# Oxidation numbers and voltaic cells

> IB Chemistry HL · R3: What are the mechanisms of chemical change?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u6-oxidation-numbers-and-voltaic-cells/

This sub-topic covers how to assign oxidation numbers to track electron transfer in redox reactions, and connects this to the structure, operation, and potential calculation of spontaneous voltaic (galvanic) cells that generate electrical energy.

**Prerequisites:** [Balancing redox half-equations](https://www.owlsprep.com/study/ib-chemistry-hl-u6-redox-half-equations/)

## Learning objectives

- Assign correct oxidation numbers to elements in compounds and ions
- Identify oxidation and reduction using oxidation number changes
- Explain the structure and function of a voltaic cell
- Calculate standard cell potential from standard electrode potentials

## Assigning Oxidation Numbers

**Oxidation Number (Oxidation State)** — The hypothetical charge an atom of an element would have if all bonds in the compound/ion were fully ionic, with all bonding electrons assigned to the more electronegative atom.

*Notation:* ON / OS

> **Oxidation Number Rules Mnemonic**
>
> 1. Free elements = 0  
> 2. Monatomic ions = match their charge  
> 3. H = +1 (except hydrides = -1)  
> 4. O = -2 (except peroxides = -1, OF₂ = +2)  
> 5. Sum = 0 for neutral compounds, sum = overall charge for ions

**Worked example:** Assign oxidation numbers to all elements in the permanganate ion $\text{MnO}_4^-$.

1. Step 1: Apply the oxidation number rule for oxygen: this is not a peroxide, so O = -2.
2. Step 2: Let $x$ = oxidation number of Mn. The sum of oxidation numbers equals the overall ion charge of -1:
3. $$x + 4(-2) = -1$$
4. Step 3: Solve for $x$:
5. $$x = -1 + 8 = +7$$
6. Final oxidation numbers: Mn = +7, O = -2.

**Check your understanding**

What is the oxidation number of S in $\text{H}_2\text{SO}_4$?

1. What is the oxidation number of S in $\text{H}_2\text{SO}_4$?

   - +2
   - +4
   - +6
   - -2

   *Why:* Calculation: 2(+1) + x + 4(-2) = 0 → x = +6

*Calculator:* forbidden

## Identifying Redox Reactions with Oxidation Numbers

Oxidation numbers let us track electron transfer to identify oxidation and reduction even for covalent species, without writing full ionic half-equations.

> **mnemonic**
>
> OIL RIG: Oxidation Is Loss of electrons → **Increase in oxidation number**. Reduction Is Gain of electrons → **Decrease in oxidation number**.

**Worked example:** Identify which element is oxidized and which is reduced in the reaction: $2\text{Mg}_{(s)} + \text{O}_2_{(g)} \rightarrow 2\text{MgO}_{(s)}$.

1. Step 1: Assign oxidation numbers to all elements:
2. Reactants: Mg = 0 (free element), O₂ = 0 (free element)
3. Products: MgO: Mg = +2, O = -2
4. Step 2: Compare changes: Mg goes from 0 → +2 (oxidation number increases, so Mg is oxidized). O goes from 0 → -2 (oxidation number decreases, so O is reduced).

*Calculator:* forbidden

## Structure and Operation of Voltaic Cells

A voltaic (galvanic) cell separates two half-reactions of a spontaneous redox reaction to force electrons to flow through an external circuit, generating usable electrical energy.

**Voltaic Cell Key Terms** — Anode: Electrode where oxidation occurs. Cathode: Electrode where reduction occurs. In voltaic cells, anode is negative, cathode is positive.

- **External circuit**: Wires connect the two electrodes, allowing electron flow from anode to cathode.
- **Salt bridge**: Inert electrolyte (e.g. KNO₃) allows ion migration to balance charge buildup. Anions flow to the anode, cations flow to the cathode.

> **Exam Reminder**
>
> Electrons never flow through the salt bridge. Only ions flow through the salt bridge.

**Worked example:** Describe the direction of all charge flow in a zinc-copper voltaic cell (Zn anode, Cu cathode).

1. 1. Oxidation at the zinc anode releases electrons: $\text{Zn}_{(s)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2e^-$
2. 2. Electrons flow through the external wire from the zinc anode to the copper cathode.
3. 3. Reduction at the copper cathode uses the incoming electrons: $\text{Cu}^{2+}_{(aq)} + 2e^- \rightarrow \text{Cu}_{(s)}$
4. 4. Positive charge builds up at the anode, so negative anions from the salt bridge migrate into the anode compartment to balance charge.
5. 5. Negative charge builds up at the cathode, so positive cations from the salt bridge migrate into the cathode compartment to balance charge.

*Calculator:* forbidden

## Calculating Standard Cell Potential

The standard cell potential ($E^\circ_{cell}$) is the maximum voltage produced by the cell under standard conditions. It can be calculated from standard reduction potentials of the two half-cells:

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$

**Spontaneity Rule** — A positive $E^\circ_{cell}$ confirms the redox reaction is spontaneous, which is always the case for a working voltaic cell.

**Worked example:** Calculate $E^\circ_{cell}$ for a zinc-copper voltaic cell, given $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V}$ and $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}$.

1. Step 1: Identify anode and cathode: the half-reaction with the more negative $E^\circ$ is oxidized (anode), so Zn is anode, Cu is cathode.
2. Step 2: Substitute into the formula:
3. $$E^\circ_{cell} = 0.34 - (-0.76) = 1.10 \text{ V}$$
4. Step 3: Confirm result is positive, which matches the spontaneous reaction expected in a voltaic cell.

> **tip**
>
> If you get a negative $E^\circ_{cell}$, you swapped anode and cathode. Swap them and recalculate.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Assign oxygen an oxidation number of -2 in hydrogen peroxide $\text{H}_2\text{O}_2$.
  - Why it fails: The -2 rule for oxygen does not apply to peroxides.
  - Correct: Oxygen has an oxidation number of -1 in peroxides.
- **Wrong:** Claiming electrons flow through the salt bridge.
  - Why it fails: Electrons only travel through the metallic external wire; charge is balanced by ion flow in the salt bridge.
  - Correct: Anions flow to the anode, cations flow to the cathode through the salt bridge, electrons flow from anode to cathode in the external circuit.
- **Wrong:** Calling the anode positive in a voltaic cell.
  - Why it fails: Oxidation produces excess electrons at the anode, giving it a negative charge (this is reversed for electrolytic cells).
  - Correct: In voltaic cells: anode = negative, cathode = positive.
- **Wrong:** Assuming the sum of oxidation numbers is always zero for all species.
  - Why it fails: The sum equals the overall charge of the species, which is non-zero for ions.
  - Correct: Sum of oxidation numbers equals the overall charge of the molecule or ion.
- **Wrong:** Calculating $E^\circ_{cell}$ as $E^\circ_{anode} - E^\circ_{cathode}$.
  - Why it fails: The formula uses standard reduction potentials for both half-cells, so cathode potential minus anode potential is required.
  - Correct: Use $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ to get a positive value for spontaneous reactions.

## Cheatsheet

| Concept | Key Fact/Rule |
| --- | --- |
| Free element ON | Always 0 |
| Oxidation | Increase in ON, occurs at anode |
| Reduction | Decrease in ON, occurs at cathode |
| Voltaic cell anode charge | Negative |
| Voltaic cell cathode charge | Positive |
| Electron flow direction | Anode → cathode (external circuit) |
| Salt bridge ion flow | Anions → anode, cations → cathode |
| ON sum rule | Equals overall charge of the species |
| E°cell formula | $E^\circ_{cathode} - E^\circ_{anode}$ |
| Spontaneous voltaic cell | Positive E°cell |

## What's next

Mastering oxidation numbers and voltaic cells lays the foundation for all further redox topics in IB Chemistry HL. This content is heavily tested across both multiple choice and extended response questions in exams, so memorizing the rules for oxidation numbers and cell components is critical for higher marks. The concepts here extend directly to non-spontaneous electrochemical processes, quantitative electrolysis calculations, and predicting cell potential under non-standard conditions, all core topics for the HL syllabus.

- [Reaction mechanism basics](https://www.owlsprep.com/study/ib-chemistry-hl-u6-reaction-mechanism-basics/)
- [AHL: Stereoisomerism](https://www.owlsprep.com/study/ib-chemistry-hl-u6-ahl-stereoisomerism/)
- [AHL: Advanced organic reaction mechanisms](https://www.owlsprep.com/study/ib-chemistry-hl-u6-ahl-advanced-organic-reaction-mechanisms/)

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