R3.2.2 Redox half-equations

Writing Oxidation and Reduction Half-Equations

What Are Half-Equations, and Why Do We Use Them?

Definition

Half-equation in a redox reaction

A half-equation focuses on one part of a redox reaction: either the oxidation or reduction process.

By breaking a redox reaction into these two components, we can better track the movement of electrons.

Example

Let’s take the reaction between zinc and copper(II) sulfate as an example:

Full Reaction:

$Zn (s) + {Cu}^{2 +} (a q) \to {Zn}^{2 +} (a q) + Cu (s)$

This reaction involves two simultaneous processes:

Zinc metal ( $Zn$ ) is oxidized to form zinc ions ( ${Zn}^{2 +}$ ).
Copper ions ( ${Cu}^{2 +}$ ) are reduced to form copper metal ( $Cu$ ).

Writing Oxidation and Reduction Half-Equations

Step-by-Step Guide:

Identify the species being oxidized and reduced:
- Oxidation: The species losing electrons.
- Reduction: The species gaining electrons.
Write the unbalanced equation for each process:
- For oxidation, place the electrons ( $e^{-}$ ) on the right-hand side of the equation.
- For reduction, place the electrons ( $e^{-}$ ) on the left-hand side.
Balance the atoms:
- Ensure the number of atoms of each element is the same on both sides of the equation.
Balance the charges:
- Add electrons to one side of the equation to balance the overall charge.
Combine the half-equations:
- Ensure the number of electrons lost in oxidation equals the number of electrons gained in reduction.
- Add the two half-equations together, canceling out the electrons.

Example

Zinc and Copper(II) Sulfate Reaction

Step 1: Identify the species being oxidized and reduced

Zinc( $Zn$ ) is oxidized: it loses electrons to form ${Zn}^{2 +}$ .
Copper ions( ${Cu}^{2 +}$ ) are reduced: they gain electrons to form $Cu$ .

Step 2: Write the unbalanced half-equations

Oxidation: $Zn \to {Zn}^{2 +}$
Reduction: ${Cu}^{2 +} \to Cu$

Step 3: Balance the atoms

Both equations already have balanced atoms.

Step 4: Balance the charges

Oxidation: $Zn \to {Zn}^{2 +} + 2 e^{-}$
Reduction: ${Cu}^{2 +} + 2 e^{-} \to Cu$

Step 5: Combine the half-equations

Add the two half-equations together:
$Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu$
The electrons cancel out, and the overall reaction is balanced.

Balancing Half-Equations in Acidic or Neutral Solutions

When redox reactions occur in aqueous solutions, water ( $H_{2} O$ ), hydrogen ions ( $H^{+}$ ), or hydroxide ions ( ${OH}^{-}$ ) may need to be included to balance oxygen and hydrogen atoms in the half-equations.

Steps for Acidic Solutions:

Balance all atoms except hydrogen and oxygen.
Balance oxygen atoms by adding $H_{2} O$ molecules.
Balance hydrogen atoms by adding $H^{+}$ ions.
Balance the charges by adding electrons ( $e^{-}$ ).

Example

Reduction of Dichromate Ions ( ${Cr}_{2} O_{7}^{2 -}$ ) in Acidic Solution

Step 1: Write the unbalanced equation

${Cr}_{2} O_{7}^{2 -} \to {Cr}^{3 +}$

Step 2: Balance chromium atoms

${Cr}_{2} O_{7}^{2 -} \to 2 {Cr}^{3 +}$

Step 3: Balance oxygen atoms by adding $H_{2} O$

${Cr}_{2} O_{7}^{2 -} \to 2 {Cr}^{3 +} + 7 H_{2} O$

Step 4: Balance hydrogen atoms by adding $H^{+}$

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} \to 2 {Cr}^{3 +} + 7 H_{2} O$

Step 5: Balance the charges by adding electrons

Left-hand side: $2 -$ (from ${Cr}_{2} O_{7}^{2 -}$ ) + $14 +$ (from $H^{+}$ ) = $12 +$
Right-hand side: $6 +$ (from $2 {Cr}^{3 +}$ )
Add 6 electrons to the left-hand side:
${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$

Tip

When balancing half-equations in acidic solutions, always use $H_{2} O$ to balance oxygen and $H^{+}$ to balance hydrogen.

Steps for Neutral Solutions:

Balance all atoms except hydrogen and oxygen.
Balance oxygen atoms by adding $H_{2} O$ .
Balance hydrogen atoms by adding $H_{2} O$ or $O H^{-}$ .
Balance charges by adding electrons ( $e^{-}$ ).

Example

Reduction of Manganese Dioxide ( $M n O_{2}$ ) in Neutral Solution

Step 1: Write the unbalanced equation
$M n O_{2} (s) \to M n^{2 +} (a q)$

Step 2: Balance manganese atoms
$M n O_{2} (s) \to M n^{2 +} (a q)$
(No changes needed as manganese is already balanced.)

Step 3: Balance oxygen atoms by adding $H_{2} O$
$M n O_{2} (s) \to M n^{2 +} (a q) + 2 H_{2} O (l)$

Step 4: Balance hydrogen atoms by adding $O H^{-}$
$M n O_{2} (s) + 4 H^{+} (a q) \to M n^{2 +} (a q) + 2 H_{2} O (l)$

Step 5: Balance charges by adding electrons

Left side: $- 4$ (from $4 O H^{-}$ )
Right side: $+ 2$ (from $M n^{2 +}$ )
Add 2 electrons to the right side:

$M n O_{2} (s) + 4 H^{+} (a q) + 2 e^{-} \to M n^{2 +} (a q) + 2 H_{2} O (l) +$

Applications of Half-Equations

1. Electrochemical Cells

Half-equations describe the reactions at the anode (oxidation) and cathode (reduction) in batteries and fuel cells. For example:

In a zinc-copper voltaic cell:
Anode: $Zn \to {Zn}^{2 +} + 2 e^{-}$
Cathode: ${Cu}^{2 +} + 2 e^{-} \to Cu$

2. Corrosion

The rusting of iron involves redox reactions:

Oxidation: $Fe \to {Fe}^{2 +} + 2 e^{-}$
Reduction: $O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O$

3. Environmental Chemistry

Redox reactions are critical in processes like water purification and the breakdown of pollutants.

Common Mistake

Forgetting to balance charges: Ensure the total charge is the same on both sides of the half-equation.
Ignoring the solution type: Use $H^{+}$ for acidic solutions and ${OH}^{-}$ for basic solutions.
Not canceling electrons: When combining half-equations, the number of electrons must match.

Reflection

Self review

Write the oxidation and reduction half-equations for the reaction between magnesium and hydrochloric acid.
Balance the following half-equation in acidic solution: ${MnO}_{4}^{-} \to {Mn}^{2 +}$ .
Why is it impossible to have oxidation without reduction in a redox reaction?

Theory of Knowledge

How do different definitions of oxidation and reduction (electron transfer, oxidation state, oxygen gain/loss) help us understand redox reactions in various contexts?
For example, how might the concept of electron transfer apply to both chemical batteries and biological energy systems like cellular respiration?

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R3.2.2 Redox half-equations

Writing Oxidation and Reduction Half-Equations

What Are Half-Equations, and Why Do We Use Them?

Full Reaction:

Writing Oxidation and Reduction Half-Equations

Step-by-Step Guide:

Zinc and Copper(II) Sulfate Reaction

Step 1: Identify the species being oxidized and reduced

Step 2: Write the unbalanced half-equations

Step 3: Balance the atoms

Step 4: Balance the charges

Step 5: Combine the half-equations

Balancing Half-Equations in Acidic or Neutral Solutions

Steps for Acidic Solutions:

Reduction of Dichromate Ions (Cr2O72−) in Acidic Solution

Step 1: Write the unbalanced equation

Step 2: Balance chromium atoms

Step 3: Balance oxygen atoms by adding H2O

Step 4: Balance hydrogen atoms by adding H+

Step 5: Balance the charges by adding electrons

Steps for Neutral Solutions:

Reduction of Manganese Dioxide (MnO2) in Neutral Solution

Applications of Half-Equations

1. Electrochemical Cells

2. Corrosion

3. Environmental Chemistry

Reflection

You've read 2/2 free chapters this week.

Introduction to Redox Half-Equations

Reduction of Dichromate Ions ( ${Cr}_{2} O_{7}^{2 -}$ ) in Acidic Solution

Step 3: Balance oxygen atoms by adding $H_{2} O$

Step 4: Balance hydrogen atoms by adding $H^{+}$

Reduction of Manganese Dioxide ( $M n O_{2}$ ) in Neutral Solution