Topics

Electrochemical (galvanic) cells

The idea

A spontaneous redox reaction releases energy. If the two half reactions happen in the same beaker, that energy is released as heat and is wasted.
Separate them into two half cells and connect them with a wire, and the electrons must travel through the wire to get from the reductant to the oxidant. That flow of electrons is an electric current.
An electrochemical cell is a spontaneous redox reaction with the two halves physically separated so the electrons have to do work on the way.

The parts

Part	What it does
Anode	where oxidation happens; the negative electrode in a galvanic cell
Cathode	where reduction happens; the positive electrode in a galvanic cell
External circuit	carries electrons from anode to cathode
Salt bridge	carries ions between the half cells, completing the circuit and keeping both solutions neutral
Voltmeter	measures the cell potential

Two memory aids that work: AN OX (ANode = OXidation) and RED CAT (REDuction = CAThode). These hold for both kinds of cell, which is why they are worth learning rather than memorising which electrode is positive.

Which electrode is which

The electrode of the more negative $E°$ is the anode, and it is oxidised.
The electrode of the more positive $E°$ is the cathode, and it is reduced.
Electrons always flow from anode to cathode through the wire.

For a zinc–copper cell:

Anode (−): $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$ , $E° = -0.76$ V
Cathode (+): $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ , $E° = +0.34$ V
Overall: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$
$E°_{cell} = (+0.34) - (-0.76) = +1.10$ V

Why the salt bridge is essential

Without it, the cell stops almost immediately. Here is why:
- As zinc dissolves, the anode compartment accumulates positive Zn²⁺ ions.
- As Cu²⁺ is deposited, the cathode compartment is left with excess negative SO₄²⁻ ions.
- A build-up of charge opposes further electron flow, and the current stops.
The salt bridge — a tube of inert electrolyte such as KNO₃ — lets ions move to cancel this build-up:
- Anions move towards the anode (towards the building positive charge).
- Cations move towards the cathode.
The electrolyte in the bridge must be inert: its ions must not react with, or precipitate, anything in either half cell. This is why KNO₃ is used — all potassium and all nitrate salts are soluble.

Observations in a working cell

The criteria require observations, so know what a running zinc–copper cell looks like:

The zinc electrode loses mass and becomes pitted as it dissolves.
The copper electrode gains mass as copper is deposited on it.
The blue colour of the copper sulfate solution fades as Cu²⁺ is consumed.
The voltmeter reads about 1.10 V at standard conditions, and the reading falls over time as the concentrations move away from standard.

Cell notation

A cell is written compactly, anode on the left:

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

A single line | is a phase boundary; a double line ‖ is the salt bridge.
If both species of a couple are in solution — such as Fe³⁺/Fe²⁺ — an inert electrode of platinum or graphite is needed to make electrical contact, and it appears at the end:

$\text{Pt}(s) \mid \text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq) \parallel \ldots$

Worked ExampleDesigning and describing a galvanic cell

A cell is constructed from a nickel electrode in 1 mol L⁻¹ nickel(II) sulfate and a silver electrode in 1 mol L⁻¹ silver nitrate, joined by a potassium nitrate salt bridge.

Identify the anode and cathode, write the half equations and overall equation, calculate E°cell, and describe what would be observed.

$\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$ , $E° = -0.25$ V $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ , $E° = +0.80$ V

Step 1 — Identify the electrodes

The more negative $E°$ is oxidised at the anode. Nickel at −0.25 V is more negative than silver at +0.80 V.

Anode (negative): nickel
Cathode (positive): silver

Step 2 — Write the half equations

Anode, oxidation:

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

Cathode, reduction:

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

Step 3 — Balance the electrons and combine

The silver half equation must be doubled so both involve two electrons — but $E°$ is not doubled:

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

Adding and cancelling the electrons:

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

Step 4 — Calculate $E°_{cell}$

$E°_{cell} = E°(\text{reduced}) - E°(\text{oxidised}) = (+0.80) - (-0.25) = +1.05 \text{ V}$

Positive, confirming the reaction is spontaneous and the cell will deliver current.

Step 5 — Describe the observations

The nickel electrode dissolves, losing mass and becoming pitted.
The nickel half cell's solution becomes more intensely green, as the concentration of Ni²⁺ rises.
Silver crystals deposit on the silver electrode, which gains mass.
The colourless silver nitrate solution becomes less concentrated in Ag⁺, though it stays colourless.
The voltmeter reads about 1.05 V, falling slowly as the cell operates.

Step 6 — Ion movement

In the external circuit, electrons flow from the nickel anode to the silver cathode.

In the salt bridge, nitrate ions move towards the nickel half cell to balance the accumulating Ni²⁺, and potassium ions move towards the silver half cell to replace the Ag⁺ removed.

Answer: nickel is the anode (oxidised) and silver the cathode (reduced); E°cell = +1.05 V.

Electrochemical (galvanic) cells

Covers AS91393

The idea

A spontaneous redox reaction releases energy. If the two half reactions happen in the same beaker, that energy is released as heat and is wasted.
Separate them into two half cells and connect them with a wire, and the electrons must travel through the wire to get from the reductant to the oxidant. That flow of electrons is an electric current.
An electrochemical cell is a spontaneous redox reaction with the two halves physically separated so the electrons have to do work on the way.

The parts

Part	What it does
Anode	where oxidation happens; the negative electrode in a galvanic cell
Cathode	where reduction happens; the positive electrode in a galvanic cell
External circuit	carries electrons from anode to cathode
Salt bridge	carries ions between the half cells, completing the circuit and keeping both solutions neutral
Voltmeter	measures the cell potential

Two memory aids that work: AN OX (ANode = OXidation) and RED CAT (REDuction = CAThode). These hold for both kinds of cell, which is why they are worth learning rather than memorising which electrode is positive.

Which electrode is which

The electrode of the more negative $E°$ is the anode, and it is oxidised.
The electrode of the more positive $E°$ is the cathode, and it is reduced.
Electrons always flow from anode to cathode through the wire.

For a zinc–copper cell:

Anode (−): $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$ , $E° = -0.76$ V
Cathode (+): $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ , $E° = +0.34$ V
Overall: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$
$E°_{cell} = (+0.34) - (-0.76) = +1.10$ V

Why the salt bridge is essential

Without it, the cell stops almost immediately. Here is why:
- As zinc dissolves, the anode compartment accumulates positive Zn²⁺ ions.
- As Cu²⁺ is deposited, the cathode compartment is left with excess negative SO₄²⁻ ions.
- A build-up of charge opposes further electron flow, and the current stops.
The salt bridge — a tube of inert electrolyte such as KNO₃ — lets ions move to cancel this build-up:
- Anions move towards the anode (towards the building positive charge).
- Cations move towards the cathode.
The electrolyte in the bridge must be inert: its ions must not react with, or precipitate, anything in either half cell. This is why KNO₃ is used — all potassium and all nitrate salts are soluble.

Observations in a working cell

The criteria require observations, so know what a running zinc–copper cell looks like:

The zinc electrode loses mass and becomes pitted as it dissolves.
The copper electrode gains mass as copper is deposited on it.
The blue colour of the copper sulfate solution fades as Cu²⁺ is consumed.
The voltmeter reads about 1.10 V at standard conditions, and the reading falls over time as the concentrations move away from standard.

Cell notation

A cell is written compactly, anode on the left:

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

A single line | is a phase boundary; a double line ‖ is the salt bridge.
If both species of a couple are in solution — such as Fe³⁺/Fe²⁺ — an inert electrode of platinum or graphite is needed to make electrical contact, and it appears at the end:

$\text{Pt}(s) \mid \text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq) \parallel \ldots$

Worked ExampleDesigning and describing a galvanic cell

A cell is constructed from a nickel electrode in 1 mol L⁻¹ nickel(II) sulfate and a silver electrode in 1 mol L⁻¹ silver nitrate, joined by a potassium nitrate salt bridge.

Identify the anode and cathode, write the half equations and overall equation, calculate E°cell, and describe what would be observed.

$\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$ , $E° = -0.25$ V $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ , $E° = +0.80$ V

Step 1 — Identify the electrodes

The more negative $E°$ is oxidised at the anode. Nickel at −0.25 V is more negative than silver at +0.80 V.

Anode (negative): nickel
Cathode (positive): silver

Step 2 — Write the half equations

Anode, oxidation:

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

Cathode, reduction:

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

Step 3 — Balance the electrons and combine

The silver half equation must be doubled so both involve two electrons — but $E°$ is not doubled:

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

Adding and cancelling the electrons:

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

Step 4 — Calculate $E°_{cell}$

$E°_{cell} = E°(\text{reduced}) - E°(\text{oxidised}) = (+0.80) - (-0.25) = +1.05 \text{ V}$

Positive, confirming the reaction is spontaneous and the cell will deliver current.

Step 5 — Describe the observations

The nickel electrode dissolves, losing mass and becoming pitted.
The nickel half cell's solution becomes more intensely green, as the concentration of Ni²⁺ rises.
Silver crystals deposit on the silver electrode, which gains mass.
The colourless silver nitrate solution becomes less concentrated in Ag⁺, though it stays colourless.
The voltmeter reads about 1.05 V, falling slowly as the cell operates.

Step 6 — Ion movement

In the external circuit, electrons flow from the nickel anode to the silver cathode.

In the salt bridge, nitrate ions move towards the nickel half cell to balance the accumulating Ni²⁺, and potassium ions move towards the silver half cell to replace the Ag⁺ removed.

Answer: nickel is the anode (oxidised) and silver the cathode (reduced); E°cell = +1.05 V.

Electrochemical (galvanic) cells

The idea

The parts

Which electrode is which

Why the salt bridge is essential

Observations in a working cell

Cell notation

Step 1 — Identify the electrodes

Step 2 — Write the half equations

Step 3 — Balance the electrons and combine

Step 4 — Calculate E°cellE°_{cell}E°cell​

Step 5 — Describe the observations

Step 6 — Ion movement

Electrochemical (galvanic) cells

The idea

The parts

Which electrode is which

Why the salt bridge is essential

Observations in a working cell

Cell notation

Step 1 — Identify the electrodes

Step 2 — Write the half equations

Step 3 — Balance the electrons and combine

Step 4 — Calculate E°cellE°_{cell}E°cell​

Step 5 — Describe the observations

Step 6 — Ion movement

Step 4 — Calculate $E°_{cell}$

Step 4 — Calculate $E°_{cell}$