Oxidation numbers and voltaic cells

IB Chemistry HL· 18 min read

1. Assigning Oxidation Numbers★★☆☆☆⏱ 5 min

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📘 Definition

Oxidation Number (Oxidation State)

ON / OS

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.

📐 Worked Example

Assign oxidation numbers to all elements in the permanganate ion $MnO_{4}^{-}$ .

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 $x$ :
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$x = -1 + 8 = +7$
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Final oxidation numbers: Mn = +7, O = -2.

✓ Quick check

What is the oxidation number of S in $H_{2} SO_{4}$ ?

What is the oxidation number of S in $H_{2} SO_{4}$ ?
- +2
- +4
- +6
- -2
Reveal answer
+6 —
Calculation: 2(+1) + x + 4(-2) = 0 → x = +6

2. Identifying Redox Reactions with Oxidation Numbers★★☆☆☆⏱ 5 min

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

📐 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:
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Reactants: Mg = 0 (free element), O₂ = 0 (free element)
3
Products: MgO: Mg = +2, O = -2
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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).

3. Structure and Operation of Voltaic Cells★★★☆☆⏱ 8 min

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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.

📘 Definition

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.

📐 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: $Zn_{(s)} \to Zn_{(a q)}^{2 +} + 2 e^{-}$
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1. Electrons flow through the external wire from the zinc anode to the copper cathode.
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1. Reduction at the copper cathode uses the incoming electrons: $Cu_{(a q)}^{2 +} + 2 e^{-} \to Cu_{(s)}$
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1. Positive charge builds up at the anode, so negative anions from the salt bridge migrate into the anode compartment to balance charge.
5
1. Negative charge builds up at the cathode, so positive cations from the salt bridge migrate into the cathode compartment to balance charge.

4. Calculating Standard Cell Potential★★★☆☆⏱ 5 min

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The standard cell potential ( $E_{ce l l}^{\circ}$ ) 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}

📘 Definition

Spontaneity Rule

A positive $E_{ce l l}^{\circ}$ confirms the redox reaction is spontaneous, which is always the case for a working voltaic cell.

📐 Worked Example

Calculate $E_{ce l l}^{\circ}$ for a zinc-copper voltaic cell, given $E^{\circ} (Zn^{2 +} / Zn) = - 0.76 V$ and $E^{\circ} (Cu^{2 +} / Cu) = + 0.34 V$ .

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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.
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Step 2: Substitute into the formula:
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$E^\circ_{cell} = 0.34 - (-0.76) = 1.10 \text{ V}$
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Step 3: Confirm result is positive, which matches the spontaneous reaction expected in a voltaic cell.

5. Common Pitfalls

Wrong move:

Assign oxygen an oxidation number of -2 in hydrogen peroxide $H_{2} O_{2}$ .

Why:

The -2 rule for oxygen does not apply to peroxides.

Correct move:

Oxygen has an oxidation number of -1 in peroxides.

Wrong move:

Claiming electrons flow through the salt bridge.

Why:

Electrons only travel through the metallic external wire; charge is balanced by ion flow in the salt bridge.

Correct move:

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 move:

Calling the anode positive in a voltaic cell.

Why:

Oxidation produces excess electrons at the anode, giving it a negative charge (this is reversed for electrolytic cells).

Correct move:

In voltaic cells: anode = negative, cathode = positive.

Wrong move:

Assuming the sum of oxidation numbers is always zero for all species.

Why:

The sum equals the overall charge of the species, which is non-zero for ions.

Correct move:

Sum of oxidation numbers equals the overall charge of the molecule or ion.

Wrong move:

Calculating $E_{ce l l}^{\circ}$ as $E_{an o d e}^{\circ} - E_{c a t h o d e}^{\circ}$ .

Why:

The formula uses standard reduction potentials for both half-cells, so cathode potential minus anode potential is required.

Correct move:

Use $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ to get a positive value for spontaneous reactions.

6. Quick Reference 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	$E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$
Spontaneous voltaic cell	Positive E°cell

7. Frequently Asked

Why do we use oxidation numbers instead of ionic charges?

Oxidation numbers work for covalent compounds, where formal ionic charges do not exist, but electron transfer still occurs during redox reactions.

Is a positive cell potential always required for a voltaic cell?

Yes. A positive $E_{ce l l}^{\circ}$ confirms the redox reaction is spontaneous, which is required to generate energy in a voltaic cell.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 1
Assign oxidation number to Mn
2022 · Paper 2
Calculate E°cell for voltaic cell
2021 · Paper 1
Salt bridge function question

Going deeper

syllabusIB Chemistry HL 2025 SyllabusTopic 6: Redox processes

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.