Electrochemical cells

CIE A-Level Chemistry· 7 min read

1. Structure and Components of an Electrochemical Cell★★☆☆☆⏱ 15 min

An electrochemical (galvanic) cell converts chemical energy from spontaneous redox reactions into electrical energy. It consists of two separate half-cells connected by an external wire and a salt bridge.

📘 Definition

Electrochemical (Galvanic) Cell

A device that generates electrical energy from spontaneous, exergonic redox reactions, with oxidation in one half-cell and reduction in the other.

Example:

A standard zinc-copper cell used to demonstrate spontaneous redox

Anode: Negative electrode where oxidation (loss of electrons) occurs
Cathode: Positive electrode where reduction (gain of electrons) occurs
Salt bridge: Soaked in inert electrolyte, connects half-cells to balance charge
External circuit: Wires connect electrodes, allowing electron flow to a load

📐 Worked Example

Draw and label all key components of a zinc-copper electrochemical cell with zinc as anode

1
Draw two separate beakers for the two half-cells
2
Add a solid zinc strip to the first beaker, label as negative anode, fill with 1 mol dm⁻³ Zn(NO₃)₂ electrolyte
3
Add a solid copper strip to the second beaker, label as positive cathode, fill with 1 mol dm⁻³ Cu(NO₃)₂ electrolyte
4
Connect the two electrodes with an external wire, insert a high-resistance voltmeter
5
Place a KNO₃-soaked salt bridge between the two beakers, each end immersed in electrolyte
6
Label flow: electrons from Zn anode to Cu cathode; anions to anode, cations to cathode

2. Standard Cell Notation★★☆☆☆⏱ 10 min

CIE exams often require standard cell notation to represent an electrochemical cell instead of a full drawing. There are clear, fixed conventions for this notation that you must follow.

Single vertical line | = phase boundary between solid electrode and aqueous electrolyte
Double vertical line || = the salt bridge between two half-cells
Anode (oxidation) half-cell is always written on the left
Cathode (reduction) half-cell is always written on the right

📐 Worked Example

Write cell notation for the reaction $Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$

1
Identify anode half-reaction: Zn oxidises to Zn²⁺, so this goes on the left
2
Write anode side with phase boundary: $Z n (s) ∣ Z n^{2 +} (a q, 1 m o l d m^{- 3})$
3
Add double line for salt bridge: $Z n (s) ∣ Z n^{2 +} (a q, 1 m o l d m^{- 3}) ∣∣$
4
Add cathode side (Cu²⁺ reduces to Cu) on the right
5
$\text{Final notation: } Zn(s) | Zn^{2+}(aq, 1\ mol\ dm^{-3}) || Cu^{2+}(aq, 1\ mol\ dm^{-3}) | Cu(s)$

3. Calculating Standard Cell Potential★★★☆☆⏱ 15 min

Standard cell potential ( $E_{ce l l}^{\circ}$ ) is the overall potential difference between the two half-cells, measured under standard conditions (1 mol dm⁻³, 298 K, 1 atm pressure).

📘 Definition

Standard Cell Potential

E^\circ_{cell}

The potential difference between two electrodes when measured under standard conditions with no current flowing

Example:

E°cell for a zinc-copper cell is +1.10 V

The most reliable formula for calculation (avoids sign errors) is:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode (reduction)}} - E^\circ_{\text{anode (oxidation)}}

📐 Worked Example

Given $E^{\circ} (Z n^{2 +} / Z n) = - 0.76 V$ and $E^{\circ} (C u^{2 +} / C u) = + 0.34 V$ , calculate E°cell for the zinc-copper cell

1
Confirm: Zn is anode (oxidation), Cu is cathode (reduction)
2
Substitute values into the formula:
3
$E^\circ_{\text{cell}} = 0.34\ V - (-0.76\ V)$
4
Simplify and calculate the final result:
5
$E^\circ_{\text{cell}} = 0.34 + 0.76 = +1.10\ V$

✓ Quick check

Test your understanding below:

Calculate E°cell for a cell with Fe anode (E° = -0.44 V) and Ag cathode (E° = +0.80 V):
- +1.24 V
- +0.36 V
- -0.36 V
- -1.24 V
Reveal answer
+1.24 V —
Correct: 0.80 - (-0.44) = +1.24 V
Where is the silver half-cell written in cell notation?
- Left of the salt bridge
- Right of the salt bridge
- Either side
- Depends on E°cell
Reveal answer
Right of the salt bridge —
Silver is the cathode (reduction), which is always written on the right of the salt bridge

4. Predicting Reaction Feasibility★★★☆☆⏱ 15 min

The sign of E°cell tells us whether a reaction is feasible (spontaneous) under standard conditions.

📘 Definition

Feasible Reaction

A reaction that can occur spontaneously without external energy input, predicted from E°cell under standard conditions

Example:

Reaction of zinc with copper(II) ions is feasible

The core rule: positive E°cell means the forward reaction is feasible under standard conditions. Negative E°cell means the forward reaction is not feasible, and the reverse reaction is feasible.

📐 Worked Example

Predict if $2 A g (s) + 2 H^{+} (a q) \to 2 A g^{+} (a q) + H_{2} (g)$ is feasible under standard conditions, given $E^{\circ} (A g^{+} / A g) = + 0.80 V$ , $E^{\circ} (H^{+} / H_{2}) = 0 V$

1
Split into half-reactions:
2
$\text{Oxidation (anode): } 2Ag(s) \rightarrow 2Ag^+(aq) + 2e^-$
3
$\text{Reduction (cathode): } 2H^+(aq) + 2e^- \rightarrow H_2(g)$
4
Calculate E°cell:
5
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0 - 0.80 = -0.80\ V$
6
Conclusion: E°cell is negative, so the forward reaction is not feasible under standard conditions.

5. Common Pitfalls

Wrong move:

Writing the anode half-cell on the right and cathode on the left in cell notation

Why:

Swapping the order leads to a negative E°cell, which incorrectly predicts the reaction is not feasible

Correct move:

Always place the oxidation (anode) half-cell on the left, reduction (cathode) on the right of the salt bridge

Wrong move:

Forgetting to flip signs when using the E reduction + E oxidation method

Why:

All E values in the data booklet are reduction potentials, so oxidation E is the negative of the given value

Correct move:

Use the $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ formula to eliminate sign errors entirely

Wrong move:

Claiming a reaction with negative E°cell can never occur under any conditions

Why:

Negative E°cell only means non-feasible under standard conditions, non-standard conditions can make it feasible

Correct move:

Always add the qualifier 'under standard conditions' when predicting feasibility from E°cell

Wrong move:

Stating electrons flow through the salt bridge or ions flow through the external wire

Why:

Electrons can only travel through the metallic wire, ions balance charge through the salt bridge

Correct move:

Remember: electrons flow anode → cathode through external wire; ions flow through salt bridge

6. Quick Reference Cheatsheet

Concept	Rule/Formula	Key Convention
Cell components	Anode = oxidation (negative); Cathode = reduction (positive)	Electrons flow anode → cathode (external wire)
Cell notation	Anode \| Anode electrolyte \|\| Cathode electrolyte \| Cathode	\| = phase boundary, \|\| = salt bridge, anode left
E°cell calculation	$E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$	All values are standard reduction potentials
Feasibility	Positive E°cell = forward reaction feasible	Only applies to standard conditions

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.

2022 · 22
Calculate E°cell, draw cell diagram
2023 · 12
Predict reaction feasibility
2021 · 33
Label cell components

Going deeper

syllabusCIE 9701 A-Level Chemistry SyllabusVerify latest learning outcomes

What's Next

Electrochemical cells are the foundation for all advanced electrochemistry topics in CIE A-Level Chemistry, including practical applications like batteries, fuel cells, and electrolysis, as well as non-standard potential calculations with the Nernst equation. Mastery of cell conventions and E°cell calculations is essential, as these concepts appear frequently in both multiple-choice and extended response questions. E°cell is also directly related to Gibbs free energy change, connecting electrochemistry to the energetics topic you learned earlier, and explaining why spontaneous reactions have positive E°cell. Next, you will build on this knowledge to explore how non-standard conditions change cell potential.