OxfordAQA A-Level Chemistry: Electrode potentials and electrochemical cells (9620) – Revision Notes

For full explanations and worked examples, use the study guide. These notes condense section 3.1.9, Electrode potentials and electrochemical cells (3.1.9.1 and 3.1.9.2), of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. The section is International A2 content, so it is International A-level only. The specification places it in Unit 3 (Inorganic 2 and Physical 2); Unit 5 (Practical and synoptic) can assess all content, including required practical 6.

When you finish, try the practice questions. Le Chatelier’s principle, which explains the effect of conditions, is in the equilibria revision notes. See also the OxfordAQA A-Level Chemistry hub, the printable checklist and the free diagnostics.

3.1.9.1 Electrode potentials and cells (International A-level only)

Key terms

• Half-cell: an electrode and its solution, such as Sn in Sn²⁺(aq), or Pt in a solution of V²⁺ and V³⁺.
• Salt bridge: filter paper soaked in an unreactive salt solution (such as KNO₃). Ions move through it; electrons do not.
• Standard hydrogen electrode (SHE): the reference half-cell, defined as 0.00 V.
• Standard electrode potential, E⦵: the EMF of a half-cell measured against the SHE at 298 K, 100 kPa and 1.00 mol dm⁻³ solutions of ions.
• Electrochemical series: half-equations listed in order of E⦵.
• EMF: the potential difference of a cell when no current flows.

The SHE at a glance

Part	Detail
Electrode	Platinum (inert; conducts electrons to the H₂/H⁺ equilibrium)
Gas	H₂ at 100 kPa
Solution	[H⁺] = 1.00 mol dm⁻³
Temperature	298 K
Half-equation	2H⁺(aq) + 2e⁻ ⇌ H₂(g)
In a cell diagram	Pt(s) | H₂(g) | H⁺(aq)

Conventions

Item	Rule
IUPAC half-equation	Written as a reduction, electrons on the left, ⇌
|	Phase boundary
||	Salt bridge
Left-hand electrode	Negative; oxidation happens here
Right-hand electrode	Positive; reduction happens here
Two aqueous species	Comma between them, Pt(s) on the outside
EMF	E⦵(right) − E⦵(left) = E⦵(positive) − E⦵(negative), always positive

Method in steps: build a cell

1. Look up both E⦵ values.
2. More negative half-cell → negative electrode → left.
3. Left side reads as an oxidation, electrode outermost: Cd(s) | Cd²⁺(aq).
4. Right side reads as a reduction, electrode outermost: Au³⁺(aq) | Au(s).
5. EMF = E⦵(right) − E⦵(left).
6. Overall equation: reverse the left half-equation, scale both to the same number of electrons, add, cancel. Never scale E⦵.

Method in steps: is it feasible?

1. Find the half-equation with the oxidising agent on the left.
2. Find the half-equation with the reducing agent on the right.
3. Feasible if E⦵(oxidising agent’s half-equation) − E⦵(reducing agent’s half-equation) is positive.
4. The more positive half-equation goes forwards; the other goes backwards.

Method in steps: unknown E⦵ from a measured EMF

1. Note which electrode is negative.
2. Write EMF = E(positive) − E(negative) and substitute the known value.
3. Solve for the unknown. Against the SHE, the reading is E⦵ with the sign of the unknown electrode.

Effect of conditions (no Nernst equation)

Change	Equilibrium shift	Electrode potential
More oxidised form (left-hand species)	Right	More positive
Less oxidised form	Left	More negative
More H⁺ where H⁺ is on the left	Right	More positive
Temperature or gas pressure changed	Shifts	Changes, so not standard

Then ask: is this the positive or the negative electrode? A more positive positive electrode, or a more negative negative electrode, increases the EMF.

Data used on these notes

Half-equation	E⦵ / V
Al³⁺ + 3e⁻ ⇌ Al	−1.66
Cr³⁺ + 3e⁻ ⇌ Cr	−0.74
Cd²⁺ + 2e⁻ ⇌ Cd	−0.40
V³⁺ + e⁻ ⇌ V²⁺	−0.26
Sn²⁺ + 2e⁻ ⇌ Sn	−0.13
2H⁺ + 2e⁻ ⇌ H₂	0.00
VO₂⁺ + 2H⁺ + e⁻ ⇌ VO²⁺ + H₂O	+1.00
Au³⁺ + 3e⁻ ⇌ Au	+1.52
S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻	+2.01

Always use the values given in the question.

What E⦵ can and cannot tell you

Can tell you	Cannot tell you
Feasibility under standard conditions	Rate of reaction
Relative strength of oxidising and reducing agents	Outcome far from standard conditions
EMF of a standard cell	Whether the activation energy is too high

Required practical 6: measure the EMF of a cell

• Emery paper on metal electrodes removes oxide.
• 1.00 mol dm⁻³ solutions for a standard value; Pt where both species are ions.
• Salt bridge ions must not react with or precipitate ions in either half-cell.
• High-resistance voltmeter: almost no current, so the reading is the EMF.
• Record which electrode is negative.
• Errors to evaluate: oxide, concentration, temperature, dry salt bridge.

3.1.9.2 Commercial applications (International A-level only)

Three types of cell

Type	Key idea
Non-rechargeable (irreversible)	Reactions not reversible in practice; discarded when used up
Rechargeable	External supply reverses the electrode reactions
Fuel cell	Reactants supplied continuously; no electrical recharging

Method in steps: commercial cell from given data

1. Positive electrode: the more positive half-equation, forwards (reduction).
2. Negative electrode: the less positive half-equation, reversed (oxidation).
3. Balance electrons and add for the overall discharge reaction.
4. EMF = E⦵(positive) − E⦵(negative).
5. Rechargeable cell on charge: both electrode reactions reversed.

Lithium cell (as given in the specification)

Positive electrode:  Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻
Negative electrode:  Li → Li⁺ + e⁻

Lithium is oxidised (0 → +1); cobalt is reduced (+4 → +3). Li⁺ ions cross the electrolyte; electrons go round the external circuit. Charging reverses both reactions.

Alkaline hydrogen–oxygen fuel cell

Negative electrode:  H₂ + 2OH⁻ → 2H₂O + 2e⁻
Positive electrode:  O₂ + 2H₂O + 4e⁻ → 4OH⁻
Overall:             2H₂ + O₂ → 2H₂O

No net use of OH⁻. Hydrogen goes to the negative electrode, oxygen to the positive.

How a current is generated

Oxidation at the negative electrode releases electrons → electrons flow through the external circuit → reduction at the positive electrode takes them up → ions move inside the cell to balance charge.

Benefits and risks

Cell	Benefit	Risk
Fuel cell	Water is the only product where it runs; no CO₂ there	Making hydrogen uses energy, possibly from fossil fuels; hydrogen–air mixtures explode; bulky to store
Rechargeable	Reused over many charge cycles, so fewer cells discarded	An overcharged or crushed lithium cell can overheat and ignite
Any cell	Portable electrical energy	Lithium and cobalt must be mined; dumped cells leak harmful compounds, so collect and recycle

Must-know distinctions

This	Not this
Negative electrode = oxidation	Negative electrode = reduction
Scale the half-equation	Scale the E⦵
Sign of E⦵ from polarity	Sign from the size of the reading
Feasible	Fast
Rechargeable: external supply reverses reactions	Fuel cell: reactants replaced by continuous supply

Quick self-test

Use the data table above.

1. State the three standard conditions for E⦵.
2. Write the IUPAC half-equation for a chromium/chromium(III) electrode.
3. Calculate the EMF of a cadmium–gold cell.
4. Write the conventional representation and EMF of a chromium–tin cell.
5. Will S₂O₈²⁻ oxidise VO²⁺ to VO₂⁺? Give the EMF.
6. Will cadmium metal reduce V³⁺ to V²⁺? Give the equation.
7. Half-cell X is the positive electrode when connected to a standard cadmium half-cell. The EMF is 0.74 V. Find E⦵ for X.
8. In the cadmium–gold cell, [Cd²⁺] is increased. Predict the effect on the EMF.
9. In the lithium cell, which species is reduced during discharge?
10. Calculate the EMF of an aluminium–tin cell and write the overall equation.
11. In the alkaline fuel cell, which gas is supplied to the negative electrode?
12. Why is a non-rechargeable cell thrown away when it stops working?

Answers

1. 298 K; 100 kPa; 1.00 mol dm⁻³ solutions of ions.
2. Cr³⁺(aq) + 3e⁻ ⇌ Cr(s)
3. 1.52 − (−0.40) = +1.92 V
4. Cr(s) | Cr³⁺(aq) || Sn²⁺(aq) | Sn(s); EMF = −0.13 − (−0.74) = +0.61 V
5. Yes: 2.01 − 1.00 = +1.01 V. S₂O₈²⁻ + 2VO²⁺ + 2H₂O → 2SO₄²⁻ + 2VO₂⁺ + 4H⁺
6. Yes: −0.26 − (−0.40) = +0.14 V. Cd + 2V³⁺ → Cd²⁺ + 2V²⁺
7. E(X) = −0.40 + 0.74 = +0.34 V
8. Cd²⁺ + 2e⁻ ⇌ Cd shifts right, so the cadmium (negative) electrode becomes less negative and the EMF decreases below 1.92 V.
9. CoO₂ (cobalt +4 → +3).
10. −0.13 − (−1.66) = +1.53 V; 2Al + 3Sn²⁺ → 2Al³⁺ + 3Sn
11. Hydrogen.
12. Its reactions cannot be reversed in practice, so once a reactant is used up the cell cannot be restored.

Where marks are usually lost

• Writing an electrode half-equation as an oxidation instead of an IUPAC reduction.
• Doubling or tripling an E⦵ value when the half-equation is scaled.
• Calculating left − right, getting a negative EMF, then dropping the minus sign.
• Leaving out Pt(s) where both species are ions, or H⁺(aq) where it appears in the half-equation.
• Quoting standard conditions as “room temperature” or “1 atm” instead of 298 K and 100 kPa.
• Taking the sign of an unknown E⦵ from the voltmeter instead of from which electrode is negative.
• Explaining a concentration change without saying which electrode it affects.
• Treating feasibility as a statement about rate.
• Saying a fuel cell is recharged, or that electrons cross the salt bridge.
• Writing acid-electrolyte half-equations for the alkaline fuel cell.

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.9 Electrode potentials and electrochemical cells (International A2). Published by OxfordAQA.