The shape of the lesson
By the end of the lesson, learners can
- The standard hydrogen electrode: Set up the standard hydrogen electrode against a half-cell in the simulator, then build reduction half-equations: balance atoms, then add electrons to balance charge.
- E ⊖ as oxidising power: Pick oxidising and reducing agents in the simulator and read the table from top to bottom: oxidising agents on the left get stronger; reducing agents on the right get stronger the other way.
- SHE = 0.00 V: Pt, H2 100 kPa, 1.00 mol dm−3 H+, 298 K.
- Pt electrode for ion/ion and gas/ion couples.
- More positive E⊖: stronger oxidising agent (left).
- More negative E⊖: stronger reducing agent (right).
One picture every idea returns to
The standard electrode potential, E⊖, of a half-cell is the potential difference between it and the standard hydrogen electrode (SHE), measured under standard conditions: 298 K, 1.00 mol dm−3 for every ion in solution and 100 kPa (Cambridge also accepts 101 kPa) for any gas. The SHE is H2 gas bubbling over platinum (coated with platinum black) in 1.00 mol dm−3 H+, and is defined as 0.00 V. A metal dips into its own ions; an ion/ion couple such as Fe3+/Fe2+, or a non-metal such as Cl2/Cl−, needs an inert platinum electrode. Half-equations are written as reductions, so the more positive E⊖, the stronger the oxidising agent on the left; the more negative, the stronger the reducing agent on the right.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Every half-cell against hydrogen.. A single half-cell has no measurable voltage on its own, so chemists compare each one with the standard hydrogen electrode. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: The standard hydrogen electrode | Set up the standard hydrogen electrode against a half-cell in the simulator, then build reduction half-equations: balance atoms, then add electrons to balance charge. Project tab 3 of the lab (Electrode potentials and electrolysis lab, standard hydrogen electrode) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: E ⊖ as oxidising power | Pick oxidising and reducing agents in the simulator and read the table from top to bottom: oxidising agents on the left get stronger; reducing agents on the right get stronger the other way. Project tab 5 of the lab (Electrode potentials and electrolysis lab, feasible or not?) and let a student drive it while the class predicts. | Mix-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in E ⊖ as oxidising power, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| E ⊖ values are always quoted for the reduction. | Do not double E ⊖ when a half-equation is multiplied up: it is a property of the couple, not of the amount. An ion/ion or gas/ion half-cell needs a platinum electrode and, for ion/ion, equal 1.00 mol dm −3 concentrations of both ions. A more positive E ⊖ means the species on the left is a stronger oxidising agent, not that the species on the right is. |
Standard electrode potentials: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Define the term standard electrode potential and state the standard conditions.
Draw and label (or describe) the apparatus used to measure E⊖ for the Fe3+/Fe2+ couple.
Explain why a platinum electrode is used in the Cl2/Cl− half-cell rather than a chlorine electrode.
Use E⊖ values to arrange Ag+, Fe3+, I2 and Cu2+ in order of increasing oxidising power. Explain your reasoning.
Explain why a salt bridge made with potassium chloride is not suitable for a cell containing an Ag+/Ag half-cell, and give a suitable alternative.
The E⊖ for Zn2+/Zn is −0.76 V. State the polarity of the zinc electrode when connected to the SHE and the direction of electron flow in the external circuit.
Answer key and marking notes
- The potential difference (e.m.f.) of a half-cell compared with the standard hydrogen electrode.
- Measured at 298 K, with all ions at 1.00 mol dm−3, and any gases at 100 kPa (or 101 kPa).
- With a high-resistance voltmeter, so no current flows.
- Standard hydrogen electrode: H2 at 100 kPa bubbling over platinum in 1.00 mol dm−3 H+.
- Second half-cell: platinum electrode in a solution containing 1.00 mol dm−3 Fe3+ and 1.00 mol dm−3 Fe2+.
- A salt bridge of KNO3(aq) joins the two solutions.
- A high-resistance voltmeter connects the two electrodes; temperature 298 K.
- Chlorine is a gas and a non-conductor, so it cannot be an electrode.
- Platinum is inert and conducts electrons between the external circuit and the Cl2/Cl− equilibrium at its surface.
- E⊖: Cu2+/Cu +0.34, I2/I− +0.54, Fe3+/Fe2+ +0.77, Ag+/Ag +0.80 V.
- The more positive E⊖, the greater the tendency of the left-hand species to gain electrons.
- Increasing oxidising power: Cu2+ < I2 < Fe3+ < Ag+.
- Cl− ions would react with Ag+ to precipitate AgCl, changing [Ag+] and the potential.
- Use potassium nitrate, KNO3(aq), whose ions do not react.
- Zinc is the negative electrode (it is oxidised: Zn → Zn2+ + 2e−).
- Electrons flow through the wire from the zinc to the platinum of the SHE.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which statement describes the standard hydrogen electrode?
- Ahydrogen gas at 100 kPa bubbled over platinum in 1.00 mol dm−3 H+(aq) at 298 K, assigned 0.00 V
- Bhydrogen gas at 100 kPa bubbled over copper in 1.00 mol dm−3 H+(aq) at 298 K, assigned 0.00 V
- Chydrogen gas at 100 kPa bubbled over platinum in pure water at 298 K, assigned 0.00 V
- Dhydrogen gas at 100 kPa bubbled over platinum in 1.00 mol dm−3 H+(aq) at 273 K, assigned 1.00 V
Which half-cell needs an inert platinum electrode?
- ACl2(g) / Cl−(aq)
- BZn2+(aq) / Zn(s)
- CAg+(aq) / Ag(s)
- DCu2+(aq) / Cu(s)
Using E⊖ values, which species is the strongest oxidising agent: Fe3+ (+0.77 V), I2 (+0.54 V), Cu2+ (+0.34 V) or Ag+ (+0.80 V)?
- AAg+
- BFe3+
- CI2
- DCu2+
A Zn2+/Zn half-cell is connected to the standard hydrogen electrode. The voltmeter reads 0.76 V and the zinc is the negative electrode. What is E⊖(Zn2+/Zn)?
- A−0.76 V
- B+0.76 V
- C0.00 V
- D−0.38 V
Why is a salt bridge of saturated potassium nitrate used rather than potassium chloride when one half-cell is Ag+/Ag?
- Achloride ions would precipitate silver chloride, lowering [Ag+] and changing the potential
- Bpotassium chloride does not conduct electricity
- Cnitrate ions are reduced at the silver electrode, which completes the circuit
- Dpotassium nitrate lets electrons flow between the two solutions
Which metal can reduce Fe3+ to Fe2+ but cannot reduce Fe2+ to Fe? (E⊖: Zn2+/Zn −0.76, Fe2+/Fe −0.44, Cu2+/Cu +0.34, Fe3+/Fe2+ +0.77, Ag+/Ag +0.80 V)
- Acopper
- Bzinc
- Csilver
- Dmagnesium
The half-equation Ag+ + e− ⇌ Ag (E⊖ = +0.80 V) is multiplied by 2 to combine it with Cu → Cu2+ + 2e−. What is E⊖ for 2Ag+ + 2e− ⇌ 2Ag?
- A+0.80 V
- B+1.60 V
- C+0.40 V
- D−0.80 V
Suppose the Cu2+/Cu half-cell were chosen as the reference electrode and given a potential of 0.00 V. What would the electrode potential of Zn2+/Zn be on this new scale?
- A−1.10 V
- B−0.76 V
- C−0.42 V
- D+1.10 V
Quiz key
The SHE is H2 at 100 kPa (101 kPa is also accepted) over platinum coated with platinum black, dipping into 1.00 mol dm−3 H+ (e.g. 1.00 mol dm−3 HCl) at 298 K, and its potential is defined as 0.00 V. Copper is not inert and would set up its own Cu2+/Cu potential. Pure water has [H+] = 10−7 mol dm−3, not 1.00. Standard temperature is 298 K, not 273 K, and the reference value is zero by definition.
Neither chlorine gas nor chloride ions can conduct electrons to the external circuit, so platinum is used: it is inert, conducts, and the Cl2/Cl− equilibrium is set up on its surface. The same applies to ion/ion couples such as Fe3+/Fe2+. Zinc, silver and copper half-cells each use the metal itself, dipping into a solution of its own ions.
E⊖ values are for reductions, so the most positive value belongs to the species on the left that most readily gains electrons: Ag+ at +0.80 V. Fe3+ is very close (+0.77 V) but slightly weaker. I2 and Cu2+ are weaker oxidising agents still. Picking the largest charge (Fe3+) confuses ionic charge with oxidising power.
The sign of E⊖ is the sign of the half-cell's electrode relative to the SHE. Zinc is negative, so E⊖ = −0.76 V: zinc releases electrons more readily than hydrogen does. +0.76 V ignores the polarity. 0.00 V is the value of the SHE itself. −0.38 V wrongly halves the reading because Zn2+ takes two electrons; E⊖ values are not divided by the number of electrons.
Ag+(aq) + Cl−(aq) → AgCl(s), a white precipitate. This would reduce [Ag+] below 1.00 mol dm−3 and the measured potential would no longer be standard. KCl(aq) does conduct, by moving ions. Nitrate ions are spectator ions in the bridge and are not reduced. Electrons never pass through the salt bridge: they flow through the external wire, while ions move in the bridge.
A metal reduces an ion if the metal's couple has a less positive E⊖ than the ion's couple. Copper (+0.34 V) is below +0.77 V, so it reduces Fe3+ (E⊖cell = +0.43 V), but above −0.44 V, so it cannot reduce Fe2+ (E⊖cell = −0.78 V). Zinc (−0.76 V) is below both, so it reduces Fe2+ to Fe as well. Silver (+0.80 V) is above +0.77 V, so it reduces neither. Magnesium (Mg2+/Mg, −2.38 V) is even more negative than zinc, so it too would reduce Fe2+ to iron.
An electrode potential is a property of the couple: the tendency per electron, not the total for a given amount. Doubling the half-equation leaves E⊖ at +0.80 V, so E⊖cell for Cu + 2Ag+ is 0.80 − 0.34 = +0.46 V. Doubling to +1.60 V is the classic error, borrowed from ΔH. Halving it has no basis, and the sign changes only if the reaction is reversed, which it is not.
On the hydrogen scale, zinc is 0.34 − (−0.76) = 1.10 V below copper. Moving the zero to copper shifts every value by −0.34 V, so Zn2+/Zn becomes −0.76 − 0.34 = −1.10 V. −0.76 V is the value on the hydrogen scale, unchanged. −0.42 V comes from adding 0.34 instead of subtracting. +1.10 V has the wrong sign: zinc is still the more negative couple.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.