve 2 electrons, we can simply add them and cancel the electrons.

\[Zn(s) + Cu^{2+}(aq) + \cancel{2e^-} \rightarrow Zn^{2+}(aq) + \cancel{2e^-} + Cu(s)\]

Overall Redox Equation:

\[Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\]

Did you know? Redox reactions are the basis of corrosion (like rust). Rusting occurs when iron is oxidised (loses electrons) and oxygen is reduced (gains electrons), often accelerated by water and salt!

A Harder Example: Combining Magnesium (Oxidation) and Dichromate (Reduction)

Let’s use the dichromate reduction from earlier and the oxidation of Magnesium (\(Mg \rightarrow Mg^{2+}\)):

1. Half-Equations:

Oxidation: \[Mg \rightarrow Mg^{2+} + \boldsymbol{2e^-}\]

Reduction: \[\boldsymbol{6e^-} + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O\]

2. Equalise Electrons:

The electrons need to be 6 on both sides. We must multiply the oxidation equation by 3.

New Oxidation: \[3Mg \rightarrow 3Mg^{2+} + \boldsymbol{6e^-}\]

3. Combine and Cancel:

\[3Mg + \cancel{6e^-} + 14H^+ + Cr_2O_7^{2-} \rightarrow 3Mg^{2+} + \cancel{6e^-} + 2Cr^{3+} + 7H_2O\]

Overall Redox Equation:

\[3Mg + 14H^+ + Cr_2O_7^{2-} \rightarrow 3Mg^{2+} + 2Cr^{3+} + 7H_2O\]

This step-by-step method ensures you account for all atoms and, most importantly, the flow of electrons!

Key Takeaway: Half-equations explicitly show electron transfer. When combining them, ensure the total number of electrons lost in oxidation exactly matches the total number of electrons gained in reduction. This is the law of conservation of charge!