Define and identify the components of electrochemical cells

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Outcome 30-B2.1k

Leave a strip of zinc metal in a blue copper(II) sulfate solution and come back an hour later. The blue colour fades, a reddish coating of copper metal appears on the zinc, and the zinc strip slowly dissolves. Something is clearly happening between the zinc and the copper ions — electrons are being transferred from one species to another. An electrochemical cell takes that same transfer of electrons and forces it to travel through a wire, turning chemical energy into electrical energy (or the reverse). In Chemistry 30, you will study how these cells are built, what each part does, and how to read and write the symbolic shorthand chemists use to describe them. This lesson covers every component you need to identify and define for outcome 30-B2.1k.

What you will learn

Define an electrochemical cell and explain why it can produce or use electrical energy.
Identify and describe the function of each component: anode, cathode, electrodes, electrolyte, and salt bridge or porous barrier.
Distinguish between the oxidation half-reaction at the anode and the reduction half-reaction at the cathode.
Correctly label the direction of electron flow and ion flow in a galvanic cell diagram.
Identify components and half-reactions in diploma-style electrochemical cell questions.

Prerequisite Review: Oxidation and Reduction

Before studying cell components, you need two ideas from earlier chemistry. First, oxidation means losing electrons, and reduction means gaining electrons. The memory aid OIL RIG (Oxidation Is Loss, Reduction Is Gain) is helpful here. Second, a redox reaction pairs these two processes: one species loses electrons while another gains them at the same time.

In the zinc-copper example from the introduction, zinc atoms each lose two electrons and become zinc ions. Copper(II) ions each gain two electrons and become copper atoms. Written as half-reactions, these are:

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

(oxidation) and

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

(reduction). In a beaker, these two half-reactions happen at the same surface and the electrons transfer directly. In an electrochemical cell, the two half-reactions are separated into different compartments so that the electrons must travel through an external wire — and that travelling current is what we call electricity.

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

Oxidation = loss of electrons; reduction = gain of electrons (OIL RIG).
A redox reaction always has both an oxidation half-reaction and a reduction half-reaction.
Separating the two half-reactions into different compartments forces electrons through an external circuit.
The overall redox equation is obtained by adding the two balanced half-reactions so that electrons cancel.

What Is an Electrochemical Cell?

An electrochemical cell is any device that converts chemical energy to electrical energy, or electrical energy to chemical energy, through a redox reaction. There are two main types. A galvanic cell (also called a voltaic cell) converts chemical energy into electrical energy spontaneously — it is the basis of every common battery. An electrolytic cell does the opposite: it uses electrical energy from an external source to force a non-spontaneous redox reaction to occur. This lesson focuses on the galvanic cell because it introduces all the fundamental components that both types share.

The key idea is that a galvanic cell separates the oxidation half-reaction and the reduction half-reaction into two distinct regions called half-cells. Because the two half-reactions are in different locations, the electrons released by oxidation cannot jump directly to the species being reduced. Instead, they travel through a wire connecting the two half-cells, and that flow of electrons is the electric current the cell produces.

\text{Chemical energy} \xrightarrow{\text{galvanic cell}} \text{Electrical energy}

A galvanic (voltaic) cell converts chemical energy to electrical energy spontaneously.
An electrolytic cell converts electrical energy to chemical energy (non-spontaneous).
Both cell types contain two half-cells, each hosting one half-reaction.
Separating half-reactions forces electrons through an external circuit, producing current.

The Five Key Components of a Galvanic Cell

Every galvanic cell contains the same five types of components. Understanding the role of each one is the core of this outcome. The diagram below describes a classic zinc-copper galvanic cell with each part labelled.

1. Electrodes: An electrode is a solid electrical conductor that makes contact with the electrolyte in each half-cell. Electrodes provide the surface on which the half-reaction takes place. In many cells the electrode itself is one of the reacting species (an active electrode), such as a zinc strip. When neither reactant nor product is a solid conductor, an inert electrode — typically platinum or graphite — is used simply to carry current without reacting.

2. Anode: The anode is the electrode where oxidation occurs. The species at the anode loses electrons. In a galvanic cell, the anode is labelled the negative terminal because it is the source of electrons that flow into the external circuit. A useful memory aid: Anode and Oxidation both start with a vowel sound.

3. Cathode: The cathode is the electrode where reduction occurs. The species at the cathode gains electrons arriving from the external circuit. In a galvanic cell, the cathode is labelled the positive terminal. Cathode and Reduction: the C and R help you remember Cathode = Reduction.

4. Electrolyte: An electrolyte is a substance that dissolves in water (or is molten) to produce ions, allowing the solution to conduct electricity. Each half-cell contains an electrolyte solution. Ions in the electrolyte carry charge through the liquid portion of the circuit. Without an electrolyte, current cannot flow through the solution. In the zinc-copper cell, zinc sulfate solution (

\text{ZnSO}_4(aq)

) is the electrolyte in the anode half-cell and copper(II) sulfate solution (

\text{CuSO}_4(aq)

) is the electrolyte in the cathode half-cell.

\text{Anode (oxidation): } \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^{-} \quad \text{Cathode (reduction): } \text{Cu}^{2+}(aq) + 2e^{-} \rightarrow \text{Cu}(s)

Electrode: solid conductor; site of the half-reaction (active or inert).
Anode: site of oxidation; negative terminal in a galvanic cell; electrons leave from here into the wire.
Cathode: site of reduction; positive terminal in a galvanic cell; electrons arrive here from the wire.
Electrolyte: ionic solution that conducts charge through the liquid part of the circuit.
Salt bridge / porous barrier: maintains electrical neutrality in both half-cells by allowing ion migration, completing the internal circuit.

Electron Flow and Ion Flow: Completing Both Circuits

A galvanic cell has two conducting pathways that must both be complete for current to flow. The external circuit carries electrons through a wire from the anode to the cathode. Electrons always flow from the anode (negative) to the cathode (positive) through the external wire. This is the conventional current direction that powers a light bulb or any other load placed in the circuit.

The internal circuit carries charge through the electrolyte solutions and through the salt bridge. In the electrolyte, positive ions (cations) migrate toward the cathode compartment, and negative ions (anions) migrate toward the anode compartment. Inside the salt bridge, anions flow toward the anode half-cell and cations flow toward the cathode half-cell. This ion movement is what keeps the charge balanced and closes the electrical circuit through the liquid.

It is important not to confuse electron flow in the wire with ion flow in the solution. Electrons carry charge in the wire; ions carry charge in the solution. Both pathways are necessary, and both must be unbroken for the cell to function. If the salt bridge is removed, ion flow stops, charge imbalance builds up rapidly, and the cell voltage drops to zero even though the chemical reaction is still thermodynamically possible.

e^{-}: \text{anode} \xrightarrow{\text{external wire}} \text{cathode}

Electrons flow through the external wire from the anode (−) to the cathode (+).
Cations in solution migrate toward the cathode; anions migrate toward the anode.
In the salt bridge, anions move toward the anode half-cell; cations move toward the cathode half-cell.
Both the external electron path and the internal ion path must be complete for current to flow.
Removing the salt bridge stops ion flow and halts the cell.

Cell Notation (Line Notation)

Chemists use a compact shorthand called cell notation (or line notation) to describe a galvanic cell without drawing the full diagram. The conventions used in Alberta Chemistry 30 are as follows. The anode half-cell is always written on the left, and the cathode half-cell is always written on the right. A single vertical line ( | ) represents a phase boundary — a boundary between two different states of matter (for example, solid electrode touching liquid electrolyte). A double vertical line ( || ) represents the salt bridge or porous barrier between the two half-cells.

For the zinc-copper cell, the notation is written as:

\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \| \text{Cu}^{2+}(aq) \mid \text{Cu}(s)

. Reading left to right: the zinc solid anode | zinc ion electrolyte || (salt bridge) || copper ion electrolyte | copper solid cathode. When an inert electrode such as platinum is used, it appears at the far left (anode side) or far right (cathode side) of the notation. Concentrations or other conditions can be noted in brackets if needed, but the basic structure always follows this left-anode, right-cathode rule.

Cell notation is a reading and writing skill for Alberta Chemistry 30. Given a cell notation, you should be able to identify the anode material, cathode material, electrolytes, and the species being oxidized and reduced. Given a description of a cell, you should be able to write the correct notation.

\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \| \text{Cu}^{2+}(aq) \mid \text{Cu}(s)

Anode half-cell is always on the left in cell notation; cathode is always on the right.
A single line | represents a phase boundary (e.g., solid | solution).
A double line || represents the salt bridge or porous barrier.
Inert electrodes (Pt, graphite) appear at the outermost left or right positions.
Cell notation allows a cell to be fully described in one compact line.

Summary of Galvanic Cell Components

Component	Location / Role	In Zn–Cu Cell Example
Anode	Electrode where oxidation occurs; negative terminal; electrons leave here	Zinc strip ( $\text{Zn}(s)$ )
Cathode	Electrode where reduction occurs; positive terminal; electrons arrive here	Copper strip ( $\text{Cu}(s)$ )
Electrolyte	Ionic solution that conducts charge through the liquid	$\text{ZnSO}_4(aq)$ and $\text{CuSO}_4(aq)$
Salt Bridge	Allows ion migration between half-cells; maintains electrical neutrality	$\text{KNO}_3(aq)$ in gel tube
External Wire	Conducts electrons from anode to cathode; where current is used	Copper wire connecting electrodes

Worked example

Identifying Cell Components from a Description

A galvanic cell is set up with a magnesium electrode in a magnesium nitrate solution in one half-cell, and a silver electrode in a silver nitrate solution in the other half-cell. A salt bridge containing potassium nitrate connects the two half-cells. Magnesium is the more active metal and is oxidized. (a) Identify the anode and cathode. (b) Write the balanced half-reaction at each electrode. (c) Write the overall balanced redox equation. (d) Write the cell notation for this cell. (e) State the direction of electron flow through the external wire.

Step 1: Identify which species is oxidized and which is reduced.
The problem states magnesium is oxidized. Oxidation occurs at the anode. Therefore the magnesium electrode is the anode. Since magnesium is oxidized, silver ions must be reduced at the cathode — the silver electrode is the cathode.
$\text{Anode: Mg}(s) \quad \text{Cathode: Ag}(s)$
Step 2: Write the oxidation half-reaction at the anode.
Magnesium atoms each lose two electrons to become magnesium ions. Write the species losing electrons on the left and the products (including electrons) on the right.
$\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^{-}$
Step 3: Write the reduction half-reaction at the cathode.
Each silver ion gains one electron to become a silver atom. The half-reaction shows one electron gained per silver ion.
$\text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s)$
Step 4: Balance electrons and combine half-reactions.
The oxidation half-reaction produces 2 electrons, but the reduction half-reaction consumes only 1 electron per silver ion. Multiply the reduction half-reaction by 2 so that electrons produced equal electrons consumed, then add the two half-reactions. The electrons cancel.
$\text{Mg}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Ag}(s)$
Step 5: Verify atom and charge balance.
Left side: 1 Mg, 2 Ag, net charge = 2+. Right side: 1 Mg, 2 Ag, net charge = 2+. Atoms and charges balance — the equation is correct.
$\text{Charge: left } 2(+1) = +2; \text{ right } +2 \checkmark$
Step 6: Write the cell notation.
Place the anode (Mg) on the left and the cathode (Ag) on the right. Use a single line for each phase boundary and a double line for the salt bridge.
$\text{Mg}(s) \mid \text{Mg}^{2+}(aq) \| \text{Ag}^{+}(aq) \mid \text{Ag}(s)$
Step 7: State the direction of electron flow.
Electrons are released at the anode and travel through the external wire to the cathode. In this cell, electrons flow from the magnesium electrode to the silver electrode through the wire.
$e^{-}: \text{Mg electrode} \xrightarrow{\text{wire}} \text{Ag electrode}$

Answer: Anode: Mg(s); Cathode: Ag(s). Oxidation half-reaction:

\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^{-}

. Reduction half-reaction:

2\text{Ag}^{+}(aq) + 2e^{-} \rightarrow 2\text{Ag}(s)

. Overall:

\text{Mg}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Ag}(s)

. Cell notation:

\text{Mg}(s) \mid \text{Mg}^{2+}(aq) \| \text{Ag}^{+}(aq) \mid \text{Ag}(s)

. Electrons flow from the Mg anode to the Ag cathode through the external wire.

Check: Confirm the overall equation: 1 Mg on each side ✓; 2 Ag on each side ✓; net charge +2 on each side ✓. In cell notation, the anode (Mg) is correctly on the left and the cathode (Ag) is on the right ✓.

Common mistakes and how to avoid them

Saying electrons flow from the cathode to the anode through the wire.

Correction: Electrons always flow from the anode (negative) to the cathode (positive) through the external wire. The anode produces electrons by oxidation; the cathode consumes them by reduction.

Confusing the role of the salt bridge with the role of the wire.

Correction: The wire carries electrons (charge carriers in a metal). The salt bridge carries ions (charge carriers in solution). Both paths are needed, but they carry different charge carriers and serve different parts of the circuit.

Writing the cathode on the left side in cell notation.

Correction: By convention, the anode half-cell is always written on the left and the cathode half-cell on the right: anode | anode electrolyte || cathode electrolyte | cathode.

Forgetting to multiply one half-reaction before adding, leaving electrons unbalanced.

Correction: The number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction before the two are added. Multiply one or both half-reactions by a whole number to make the electron counts match.

Thinking the salt bridge mixes the two electrolyte solutions together.

Correction: The salt bridge allows only slow ion migration to maintain charge neutrality. It does not allow bulk mixing of the two half-cell solutions — that would allow direct electron transfer and eliminate the electrical current.

Lesson summary

An electrochemical cell converts chemical energy to electrical energy (galvanic cell) or electrical energy to chemical energy (electrolytic cell) through a redox reaction.
Every galvanic cell has five key components: anode, cathode, electrolytes, a salt bridge (or porous barrier), and an external conductor (wire).
The anode is the site of oxidation and the negative terminal; electrons leave the anode through the external wire.
The cathode is the site of reduction and the positive terminal; electrons arrive at the cathode from the external wire.
The salt bridge allows ion migration between half-cells, maintaining electrical neutrality and completing the internal circuit without mixing the solutions.
Cell notation always places the anode on the left and the cathode on the right, using | for phase boundaries and || for the salt bridge.

Check your understanding

Question 1

In a galvanic cell, which statement correctly describes the anode?

It is the site of reduction and the positive terminal.
It is the site of oxidation and the negative terminal.
It is where electrons arrive from the external wire.
It is always made of an inert material such as platinum.

Show answer and explanation

It is the site of oxidation and the negative terminal.

The anode is the site of oxidation (loss of electrons) and is the negative terminal of a galvanic cell because it releases electrons into the external wire. Reduction occurs at the cathode, not the anode.

Question 2

A galvanic cell uses an iron electrode in iron(II) sulfate solution and a tin electrode in tin(II) sulfate solution. Iron is oxidized. What is the correct cell notation?

$\text{Sn}(s) \mid \text{Sn}^{2+}(aq) \| \text{Fe}^{2+}(aq) \mid \text{Fe}(s)$
$\text{Fe}^{2+}(aq) \mid \text{Fe}(s) \| \text{Sn}(s) \mid \text{Sn}^{2+}(aq)$
$\text{Fe}(s) \mid \text{Fe}^{2+}(aq) \| \text{Sn}^{2+}(aq) \mid \text{Sn}(s)$
$\text{Sn}^{2+}(aq) \mid \text{Sn}(s) \| \text{Fe}(s) \mid \text{Fe}^{2+}(aq)$

Show answer and explanation

\text{Fe}(s) \mid \text{Fe}^{2+}(aq) \| \text{Sn}^{2+}(aq) \mid \text{Sn}(s)

Iron is oxidized, so it is the anode and goes on the left. Tin ions are reduced, so tin is the cathode and goes on the right. The solid electrode is separated from its electrolyte by a single line, and the double line represents the salt bridge:

\text{Fe}(s) \mid \text{Fe}^{2+}(aq) \| \text{Sn}^{2+}(aq) \mid \text{Sn}(s)

.

Question 3

Why is a salt bridge essential for a galvanic cell to continue operating?

It provides a path for electrons to travel between the two half-cells.
It prevents any ions from moving between the two half-cells.
It allows ion migration that maintains electrical neutrality in both half-cells.
It supplies additional reactant ions to replace those consumed at the cathode.

Show answer and explanation

It allows ion migration that maintains electrical neutrality in both half-cells.

As the cell operates, ions accumulate in one half-cell and are consumed in the other. The salt bridge allows ions to migrate and balance these charges, keeping both solutions electrically neutral. Without it, charge buildup would stop the cell. Electrons travel only through the external wire, not through the salt bridge.

Question 4

In the cell

\text{Ni}(s) \mid \text{Ni}^{2+}(aq) \| \text{Pb}^{2+}(aq) \mid \text{Pb}(s)

, what is the half-reaction occurring at the right-hand electrode?

$\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^{-}$
$\text{Ni}^{2+}(aq) + 2e^{-} \rightarrow \text{Ni}(s)$
$\text{Pb}^{2+}(aq) + 2e^{-} \rightarrow \text{Pb}(s)$
$\text{Pb}(s) \rightarrow \text{Pb}^{2+}(aq) + 2e^{-}$

Show answer and explanation

\text{Pb}^{2+}(aq) + 2e^{-} \rightarrow \text{Pb}(s)

The right-hand electrode in cell notation is always the cathode, where reduction occurs. Lead(II) ions gain two electrons each to form solid lead:

\text{Pb}^{2+}(aq) + 2e^{-} \rightarrow \text{Pb}(s)

. The nickel electrode on the left is the anode, where oxidation produces

\text{Ni}^{2+}

ions.

Key terms

Electrochemical cell: A device that interconverts chemical energy and electrical energy through a redox reaction.
Galvanic cell: An electrochemical cell in which a spontaneous redox reaction produces electrical energy; also called a voltaic cell.
Electrolytic cell: An electrochemical cell in which electrical energy from an external source drives a non-spontaneous redox reaction.
Anode: The electrode where oxidation occurs; the negative terminal in a galvanic cell; electrons leave from here into the external wire.
Cathode: The electrode where reduction occurs; the positive terminal in a galvanic cell; electrons arrive here from the external wire.
Electrolyte: A substance that dissolves in water (or is molten) to form ions, making the solution electrically conducting.
Salt bridge: A tube containing an ionic solution (in gel or liquid form) that connects two half-cells, allowing ion migration to maintain electrical neutrality without mixing the solutions.
Cell notation: A compact symbolic representation of an electrochemical cell; the anode half-cell is on the left, the cathode is on the right, | represents a phase boundary, and || represents the salt bridge.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Chemistry (CHEM30), expectation 30-B2.1k. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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30-B2.1k · Define and identify the components of electrochemical cells

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Outcome 30-B2.1k

What you will learn

Prerequisite Review: Oxidation and Reduction

What Is an Electrochemical Cell?

The Five Key Components of a Galvanic Cell

Electron Flow and Ion Flow: Completing Both Circuits

Cell Notation (Line Notation)

Summary of Galvanic Cell Components

Identifying Cell Components from a Description

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review