Related topics: Thermodynamics · Transition Metals
Where this sits
This topic applies the redox chemistry (3.1.7) from AS (oxidation states, half-equations and electron transfer) to electrochemical cells. Instead of electrons passing directly between species, the two half-reactions are separated so the electrons travel through a wire, producing a measurable voltage. Each half cell contributes its own electrode potential to that voltage, and comparing electrode potentials lets you predict whether any redox reaction is feasible.
Half cells & the standard hydrogen electrode
Dip a metal into a solution of its own ions and an equilibrium is set up between the metal and its aqueous ions. This is a half cell, and the metal strip is the electrode.
Zn2+(aq) + 2e− ⇌ Zn(s)
The position of that equilibrium sets up a potential difference between the metal and the solution: the electrode potential. A single electrode potential cannot be measured on its own, so every value is measured relative to a reference: the standard hydrogen electrode (SHE), defined as exactly 0.00 V.
The SHE is a piece of inert platinum (coated in platinum black) dipping into 1.00 mol dm−3 H+(aq), with hydrogen gas at 100 kPa bubbled over it at 298 K. It sets up the equilibrium 2H+(aq) + 2e− ⇌ H2(g), whose potential is defined as exactly 0.00 V, so every other E⦵ is a reading taken against it.
Key definition
The standard electrode potential, E⦵, of a half cell is the EMF measured when it is connected to a standard hydrogen electrode under standard conditions: 298 K, 100 kPa and 1.00 mol dm−3 solutions of ions.
Half-equations are written by IUPAC convention as reductions, with the electrons on the left, for example Cu2+(aq) + 2e− ⇌ Cu(s), E⦵ = +0.34 V.
Half cells with no metal: using a platinum electrode
Not every half cell contains a metal that can act as the electrode. When both species are ions in the same solution, you dip inert platinum into a solution containing both ions at 1.00 mol dm−3. The platinum carries the electrons but takes no part in the reaction.
Fe3+(aq) + e− ⇌ Fe2+(aq) E⦵ = +0.77 V
If the half-equation also needs H+ (as many ion/ion couples do), then the solution must be acidified to 1.00 mol dm−3 H+ as well:
VO2+(aq) + 2H+(aq) + e− ⇌ VO2+(aq) + H2O(l) E⦵ = +1.00 V
The standard hydrogen electrode is itself this kind of half cell (a gas and an ion, with nothing metallic to conduct), which is exactly why it needs a platinum electrode too.
Recap writing and balancing half-equations and oxidation states from AS redox.
Exam questions
Q1
[3 marks]
State the substances and conditions needed in a standard hydrogen electrode.
Show answer
H2(g) at 100 kPa. 1 mark
1 mol dm−3 H+(aq): e.g. HCl or HNO3 (or 0.5 mol dm−3 H2SO4). 1 mark
A platinum electrode, at 298 K. 1 mark
Allow 1 bar for the pressure. Do not accept 1 atm or 101 kPa.
Q2
[1 mark]
Which change to a hydrogen electrode has no effect on the electrode potential? Tick (✓) one box.
The electrode potential is fixed by the position of the electrode equilibrium, which depends on [H+], the H2 pressure and the temperature. Platinum is an inert conductor, so its surface area changes nothing.
Q3
[1 mark]
State two conditions needed for the following half-cell to have E⦵ = 0.00 V
H+(aq) + e− → ½H2(g)
Show answer
Any two of: 298 K; [H+] = 1 mol dm−3; 100 kPa. 1 mark
Ignore 1 atm.
Q4
[7 marks]
The diagram represents the cell used to measure the standard electrode potential for the Fe3+/Fe2+ electrode.
(a) Name the piece of apparatus labelled A, and state its purpose. (b) Name the substance used as electrode B. (c) Identify C, D and E, including the essential conditions for each.
Show answer
A is the salt bridge. 1 mark
Its purpose is to complete the circuit / allow ions to move between the solutions / balance the charge. 1 mark
B is platinum. 1 mark
C: HCl at 1 mol dm−3 (allow 1 mol dm−3 H+, or 0.5 mol dm−3 H2SO4). 1 mark
D: hydrogen, H2, at 100 kPa. 1 mark
E: FeCl2 and FeCl3, both at 1 mol dm−3 (allow any identified Fe(II) and Fe(III) compounds at appropriate concentrations). 1 mark
All at 298 K. 1 mark
Allow 1 bar for the pressure. Do not accept 1 atm or 101 kPa. Do not accept H for hydrogen. Incorrect state symbols are not credited.
Q5
[2 marks]
The standard electrode potential, E⦵, for the Fe3+/Fe2+ electrode is +0.77 V. Give the ionic equation for the overall reaction in the cell in the diagram above. State the change that needs to be made to the apparatus in the diagram to allow the cell reaction to go to completion.
Show answer
H2 + 2Fe3+ → 2H+ + 2Fe2+
1 mark
Replace the voltmeter with a lamp / wire / ammeter (or remove the voltmeter), so that current can flow. 1 mark
Ignore state symbols; allow multiples or fractions.
Q6
[6 marks]
It is difficult to ensure consistency with the setup of a standard hydrogen electrode, so a Cu2+(aq)/Cu(s) electrode (E⦵ = +0.34 V) can be used as a secondary standard. A suitable solution containing acidified TiO2+(aq) is formed when titanium(IV) oxysulfate (TiOSO4, Mr = 159.9) is dissolved in 0.50 mol dm−3 sulfuric acid to make 50 cm3 of solution.
Describe an experiment to show that the standard electrode potential for the TiO2+(aq)/Ti(s) electrode is −0.88 V. Your answer should include how to prepare the solution, how to connect the electrodes, the measurements taken, and how they are used to calculate the standard electrode potential.
Show answer
Stage 1 — preparing the solution. Weigh out 7.995 g (8.00 g) of TiOSO4; dissolve it in the 0.50 mol dm−3 sulfuric acid; transfer to a volumetric flask with washings and make up to the mark (50 cm3 of 1 mol dm−3).
Stage 2 — setting up the cell. Dip the titanium strip into the acidified TiO2+(aq); connect the two solutions with a salt bridge (filter paper or a tube containing a saturated solution of a suitable salt); connect the two metals through a high-resistance voltmeter.
Stage 3 — measurement and calculation. Record the EMF of the cell. Use E⦵cell = E⦵(right-hand electrode) − E⦵(left-hand electrode), i.e. E⦵cell = E⦵(copper) − E⦵(titanium), so E⦵(titanium) = E⦵(copper) − E⦵cell. The reading should be +1.22 V with copper as the right-hand electrode (or −1.22 V with copper on the left), giving E⦵ = −0.88 V.
This is a levels-of-response question: 5–6 marks needs all three stages covered correctly and coherently, with clear practical detail.
Source: AQA A-Level Chemistry past papers.
Representing cells & Required Practical 8
Join two half cells with a wire and a salt bridge and you have an electrochemical cell. A standard shorthand, the cell notation, records it:
Zn(s) | Zn2+(aq) ‖ Cu2+(aq) | Cu(s)
A single line | is a phase boundary (solid–solution); the double line ‖ is the salt bridge. By convention the more negative electrode is written on the left, where oxidation occurs; reduction occurs at the right-hand (more positive) electrode.
Remember it as ROOR
Reading the notation left to right, the species always run Reduced | Oxidised ‖ Oxidised | Reduced: ROOR. The reduced forms (the metals) sit on the outside against their electrodes, and the oxidised forms (the ions) face each other across the salt bridge: Zn(s) | Zn2+(aq) ‖ Cu2+(aq) | Cu(s).
Cells with an inert electrode
When a half cell has no metal of its own, the inert electrode (Pt or C) goes on the outside of the notation, and species sharing the same solution are separated by a comma rather than a line, because there is no phase boundary between them:
Fe(s) | Fe2+(aq) ‖ VO2+(aq), H+(aq), V3+(aq) | Pt(s)
Build it from the outside in: the electrode, the species it touches, the salt bridge, then the same again in reverse. A fuel cell is written the same way, with carbon or platinum at each end:
C(s) | C6H12O6(aq), H+(aq) | CO2(g) ‖ O2(g) | H+(aq), H2O(l) | Pt(s)
The salt bridge and electron flow
- The salt bridge completes the circuit by letting ions move between the solutions to balance the charge. It contains an inert, soluble ionic solution such as KNO3 or KCl.
- Ions move through the salt bridge; electrons do not. Electrons flow through the external wire, from the negative electrode to the positive electrode.
Required practical 8: measuring the EMF of an electrochemical cell
Set up two half cells joined by a salt bridge, connect a high-resistance voltmeter across the electrodes, and read the EMF. A high-resistance voltmeter is used so that negligible current flows and the cell stays at equilibrium (standard) conditions while the reading is taken.
Swap that voltmeter for a bulb, a wire or an ammeter and current does flow: the cell reaction then runs to completion, and the EMF falls as the concentrations change. That is the change to make if a question asks how to let the cell reaction actually proceed.
The reading is easy; the marks are in the set-up:
- Clean and degrease the electrodes: emery paper for the oxide layer, propanone for the grease. Either left on gives a low, drifting reading.
- The salt bridge completes the circuit by letting ions move and keeping each solution neutral. It must not react with either half-cell, so potassium nitrate is the usual choice, never a chloride bridge with silver ions.
- Standard conditions or it is not a standard EMF: every solution 1.00 mol dm−3, 298 K, 100 kPa for any gas. Diluting one half-cell shifts its equilibrium and changes the reading.
- Electrode size makes no difference. A bigger piece of metal changes how fast charge could flow, not the potential difference. A favourite one-mark question.
- Record the sign. A negative reading is a real result: it says the second metal is the negative electrode.
This is Required Practical 8. The full apparatus, method and safety are collected in the required practicals guide.
Exam questions
Q7
[2 marks]
State the purpose of the salt bridge. Identify an ionic compound that could be used in the salt bridge of a cell containing a magnesium half cell and a standard hydrogen electrode.
Show answer
To complete the circuit / allow ions to move between the half cells / maintain electrical neutrality. 1 mark
Potassium nitrate or sodium nitrate: any soluble ionic compound that does not react with the acid, the magnesium ions or the chloride ions. 1 mark
Do not accept “allows electrons to flow”.
Q8
[2 marks]
In a glucose–oxygen fuel cell the negative electrode is made of carbon and the positive electrode is made of platinum. Give the conventional representation for the glucose–oxygen fuel cell.
Show answer
C(s) | C6H12O6(aq), H+(aq) | CO2(g) ‖ O2(g) | H+(aq), H2O(l) | Pt(s)
Correct order of species, carbon on the far left and platinum on the far right. 1 mark
Representation completely correct. 1 mark
Ignore state symbols. Zero marks if electrons are included.
Q9
[1 mark]
A cell is used to measure the standard electrode potential of the half-cell Mg2+(aq) + 2e− → Mg(s): a magnesium electrode in magnesium chloride solution, joined by a salt bridge to a standard hydrogen electrode. The voltmeter is replaced by a bulb. Give an equation for the overall reaction that occurs when the cell is operating.
Show answer
Mg + 2H+ → Mg2+ + H2
Allow Mg + 2HCl → MgCl2 + H2. 1 mark
Ignore state symbols; allow multiples.
Source: AQA A-Level Chemistry past papers.
Calculating the EMF of a cell
The cell’s EMF is the difference between the two standard electrode potentials. Subtract the more negative from the more positive so the answer comes out positive:
E⦵cell = E⦵(positive electrode) − E⦵(negative electrode)
Worked example: the zinc–copper cell
Zn2+/Zn has E⦵ = −0.76 V; Cu2+/Cu has E⦵ = +0.34 V. Copper is the more positive electrode.
E⦵cell = (+0.34) − (−0.76) = +1.10 V
Standard electrode potentials are listed together as an electrochemical series (simply a list of electrode potentials in numerical order), running from the most negative (best reducing agents) to the most positive (best oxidising agents).
When the conditions are not standard
Every E⦵ assumes 1.00 mol dm−3 at 298 K. Change a concentration and the electrode equilibrium shifts, so that electrode potential (and the cell EMF) changes with it. Apply Le Chatelier to the electrode equilibrium:
Mg2+(aq) + 2e− ⇌ Mg(s) E⦵ = −2.38 V
Dilute the Mg2+ solution and the equilibrium shifts to the left, so the magnesium releases electrons even more readily and its electrode potential becomes more negative. Measured against the standard hydrogen electrode, the EMF of the cell therefore increases.
The rule for concentration changes
- Increase the concentration of a species on the left of the half-equation → equilibrium shifts right → E becomes more positive.
- Decrease it → equilibrium shifts left → E becomes more negative.
- Whenever a measured value differs from the book value, the reason is non-standard conditions. Name the one that has changed.
The same reasoning explains what happens as a cell runs: the species being oxidised is used up and its ion concentration rises, while the ion being reduced is consumed, so the EMF gradually falls to zero as the cell goes flat.
Exam questions
Q10
[1 mark]
The E⦵ values for two electrodes are shown.
Fe2+(aq) + 2e− → Fe(s) E⦵ = −0.44 V
Cu2+(aq) + 2e− → Cu(s) E⦵ = +0.34 V
What is the EMF of the cell Fe(s) | Fe2+(aq) ‖ Cu2+(aq) | Cu(s)? Tick (✓) one box.
Copper is the positive electrode: E⦵cell = (+0.34) − (−0.44) = +0.78 V.
Q11
[1 mark]
A cell with EMF = +2.15 V is made from two electrodes. The half-equations for the two electrodes are shown.
Positive electrode: PbO2(s) + 3H+(aq) + HSO4−(aq) + 2e− → PbSO4(s) + 2H2O(l)
Negative electrode: PbSO4(s) + H+(aq) + 2e− → Pb(s) + HSO4−(aq) E⦵ = −0.46 V
What is the standard electrode potential of the PbO2/PbSO4 electrode? Tick (✓) one box.
EMF = E⦵(positive) − E⦵(negative): 2.15 = E⦵(PbO2/PbSO4) − (−0.46), so E⦵ = 2.15 − 0.46 = +1.69 V.
Q12
[1 mark]
In the 1970s lithium–iodine cells became a common power source for heart pacemakers. Lithium iodide is the final product of the cell reaction. Use the data below to calculate the cell EMF of a standard lithium–iodine cell.
Li+(aq) + e− → Li(s) E⦵ = −3.04 V
½I2(s) + e− → I−(aq) E⦵ = +0.54 V
Show answer
0.54 − (−3.04) = 3.58 V
1 mark
Q13
[1 mark]
An EMF value for a commercial lithium–iodine cell is 2.80 V. Suggest why this value is different from the value calculated in Q12.
Show answer
Non-standard conditions (allow non-aqueous conditions / different conditions). 1 mark
Q14
[2 marks]
In the magnesium cell of Q9, E⦵ for Mg2+(aq) + 2e− → Mg(s) is −2.38 V. (a) State how, if at all, the EMF of the cell changes if the surface area of the platinum electrode is increased. (b) Water is added to the beaker containing the magnesium chloride solution. What is the effect on the magnitude of the EMF of the cell?
Show answer
(a) No change: platinum is an inert conductor, so its surface area does not affect the electrode potential. 1 mark
(b) The EMF increases. 1 mark
Diluting the solution lowers [Mg2+], so the equilibrium Mg2+ + 2e− ⇌ Mg shifts to the left and the magnesium electrode becomes more negative, a bigger difference from the standard hydrogen electrode.
Source: AQA A-Level Chemistry past papers.
Predicting the feasibility of reactions
Standard electrode potentials rank how readily each species is reduced. The more positive the E⦵, the more readily that species is reduced (so it is the better oxidising agent); the more negative the E⦵, the more readily it is oxidised (the better reducing agent).
Questions also ask for the weakest agent, which is the same ranking read the other way. The weakest oxidising agent is the left-hand species of the most negative couple; the weakest reducing agent is the right-hand species of the most positive couple. Watch which side of the half-equation you are quoting: oxidising agents come from the left, reducing agents from the right.
To combine two half cells: the more positive one runs as reduction (forward), the more negative one reverses to oxidation. The reaction is feasible when E⦵cell is positive.
Worked example: will zinc react with Cu2+?
Cu2+/Cu (+0.34 V) is more positive than Zn2+/Zn (−0.76 V), so Cu2+ is reduced and Zn is oxidised.
E⦵cell = (+0.34) − (−0.76) = +1.10 V > 0 ⇒ feasible
So Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) proceeds.
Worked example: why copper resists most acids but dissolves in nitric acid
For an acid to attack copper it must oxidise Cu to Cu2+, so the acid’s couple has to be more positive than +0.34 V.
| Electrode reaction | E⦵ / V |
|---|---|
| 2H+(aq) + 2e− ⇌ H2(g) | 0.00 |
| Cu2+(aq) + 2e− ⇌ Cu(s) | +0.34 |
| NO3−(aq) + 4H+(aq) + 3e− ⇌ NO(g) + 2H2O(l) | +0.96 |
With most acids the oxidising agent is H+: E⦵cell = (0.00) − (+0.34) = −0.34 V. Negative, so H+ cannot oxidise copper. Copper is left untouched by hydrochloric or dilute sulfuric acid.
In nitric acid the nitrate ion does the oxidising: E⦵cell = (+0.96) − (+0.34) = +0.62 V. Positive, so the reaction is feasible and the copper dissolves.
2NO3− + 8H+ + 3Cu → 2NO + 4H2O + 3Cu2+
Worked example: how far down will the reduction go?
Add zinc (Zn2+/Zn, −0.76 V) to a vanadium(V) solution and it is reduced step by step. Test each step separately against zinc:
V3+ + e− ⇌ V2+ (−0.26 V): E⦵cell = (−0.26) − (−0.76) = +0.50 V ⇒ feasible
V2+ + 2e− ⇌ V (−1.20 V): E⦵cell = (−1.20) − (−0.76) = −0.44 V ⇒ not feasible
So zinc takes vanadium down to V2+ and stops there. To reduce it all the way to the metal you need a reducing agent more negative than −1.20 V. Magnesium (−2.38 V) would do it.
Precision points
- A positive E⦵cell shows a reaction is thermodynamically feasible only. It says nothing about the rate, which may be very slow.
- The prediction assumes standard conditions; changing concentration or temperature shifts the electrode potential and can change the outcome.
Exam questions
Q15
[3 marks]
Use the data in the table to explain why copper does not react with most acids but does react with nitric acid. Give an equation for the reaction between copper and nitric acid.
| Electrode reaction | E⦵ / V |
|---|---|
| 2H+(aq) + 2e− → H2(g) | 0.00 |
| Cu2+(aq) + 2e− → Cu(s) | +0.34 |
| NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l) | +0.96 |
Show answer
E⦵ for H+/H2 is less positive than E⦵ for Cu2+/Cu, so H+ cannot oxidise Cu to Cu2+: E⦵cell for Cu + 2H+ → H2 + Cu2+ is negative (−0.34 V). 1 mark
E⦵ for NO3−/NO is more positive than E⦵ for Cu2+/Cu, so NO3− can oxidise Cu to Cu2+: E⦵cell is positive (+0.62 V). 1 mark
2NO3− + 8H+ + 3Cu → 2NO + 4H2O + 3Cu2+
1 mark
Allow multiples or fractions. Ignore state symbols.
Q16
[2 marks]
Use the data in the table below to explain why Zn reduces an aqueous solution of VO2+ ions to V2+ ions, but does not reduce it any further.
| Electrode reaction | E⦵ / V |
|---|---|
| VO2+(aq) + 2H+(aq) + e− → VO2+(aq) + H2O(l) | +1.00 |
| VO2+(aq) + 2H+(aq) + e− → V3+(aq) + H2O(l) | +0.34 |
| V3+(aq) + e− → V2+(aq) | −0.26 |
| Zn2+(aq) + 2e− → Zn(s) | −0.76 |
| V2+(aq) + 2e− → V(s) | −1.20 |
| Mg2+(aq) + 2e− → Mg(s) | −2.38 |
Show answer
E⦵ for V3+/V2+ (−0.26 V) is more positive than E⦵ for Zn2+/Zn (−0.76 V), so the reduction to V2+ is feasible: E⦵cell = +0.50 V. 1 mark
E⦵ for V2+/V (−1.20 V) is more negative than E⦵ for Zn2+/Zn, so reduction to V is not feasible: E⦵cell = −0.44 V. 1 mark
Q17
[1 mark]
Identify the species in the table in Q16 that can reduce an aqueous solution of VO2+ to V.
Show answer
Mg (E⦵ = −2.38 V, the only value more negative than −1.20 V). 1 mark
Mg only.
Q18
[1 mark]
Identify the weakest reducing agent in the table below.
| Electrode reaction | E⦵ / V |
|---|---|
| [Fe(H2O)6]2+(aq) + 2e− → Fe(s) + 6H2O(l) | −0.44 |
| H+(aq) + e− → ½H2(g) | 0.00 |
| [Co(NH3)6]3+(aq) + e− → [Co(NH3)6]2+(aq) | +0.11 |
| [Fe(H2O)6]3+(aq) + e− → [Fe(H2O)6]2+(aq) | +0.77 |
| VO2+(aq) + 2H+(aq) + e− → VO2+(aq) + H2O(l) | +1.00 |
| [Co(H2O)6]3+(aq) + e− → [Co(H2O)6]2+(aq) | +1.81 |
Show answer
[Co(H2O)6]2+: it is the reduced species of the couple with the most positive E⦵, so it is the least readily oxidised. 1 mark
Do not penalise the absence of brackets.
Q19
[2 marks]
Use data from the table in Q18 to explain why [Co(H2O)6]3+(aq) will undergo a redox reaction with [Fe(H2O)6]2+(aq). Give an equation for this reaction.
Show answer
E⦵ for Co3+/Co2+ is greater than E⦵ for Fe3+/Fe2+ (1.81 > 0.77), so E⦵cell = +1.04 V. 1 mark
[Co(H2O)6]3+ + [Fe(H2O)6]2+ → [Co(H2O)6]2+ + [Fe(H2O)6]3+
1 mark
The explanation must refer to E⦵ values.
Q20
[1 mark]
Some electrode potential data are shown.
Zn2+(aq) + 2e− → Zn(s) E⦵ = −0.76 V
Pb2+(aq) + 2e− → Pb(s) E⦵ = −0.13 V
Which is a correct statement about this cell? Tick (✓) one box.
Zinc is the more negative electrode, so it is oxidised and electrons flow from zinc to lead through the wire. Pb2+ is reduced, so its concentration falls; the EMF is (−0.13) − (−0.76) = +0.63 V; and zinc dissolves rather than being deposited.
Source: AQA A-Level Chemistry past papers.
Commercial & fuel cells
Electrochemical cells power everyday devices. They fall into three types: non-rechargeable (the reaction cannot be reversed), rechargeable (the reaction is reversed by an external current), and fuel cells.
A non-rechargeable (primary) cell
A non-rechargeable cell runs one way only: once a reactant is used up the cell is spent, so it is discarded or recycled rather than recharged. The everyday zinc–manganese(IV) oxide battery is a good example. Given the two electrode reactions, you can deduce how it works:
2MnO2(s) + 2NH4+(aq) + 2e− ⇌ Mn2O3(s) + 2NH3(aq) + H2O(l) E⦵ = +0.52 V
Zn2+(aq) + 2e− ⇌ Zn(s) E⦵ = −0.76 V
MnO2 is in the more positive couple, so it is reduced (MnO2 is the oxidising agent), while zinc is oxidised at the negative electrode:
negative electrode: Zn → Zn2+ + 2e−
positive electrode: 2MnO2 + 2NH4+ + 2e− → Mn2O3 + 2NH3 + H2O
overall: Zn + 2MnO2 + 2NH4+ → Zn2+ + Mn2O3 + 2NH3 + H2O E⦵cell = (+0.52) − (−0.76) = +1.28 V
Nothing drives those changes backwards, so once the MnO2 is used up the cell is finished.
The lithium (rechargeable) cell
The simplified electrode reactions are:
negative electrode: Li → Li+ + e−
positive electrode: Li+ + CoO2 + e− → Li+[CoO2]−
Lithium cells must use a non-aqueous electrolyte. Lithium sits far below water in the series (Li+/Li is −3.04 V against −0.83 V for 2H2O + 2e− ⇌ H2 + 2OH−), so in an aqueous electrolyte the lithium would simply reduce the water to hydrogen instead of driving the cell.
On recharging, an external current drives both reactions in reverse. That is the general rule: recharging runs the discharge reaction backwards, so if you are given the discharge equation you simply reverse it. A rechargeable silver–zinc cell discharges by Zn + Ag2O → ZnO + 2Ag, so while it is charging:
recharging: 2Ag + ZnO → Zn + Ag2O
A cell can only be recharged if its electrode reactions are reversible. That is why a lithium cell recharges but a zinc–MnO2 cell does not.
These portable cells bring clear benefits (a compact, on-demand supply of electricity for phones, laptops and medical devices) but also risks to weigh. Non-rechargeable cells are cheap and reliable, yet are thrown away after a single use and add to waste; rechargeable lithium cells are reused hundreds of times but contain reactive, sometimes toxic materials, so they must be recycled or disposed of carefully rather than sent to landfill.
The alkaline hydrogen–oxygen fuel cell
A fuel cell is supplied continuously with fuel (hydrogen) and oxygen, so it produces a current without needing to be recharged.
negative electrode: H2 + 2OH− → 2H2O + 2e−
positive electrode: ½O2 + H2O + 2e− → 2OH−
overall: H2 + ½O2 → H2O
Fuel cells are efficient, and because the only product is water they produce no pollutants at the point of use. But hydrogen is difficult to store (a flammable gas, often stored under high pressure), and it is usually manufactured from hydrocarbons or by electrolysis, processes that themselves consume energy and may release CO2.
The acidic and alkaline hydrogen–oxygen cells have the same EMF (+1.23 V), because the overall reaction is the same in both: 2H2 + O2 → 2H2O. Only the mobile ion, and the side the water leaves, are different.
Deducing the half-equations for an unfamiliar fuel cell
Fuel cells in exams are often ones you have never met: methanol, glucose. You are told in words what reacts at each electrode, and you build the half-equation by balancing in this order: the main atoms, then oxygen using H2O, then hydrogen using H+, then the charge using electrons.
In a methanol–oxygen fuel cell, methanol reacts with water at the negative electrode to give carbon dioxide and hydrogen ions, while oxygen reacts with hydrogen ions at the positive electrode:
negative electrode: CH3OH + H2O → CO2 + 6H+ + 6e−
positive electrode: O2 + 4H+ + 4e− → 2H2O
overall: CH3OH + 1½O2 → CO2 + 2H2O
Given the cell EMF and one electrode potential, the other follows by rearranging E⦵cell = E⦵(positive) − E⦵(negative). Here the EMF is +1.20 V and the CO2/CH3OH electrode is +0.03 V, so E⦵(O2/H2O) = 1.20 + 0.03 = +1.23 V.
Methanol has one practical advantage over hydrogen in a car: it is a liquid, so it is much easier to store and transport, and a given volume releases more energy than the same volume of hydrogen gas.
Run the cell below with each electrolyte. The alkaline cell is the one AQA examines at A-level; the acidic version is the GCSE cell. Switching the electrolyte switches the mobile ion and the side the water leaves.
Choose an electrolyte to start the fuel cell. Then watch the ions cross the electrolyte, electrons drive the lamp, and water leave at the bottom.
H2 hydrogen
O2 oxygen
H⁺ / OH⁻ ions
electrons
H2O water
Exam questions
Q21
[1 mark]
The half-equations for two electrodes that combine to make a non-rechargeable cell are
Zn2+(aq) + 2e− → Zn(s) E⦵ = −0.76 V
2MnO2(s) + 2NH4+(aq) + 2e− → Mn2O3(s) + 2NH3(aq) + H2O(l) E⦵ = +0.52 V
Identify the oxidising agent in this cell.
Show answer
MnO2: it is in the more positive couple, so it is the species reduced. 1 mark
Q22
[1 mark]
The standard electrode potentials for two half-equations for a rechargeable silver–zinc cell are
Ag2O(s) + H2O(l) + 2e− → 2Ag(s) + 2OH−(aq) E⦵ = +0.34 V
ZnO(s) + H2O(l) + 2e− → Zn(s) + 2OH−(aq) E⦵ = −1.26 V
Give an equation for the overall reaction that occurs when the cell is recharging.
Show answer
2Ag + ZnO → Zn + Ag2O
The discharge reaction reversed. 1 mark
Ignore state symbols.
Q23
[2 marks]
The EMF of an alkaline hydrogen–oxygen fuel cell is +1.23 V. The standard electrode potential for one of the electrodes is
2H2O(l) + 2e− → 2OH−(aq) + H2(g) E⦵ = −0.83 V
Give the half-equation for the other electrode and calculate its standard electrode potential.
Show answer
O2(g) + 2H2O(l) + 4e− → 4OH−(aq)
1 mark
E⦵positive = E⦵cell + E⦵negative = 1.23 + (−0.83) = +0.40 V. 1 mark
Ignore state symbols; allow multiples.
Q24
[1 mark]
Suggest why the EMF values of the acidic and alkaline hydrogen–oxygen fuel cells are the same.
Show answer
The overall reaction is the same in both: 2H2 + O2 → 2H2O. 1 mark
Q25
[2 marks]
In a methanol–oxygen fuel cell, the overall reaction is
CH3OH(l) + 1½O2(g) → CO2(g) + 2H2O(l) EMF = +1.20 V
At the positive electrode, oxygen reacts with hydrogen ions to form water. At the negative electrode, methanol reacts with water to produce carbon dioxide and hydrogen ions. Give a half-equation for each electrode.
Show answer
positive: O2 + 4H+ + 4e− → 2H2O
1 mark
negative: CH3OH + H2O → CO2 + 6H+ + 6e−
1 mark
Q26
[1 mark]
The standard electrode potential for the CO2/CH3OH electrode is +0.03 V. Calculate the standard electrode potential for the O2/H2O electrode.
Show answer
1.20 + 0.03 = +1.23 V 1 mark
Q27
[1 mark]
Suggest one advantage of using methanol, rather than hydrogen, in a fuel cell for use in cars.
Show answer
Methanol is a liquid, so it can be stored or transported easily. (Or: more energy can be produced from 1 cm3 of liquid methanol than from 1 cm3 of hydrogen gas.) 1 mark
Ignore references to safety and cost. Do not accept “no greenhouse gas emissions”.
Q28
[2 marks]
Use the data below to explain why an aqueous electrolyte is not used for a lithium cell.
Li+(aq) + e− → Li(s) E⦵ = −3.04 V
2H2O(l) + 2e− → H2(g) + 2OH−(aq) E⦵ = −0.83 V
Show answer
Lithium would react with the water in the electrolyte. 1 mark
E⦵ for Li+/Li is more negative than E⦵ for water, so Li reduces water to hydrogen (EMF = 2.21 V). 1 mark
Q29
[1 mark]
State why lithium cells can be recharged.
Show answer
The electrode reactions can be reversed by applying a reverse potential. 1 mark
Q30
[1 mark]
A rechargeable silver–zinc cell contains a porous separator between the two electrode compartments. Suggest the function of the porous separator.
Show answer
It allows ions to move / transfer between the compartments, completing the circuit: it acts as a salt bridge. 1 mark
Ignore “allows current or charge to flow”. Do not accept “allows electrons to flow”.
Source: AQA A-Level Chemistry past papers.
The cell-reasoning checklist
Electrochemistry questions reward a few consistent habits.
Before you answer
- The more positive electrode is the positive pole (reduction); the more negative is the negative pole (oxidation).
- EMF = E⦵(positive) − E⦵(negative), a positive value.
- Electrons flow through the wire (negative → positive); ions flow through the salt bridge.
- A positive E⦵cell means feasible, not necessarily fast.
- A half cell of two ions needs an inert platinum electrode, and a comma, not a line, between them in the notation.
- If a measured value differs from the book value, say which non-standard condition changed.
- Recharging reverses the discharge equation: write it backwards.
- Reducing agents come from the right of a half-equation, oxidising agents from the left.
Precision points
- Examiner reports flag the belief that electrons flow through the salt bridge. They do not; ions do.
- They also flag reversed half-equations and the idea that a fuel cell is recharged. It is supplied with fuel, not recharged.
- Name the salt-bridge requirement precisely: an inert, soluble ionic solution.
Interactive: spot the mistake
Ten answers of the kind that lose marks on this topic. Find what is wrong with each one, but read carefully, because one of them is already right.
Capstone question
Q31
[3 marks]
Two half-cells Fe2+(aq)/Fe(s) and VO2+(aq)/V3+(aq) are connected.
Fe2+(aq) + 2e− → Fe(s) E⦵ = −0.44 V
VO2+(aq) + 2H+(aq) + e− → V3+(aq) + H2O(l) E⦵ = +0.34 V
Calculate the EMF of this cell. Give the conventional representation for this cell. Give a half-equation for the reaction that occurs at the negative electrode.
Show answer
EMF = (+0.34) − (−0.44) = +0.78 V 1 mark
Fe(s) | Fe2+(aq) ‖ VO2+(aq), H+(aq), V3+(aq) | Pt(s)
1 mark
Fe(s) → Fe2+(aq) + 2e−
1 mark
Allow the representation without H+(aq). Ignore state symbols.
Source: AQA A-Level Chemistry past papers.
Electrode potentials return in the redox chemistry of transition metals (3.2.5) later in the A2 course, where E⦵ values predict which oxidation-state changes are feasible.
3.1.11 Electrode potentials & cells: Quick-reference summary
- Half cell: a metal in a solution of its ions sets up an equilibrium; its electrode potential is measured against the standard hydrogen electrode (0.00 V) at 298 K, 100 kPa, 1.00 mol dm−3. Half-equations are written as reductions (electrons on the left).
- Cell notation: most-negative electrode on the left, e.g. Zn(s) | Zn2+(aq) ‖ Cu2+(aq) | Cu(s). Single line = phase boundary; double line = salt bridge. With no metal in a half cell, use an inert Pt (or C) electrode on the outside and a comma between species sharing a solution.
- Non-standard conditions: apply Le Chatelier to the electrode equilibrium: raising the concentration of a left-hand species makes E more positive, lowering it makes E more negative (so diluting Mg2+ raises the cell EMF).
- Salt bridge completes the circuit with mobile ions (inert, soluble, e.g. KNO3): ions move through it, not electrons. Electrons flow in the external wire from the negative to the positive electrode.
- EMF = E⦵(positive electrode) − E⦵(negative electrode), e.g. Zn/Cu = +0.34 − (−0.76) = +1.10 V.
- Feasibility: the more positive half-reaction proceeds as reduction, the more negative as oxidation; a reaction is feasible when E⦵cell > 0 (thermodynamics only, not rate).
- Commercial cells: non-rechargeable (single-use, e.g. silver oxide) and rechargeable (reversed by an external current, e.g. lithium). Deduce the electrode reactions and EMF from the given E⦵ data; weigh benefits (portable power) against risks (waste, reactive or toxic materials).
- Fuel cell (alkaline H2–O2): runs continuously on supplied fuel, not recharged; overall H2 + ½O2 → H2O.