Chemistry · Simulation and virtual lab Free
Electrochemical cells: electrode potentials, salt bridge and the Nernst equation
Build cells from magnesium, zinc, iron, lead, copper and silver, read the voltage and see which way the electrons flow.
- You will learn to
- Measure cell voltages and identify the anode, the cathode and the direction of electron flow.
- Calculate E°cell from standard electrode potentials and write the cell reaction.
- Explain the role of the salt bridge and the effect of concentration (Nernst equation).
- You will measure
- Voltage of each cell (V), at different concentrations.
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: metal strips, 1 M metal salt solutions, beakers, filter-paper salt bridge soaked in KNO₃, high-resistance voltmeter, leads.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To measure the voltage of electrochemical cells made from different metals, to identify the anode and cathode, to relate the cell voltage to standard electrode potentials, and to investigate the effect of concentration (Nernst equation).
What to do
- Build Zn | Zn²⁺ || Cu²⁺ | Cu with 1 M solutions. Read the voltage and record. Which metal is the anode?
- Write the two half-equations shown. In which direction do the electrons flow in the wire?
- Remove the salt bridge. What happens to the reading? Explain.
- Try Mg | Cu, Fe | Cu and Zn | Ag. Compare each reading with E°(right) − E°(left).
- Swap the electrodes (put Cu on the left). What happens to the sign of the reading?
- Keep Zn | Cu and change the copper(II) concentration to 0.1, 0.01 and 0.001 M. Record and look at the graph.
Results
Questions
What is the function of the salt bridge?
It completes the circuit by letting ions move between the two solutions, keeping each solution electrically neutral, without mixing them.
Which electrode is the anode in a Zn–Cu cell, and what happens there?
Zinc is the anode (negative electrode): oxidation, Zn → Zn²⁺ + 2e⁻.
Calculate E°cell for Zn | Zn²⁺ || Cu²⁺ | Cu. (E°: Zn −0.76 V, Cu +0.34 V)
E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = +1.10 V.
Why does a Mg–Ag cell give a larger voltage than a Fe–Cu cell?
Magnesium and silver are much further apart in the electrochemical series (E° −2.37 V and +0.80 V, a difference of 3.17 V) than iron and copper (0.78 V).
What does a negative voltmeter reading mean in this simulation?
The voltmeter is connected the other way round: the left electrode is the positive one (cathode).
What happens to the voltage of a Zn–Cu cell when the Cu²⁺ concentration is reduced?
It falls, because the reduction Cu²⁺ + 2e⁻ → Cu becomes less favourable (about 30 mV less for each tenfold dilution).
For teachers: expected results and how the model works
Electrode potentials used (V, 298 K): Mg²⁺/Mg −2.37, Zn²⁺/Zn −0.76, Fe²⁺/Fe −0.44, Pb²⁺/Pb −0.13, Cu²⁺/Cu +0.34, Ag⁺/Ag +0.80. Each electrode potential is E = E° + (0.0592 ÷ n) log[ion]; the voltmeter reads E(right) − E(left) with an ideal high-resistance voltmeter (no current drawn), so a negative reading means the left electrode is positive. Check values with 1 M solutions: Zn–Cu 1.100 V, Mg–Ag 3.170 V, Fe–Cu 0.780 V, Pb–Ag 0.930 V. Zn–Cu with 0.01 M Cu²⁺: 1.041 V. Two identical electrodes at different concentrations give a concentration cell (e.g. Cu 1 M and Cu 0.001 M: 0.089 V). Removing the salt bridge gives 0 V. Real cells give slightly lower readings because of internal resistance and oxide layers.
Exam practice Exam set
National-exam style questions on this simulation (15 marks). Try each one on paper first, then open the marking scheme.
A cell is made from Fe | Fe²⁺ (1 M) and Ag⁺ (1 M) | Ag. E°: Fe²⁺/Fe −0.44 V, Ag⁺/Ag +0.80 V. (a) Write the cell diagram. (b) Calculate E°cell. (c) Write the overall equation. (6 marks)
Marking scheme: (a) Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s) (2). (b) 0.80 − (−0.44) = +1.24 V (2). (c) Fe + 2Ag⁺ → Fe²⁺ + 2Ag (2).
Use the Nernst equation E = E° − (0.0592 ÷ n) log Q to find the voltage of Zn | Zn²⁺ (1.0 M) || Cu²⁺ (0.010 M) | Cu. (E° = 1.10 V) (4 marks)
Marking scheme: Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2 (1); Q = [Zn²⁺] ÷ [Cu²⁺] = 1.0 ÷ 0.010 = 100 (1); E = 1.10 − 0.0296 × 2 (1) = 1.04 V (1).
Explain, using electrode potentials, why iron roofing sheets are coated with zinc (galvanised). (3 marks)
Marking scheme: Zinc has a more negative electrode potential than iron (1), so zinc is oxidised in preference to iron (1), protecting the iron even if the coating is scratched (sacrificial protection) (1).
State two conditions that must be met for the voltage of a cell to equal E°cell. (2 marks)
Marking scheme: Solutions of concentration 1 mol/dm³ (1); temperature 298 K and gases at 100 kPa / no current drawn (1).
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Electrochemical cells: electrode potentials, salt bridge and the Nernst equation
Aim: To measure the voltage of electrochemical cells made from different metals, to identify the anode and cathode, to relate the cell voltage to standard electrode potentials, and to investigate the effect of concentration (Nernst equation).
Apparatus (real laboratory): metal strips, 1 M metal salt solutions, beakers, filter-paper salt bridge soaked in KNO₃, high-resistance voltmeter, leads.
What to do (with the simulation on www.donatstem.co.rw)
- Build Zn | Zn²⁺ || Cu²⁺ | Cu with 1 M solutions. Read the voltage and record. Which metal is the anode?
- Write the two half-equations shown. In which direction do the electrons flow in the wire?
- Remove the salt bridge. What happens to the reading? Explain.
- Try Mg | Cu, Fe | Cu and Zn | Ag. Compare each reading with E°(right) − E°(left).
- Swap the electrodes (put Cu on the left). What happens to the sign of the reading?
- Keep Zn | Cu and change the copper(II) concentration to 0.1, 0.01 and 0.001 M. Record and look at the graph.
Results
| # | Cell | Voltage (V) | Anode | Cathode | E° right − E° left (V) |
|---|
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
| 7 | | | | | |
| 8 | | | | | |
| 9 | | | | | |
| 10 | | | | | |
Graph
For the Zn–Cu cell, plot the voltage (V) against log₁₀ of the Cu²⁺ concentration. Find the gradient and compare with 0.0592 ÷ 2.
Questions
- What is the function of the salt bridge? [2]
- Which electrode is the anode in a Zn–Cu cell, and what happens there? [2]
- Calculate E°cell for Zn | Zn²⁺ || Cu²⁺ | Cu. (E°: Zn −0.76 V, Cu +0.34 V) [2]
- Why does a Mg–Ag cell give a larger voltage than a Fe–Cu cell? [2]
- What does a negative voltmeter reading mean in this simulation? [2]
- What happens to the voltage of a Zn–Cu cell when the Cu²⁺ concentration is reduced? [2]
- A cell is made from Fe | Fe²⁺ (1 M) and Ag⁺ (1 M) | Ag. E°: Fe²⁺/Fe −0.44 V, Ag⁺/Ag +0.80 V. (a) Write the cell diagram. (b) Calculate E°cell. (c) Write the overall equation. [6]
- Use the Nernst equation E = E° − (0.0592 ÷ n) log Q to find the voltage of Zn | Zn²⁺ (1.0 M) || Cu²⁺ (0.010 M) | Cu. (E° = 1.10 V) [4]
- Explain, using electrode potentials, why iron roofing sheets are coated with zinc (galvanised). [3]
- State two conditions that must be met for the voltage of a cell to equal E°cell. [2]
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Electrochemical cells: electrode potentials, salt bridge and the Nernst equation
Expected results
Electrode potentials used (V, 298 K): Mg²⁺/Mg −2.37, Zn²⁺/Zn −0.76, Fe²⁺/Fe −0.44, Pb²⁺/Pb −0.13, Cu²⁺/Cu +0.34, Ag⁺/Ag +0.80. Each electrode potential is E = E° + (0.0592 ÷ n) log[ion]; the voltmeter reads E(right) − E(left) with an ideal high-resistance voltmeter (no current drawn), so a negative reading means the left electrode is positive. Check values with 1 M solutions: Zn–Cu 1.100 V, Mg–Ag 3.170 V, Fe–Cu 0.780 V, Pb–Ag 0.930 V. Zn–Cu with 0.01 M Cu²⁺: 1.041 V. Two identical electrodes at different concentrations give a concentration cell (e.g. Cu 1 M and Cu 0.001 M: 0.089 V). Removing the salt bridge gives 0 V. Real cells give slightly lower readings because of internal resistance and oxide layers.
Answers
- What is the function of the salt bridge? [2]
It completes the circuit by letting ions move between the two solutions, keeping each solution electrically neutral, without mixing them.
- Which electrode is the anode in a Zn–Cu cell, and what happens there? [2]
Zinc is the anode (negative electrode): oxidation, Zn → Zn²⁺ + 2e⁻.
- Calculate E°cell for Zn | Zn²⁺ || Cu²⁺ | Cu. (E°: Zn −0.76 V, Cu +0.34 V) [2]
E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = +1.10 V.
- Why does a Mg–Ag cell give a larger voltage than a Fe–Cu cell? [2]
Magnesium and silver are much further apart in the electrochemical series (E° −2.37 V and +0.80 V, a difference of 3.17 V) than iron and copper (0.78 V).
- What does a negative voltmeter reading mean in this simulation? [2]
The voltmeter is connected the other way round: the left electrode is the positive one (cathode).
- What happens to the voltage of a Zn–Cu cell when the Cu²⁺ concentration is reduced? [2]
It falls, because the reduction Cu²⁺ + 2e⁻ → Cu becomes less favourable (about 30 mV less for each tenfold dilution).
- A cell is made from Fe | Fe²⁺ (1 M) and Ag⁺ (1 M) | Ag. E°: Fe²⁺/Fe −0.44 V, Ag⁺/Ag +0.80 V. (a) Write the cell diagram. (b) Calculate E°cell. (c) Write the overall equation. [6]
(a) Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s) (2). (b) 0.80 − (−0.44) = +1.24 V (2). (c) Fe + 2Ag⁺ → Fe²⁺ + 2Ag (2).
- Use the Nernst equation E = E° − (0.0592 ÷ n) log Q to find the voltage of Zn | Zn²⁺ (1.0 M) || Cu²⁺ (0.010 M) | Cu. (E° = 1.10 V) [4]
Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2 (1); Q = [Zn²⁺] ÷ [Cu²⁺] = 1.0 ÷ 0.010 = 100 (1); E = 1.10 − 0.0296 × 2 (1) = 1.04 V (1).
- Explain, using electrode potentials, why iron roofing sheets are coated with zinc (galvanised). [3]
Zinc has a more negative electrode potential than iron (1), so zinc is oxidised in preference to iron (1), protecting the iron even if the coating is scratched (sacrificial protection) (1).
- State two conditions that must be met for the voltage of a cell to equal E°cell. [2]
Solutions of concentration 1 mol/dm³ (1); temperature 298 K and gases at 100 kPa / no current drawn (1).