The shape of the lesson
By the end of the lesson, learners can
- Which way does E move?: Treat each half-equation as an equilibrium and apply Le Chatelier.
- Putting numbers in the Nernst equation: Use E = E ⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]).
- E = E⊖ + (0.059 ÷ z) log ([ox] ÷ [red]) at 298 K.
- Solids (metals) are left out of the fraction.
- More oxidised form: E more positive.
- Tenfold change: 0.059 ÷ z volts.
One picture every idea returns to
For any half-cell written as a reduction, raising the concentration of the oxidised form pushes the equilibrium to the right, so E becomes more positive; raising the reduced form makes it less positive. The Nernst equation puts a number on this at 298 K: E = E⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]), where z is the number of electrons in the half-equation. Solids are left out, so for a metal E = E⊖ + (0.059 ÷ z) log [Mz+]. For Cu2+ at 0.010 mol dm−3: E = 0.34 + 0.0295 × log 0.010 = 0.34 − 0.059 = +0.281 V. A tenfold dilution changes E by only 0.059 ÷ z volts, which is why E⊖ predictions are reliable unless E⊖cell is small.
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, Change the concentration, shift the potential.. Standard electrode potentials only hold at 1.00 mol dm −3. 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: Which way does E move? | Treat each half-equation as an equilibrium and apply Le Chatelier. Project tab 6 of the lab (Electrode potentials and electrolysis lab, the Nernst equation) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Putting numbers in the Nernst equation | Use E = E ⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]). Project tab 6 of the lab (Electrode potentials and electrolysis lab, the Nernst equation) and let a student drive it while the class predicts. | Rally Coach |
| 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, 15 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 Putting numbers in the Nernst equation, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Use log to base 10, not ln, with the 0.059 V form. | Divide 0.059 by z, the electrons in the half-equation (2 for Cu 2+, 1 for Ag + and Fe 3+). Put the oxidised species on the top of the fraction, and raise concentrations to their coefficients (2Cl − gives [Cl − ] 2). Diluting the oxidised form makes E less positive, never more. |
The Nernst equation: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Calculate the electrode potential of a Ag+/Ag half-cell in which [Ag+] = 0.050 mol dm−3. (E⊖ = +0.80 V)
Calculate Ecell for the cell Zn | Zn2+ (0.50 mol dm−3) || Ag+ (0.20 mol dm−3) | Ag.
Calculate E for the Fe3+/Fe2+ half-cell when [Fe3+] = 0.10 mol dm−3 and [Fe2+] = 1.0 mol dm−3. Would Fe3+ still oxidise I− under standard conditions for the iodine half-cell?
Explain, in terms of the equilibrium Cl2 + 2e− ⇌ 2Cl−, why chlorine rather than oxygen is the main anode product when concentrated sodium chloride solution is electrolysed.
Calculate E for the Cl2/Cl− half-cell when [Cl−] = 2.0 mol dm−3 and chlorine is at 100 kPa. (E⊖ = +1.36 V)
A cell is set up from two Cu2+/Cu half-cells, one with [Cu2+] = 1.00 mol dm−3 and one with 0.010 mol dm−3. Calculate the e.m.f., identify the positive electrode and state what happens to the e.m.f. as the cell runs.
Answer key and marking notes
- z = 1: E = 0.80 + 0.059 log 0.050.
- E = 0.80 + 0.059 × (−1.301) = +0.723 V.
- E(Ag) = 0.80 + 0.059 log 0.20 = 0.80 − 0.0412 = +0.759 V.
- E(Zn) = −0.76 + 0.0295 log 0.50 = −0.76 − 0.0089 = −0.769 V.
- Ecell = 0.759 − (−0.769) = +1.53 V.
- E = 0.77 + 0.059 log (0.10 ÷ 1.0) = +0.711 V.
- Ecell = 0.711 − 0.54 = +0.171 V, still positive, so yes.
- A high [Cl−] shifts the equilibrium to the left (reduced form increased).
- So E(Cl2/Cl−) becomes less positive than +1.36 V, approaching or below E(O2/H2O).
- Cl− is therefore oxidised in preference to water, giving chlorine.
- E = 1.36 + (0.059 ÷ 2) log (1 ÷ 2.02) = 1.36 − 0.059 log 2.0.
- E = 1.36 − 0.0178 = +1.34 V.
- E(dilute) = 0.34 + 0.0295 log 0.010 = +0.281 V; E(concentrated) = +0.340 V.
- Ecell = 0.340 − 0.281 = 0.059 V; the 1.00 mol dm−3 electrode is positive (Cu2+ is reduced there).
- As it runs, the concentrated side becomes more dilute and the dilute side more concentrated, so the e.m.f. falls towards zero.
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.
The concentration of Ag+ in a Ag+/Ag half-cell is reduced from 1.00 to 0.100 mol dm−3. What happens to the electrode potential?
- Ait becomes less positive, because the equilibrium Ag+ + e− ⇌ Ag shifts to the left
- Bit becomes more positive, because the equilibrium shifts to the right
- Cit stays at +0.80 V, because E⊖ is a constant
- Dit becomes less positive, because silver metal is used up
Use the Nernst equation to calculate E for a Ag+/Ag half-cell with [Ag+] = 0.010 mol dm−3. (E⊖ = +0.80 V)
- A+0.682 V
- B+0.918 V
- C+0.741 V
- D+0.800 V
Which expression is the Nernst equation for the half-cell Cl2 + 2e− ⇌ 2Cl− (chlorine at 100 kPa)?
- AE = E⊖ + (0.059 ÷ 2) log (1 ÷ [Cl−]2)
- BE = E⊖ + (0.059 ÷ 2) log [Cl−]2
- CE = E⊖ + 0.059 log (1 ÷ [Cl−]2)
- DE = E⊖ + (0.059 ÷ 2) log (1 ÷ [Cl−])
The graph shows E for a half-cell X2+(aq)/X(s) at different concentrations of X2+. What is E⊖(X2+/X), and which metal could X be?
- A−0.44 V; iron
- B−0.499 V; iron
- C−0.44 V; zinc
- D−0.54 V; zinc
For MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O, what is the effect of lowering [H+] (raising the pH) on E and on the oxidising power of manganate(VII)?
- AE becomes less positive, so manganate(VII) is a weaker oxidising agent
- BE becomes more positive, so manganate(VII) is a stronger oxidising agent
- CE is unchanged, because H+ is not oxidised or reduced
- DE becomes less positive, so manganate(VII) is a stronger oxidising agent
Calculate Ecell for Zn | Zn2+ (0.10 mol dm−3) || Cu2+ (1.00 mol dm−3) | Cu.
- A+1.13 V
- B+1.10 V
- C+1.07 V
- D+1.16 V
From the gradient of the graph, how many electrons are involved in the half-equation X2+ + ne− ⇌ X, and does this match the formula of the ion?
- Agradient = 0.0295 V per unit of log, so z = 0.059 ÷ 0.0295 = 2, matching X2+
- Bgradient = 0.059 V per unit of log, so z = 1
- Cgradient = 0.0295 V per unit of log, so z = 0.0295
- Dgradient = 0.118 V per unit of log, so z = 4
What ratio [Fe3+] : [Fe2+] gives E = +0.888 V for the Fe3+/Fe2+ half-cell? (E⊖ = +0.77 V)
- A100 : 1
- B2 : 1
- C10 : 1
- D1 : 100
Quiz key
Removing Ag+ (the oxidised form) shifts Ag+ + e− ⇌ Ag to the left, so the electrode releases electrons more readily and E falls: 0.80 + 0.059 log 0.100 = +0.741 V. Shifting right would need more Ag+. E⊖ is constant, but E is not: E⊖ applies only at 1.00 mol dm−3. The amount of solid silver has no effect on E, because solids do not appear in the Nernst expression.
E = 0.80 + 0.059 × log 0.010 = 0.80 + 0.059 × (−2) = +0.682 V. +0.918 V adds 0.118 instead of subtracting it (a sign error in the log). +0.741 V uses z = 2, but silver ions gain only one electron. +0.800 V ignores the change of concentration.
The oxidised form is Cl2 (at standard pressure, so it counts as 1) and the reduced form is Cl− with coefficient 2, and z = 2. So E = E⊖ + (0.059 ÷ 2) log (1 ÷ [Cl−]2), which simplifies to E⊖ − 0.059 log [Cl−]: more chloride makes E less positive. Putting [Cl−]2 on top inverts the fraction and would predict the opposite effect. Using 0.059 without dividing by z = 2 doubles the correction. Leaving out the square ignores the coefficient 2 in front of Cl− and halves the correction.
At [X2+] = 1.00 mol dm−3, log [X2+] = 0, so E⊖ is the value at the right-hand end of the line: −0.44 V, matching Fe2+/Fe in the Data Booklet. −0.499 V is E at 0.010 mol dm−3, not the standard value. Zinc's E⊖ is −0.76 V, so it does not match. −0.54 V is just the bottom of the axis.
Fewer H+ ions shift the equilibrium to the left, so E falls below +1.52 V and manganate(VII) accepts electrons less readily: it is a weaker oxidant in less acidic solution. That is why it is used with excess dilute sulfuric acid in titrations. The second option reverses the direction. H+ is not redox-active here, but it appears in the equilibrium, so its concentration still affects E. A less positive E always means a weaker oxidising agent, not a stronger one.
E(Zn) = −0.76 + (0.059 ÷ 2) log 0.10 = −0.76 − 0.0295 = −0.7895 V. Ecell = 0.34 − (−0.7895) = +1.13 V: diluting the ions at the negative electrode makes it more negative and increases Ecell. +1.10 V is the standard value, ignoring the dilution. +1.07 V moves Ecell the wrong way. +1.16 V uses z = 1 for zinc.
Gradient = 0.059 ÷ 2 = 0.0295 V per unit change in log [X2+]. In the Nernst equation the gradient is 0.059 ÷ z, so z = 0.059 ÷ 0.0295 = 2, consistent with X2+ + 2e− → X. A gradient of 0.059 V would mean z = 1 (an X+ ion); the graph shows half of that. z is the ratio 0.059 ÷ gradient, not the gradient itself. 0.118 is the change for two log units if z were 1, and does not appear on this graph.
0.888 − 0.77 = 0.118 V = 0.059 log ([Fe3+] ÷ [Fe2+]), so log (ratio) = 2 and the ratio is 100 : 1. 2 : 1 confuses the log with the ratio itself. 10 : 1 would give 0.829 V. 1 : 100 would make E less positive (0.652 V), not more.
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.