F. Electrochemistry · ElectrochemistryBuilding a Galvanic Cell and Measuring Its Potential
F2.4 F2.5 A1.2 A1.4 A1.5 A1.6 A1.8 A1.10 A1.13About 75 minutes
Learning goals- I can build a galvanic cell safely and measure its cell potential with a voltmeter.
- I can use the sign of the voltmeter reading to identify the anode and cathode.
- I can compare measured cell potentials with E°cell values and explain the differences.
- I can predict the voltage of cells connected in series.
Words to know
voltmeter / multimeter
An instrument that measures potential difference. A digital multimeter shows a positive reading when its red (+) lead is on the positive electrode.
percent difference
|measured − expected| ÷ expected × 100%: a way to compare an experimental result with an accepted or calculated value.
series connection
Cells connected positive terminal to negative terminal, one after another. Their voltages add.
microscale cell
A small cell built in a well plate or on filter paper, using only a few drops of solution, which reduces waste and cost.
From the table to the bench
So far you have predicted cell potentials on paper. In this investigation you build real cells, measure them, and compare. You will find that the measured voltages are close to, but rarely exactly equal to, the E°cell values. Explaining why is the real skill, and it is what a materials scientist or battery engineer does every day.
The investigation (school lab)
Question: How do the measured potentials of several galvanic cells compare with their calculated standard cell potentials?
Materials: strips of copper, zinc, iron and magnesium; sandpaper; 1.0 mol/L solutions of Cu(NO₃)₂, Zn(NO₃)₂, Fe(NO₃)₂ (or FeSO₄) and Mg(NO₃)₂; strips of filter paper soaked in 1.0 mol/L KNO₃ (salt bridges); small beakers or a well plate; a digital multimeter with alligator-clip leads; distilled water; a thermometer.
Safety (A1.4): wear goggles and gloves. Copper(II) and zinc solutions are harmful if swallowed and toxic to aquatic life: collect all solutions in the labelled waste container. Sanding magnesium makes fine dust; sand over paper and wipe up. Read the safety data sheets (SDS) and WHMIS labels before you start.
Procedure:
- Sand each metal strip until shiny. Rinse with distilled water and dry.
- Put each metal in a container of its own ion solution (Cu in Cu²⁺, and so on).
- Connect two half-cells with a fresh KNO₃ salt-bridge strip.
- Clip the leads to the two metals. If the reading is negative, swap the leads. When the reading is positive, the red lead is on the cathode.
- Record the first steady reading (within about 10 s), to 2 decimal places. Record the temperature.
- Use a new salt bridge for each pair, so ions from one cell do not contaminate the next.
| Cell | Cathode (red lead) | E°cell (V) | Measured (V) | % difference |
|---|
| Zn–Cu | Cu | 1.10 | 1.04 | 5.5% |
| Fe–Cu | Cu | 0.79 | 0.62 | 22% |
| Mg–Cu | Cu | 2.71 | 1.95 | 28% |
| Zn–Fe | Fe | 0.31 | 0.28 | 10% |
Worked exampleExample 1: Reading the meter
A student clips the red lead to the iron strip and the black lead to the zinc strip of a Zn–Fe cell. The meter reads +0.28 V. Which electrode is the cathode? Write the overall reaction.
- A positive reading means the red lead is on the positive electrode, which is the cathode in a galvanic cell. So iron is the cathode.
- Iron is the cathode: Fe²⁺ + 2e⁻ → Fe. Zinc is the anode: Zn → Zn²⁺ + 2e⁻.
- This agrees with E° values: Fe²⁺/Fe (−0.45 V) is higher than Zn²⁺/Zn (−0.76 V).
Iron is the cathode. Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s).
Worked exampleExample 2: Percent difference
Calculate the percent difference for the Fe–Cu cell in the sample data (E°cell = 0.79 V, measured 0.62 V).
- Percent difference = |measured − expected| ÷ expected × 100%.
- = |0.62 − 0.79| ÷ 0.79 × 100% = 0.17 ÷ 0.79 × 100%.
- = 21.5%, which rounds to 22% (two significant figures, matching the data).
Why the measured values are lower (A1.8)
A good lab report does not just say "human error". It names specific causes and how each one would change the result.
- Non-standard conditions. The solutions may not be exactly 1.0 mol/L, and the room is about 22 °C, not 25 °C. Concentration and temperature change cell potentials.
- Oxide coatings. Magnesium and aluminium are covered by a thin oxide layer that re-forms quickly after sanding. It hinders the reaction, so the magnesium cell reads well below 2.71 V. This is the biggest error in the sample data.
- Iron(II) solutions oxidize in air to iron(III), so the Fe²⁺ concentration is lower than labelled.
- The meter draws a tiny current, and the cell's internal resistance (mostly in the paper salt bridge) lowers the reading.
- Contamination: a salt bridge reused between cells carries the wrong ions.
A consistency check: E° values add up exactly: E°(Zn–Fe) + E°(Fe–Cu) = 0.31 + 0.79 = 1.10 V = E°(Zn–Cu). The measured values do not quite add up (0.28 + 0.62 = 0.90 V, not 1.04 V), which tells you that the errors are different for each cell; the iron half-cell is the likely culprit, since it appears in both of the cells that add up to too little.
Cells in series
A single galvanic cell rarely gives enough voltage. Connect cells in series, positive of one to negative of the next, and their voltages add. Six lead-acid cells make a 12 V car battery. A lemon cell gives under 1 V, but two or three in series can light a red LED, which needs roughly 2 V to turn on.
Series connection raises voltage but not the current a cell can push. Lemon cells have a very high internal resistance, so they can light a small LED but not a phone.
Worked exampleExample 3: How many lemons?
Each lemon cell gives 0.90 V. A red LED needs at least 1.8 V, and a blue LED about 3.0 V. How many lemons in series does each need?
- Red: 1.8 ÷ 0.90 = 2.0, so 2 lemons (3 to be safe, since voltage drops when current flows).
- Blue: 3.0 ÷ 0.90 = 3.3, so you need 4 lemons (round up).
Red LED: 2 to 3 lemons. Blue LED: 4 or more.
Just for funIf you connected about 1000 lemons in series you would have a battery of roughly 900 V and a fruit bowl that requires a safety training course.
Hands-on activityLemon battery that lights an LED
You need: 3 or 4 lemons, 3 or 4 galvanized (zinc-coated) nails or screws, 3 or 4 pieces of bare copper wire (about 8 cm each), 5 alligator-clip leads (or insulated wire), one red LED (from an electronics store or an old kit; it is a small part, so keep it away from young children), and a multimeter if you have one. A grown-up helps with the knife.
- Roll each lemon to release juice. A grown-up makes two small slits about 3 cm apart in each lemon.
- In each lemon, push a nail into one slit and a copper wire into the other. Do not let them touch.
- Measure one lemon: red lead on copper, black on zinc. Record the voltage.
- Connect the lemons in series: copper of lemon 1 to zinc of lemon 2, copper of lemon 2 to zinc of lemon 3.
- Measure across the whole chain (zinc of the first lemon to copper of the last). Compare with 3 × (one lemon).
- Connect the LED: its longer leg goes to the copper end (positive), the shorter leg to the zinc end. Dim the lights. Does it glow? If not, add a fourth lemon, or flip the LED.
- Analyse: Was the chain voltage exactly 3 times one lemon? Why might it be less?
What to look for: a single lemon usually gives somewhere around 0.8 to 1.0 V, and the chain voltage close to the sum. The LED glows faintly, because lemon cells can only supply a tiny current. Throw the lemons away afterwards; don't eat them (zinc and copper dissolve into the juice).
Your turn
1In a Mg–Cu cell, which metal should the red lead be attached to for a positive reading?
Copper, the cathode (higher E°).
2A meter reads −0.45 V when the red lead is on nickel and the black lead is on silver. Which electrode is the cathode?
The reading is negative, so the red lead is on the negative electrode (the anode). Silver is the cathode.
3Calculate the percent difference: E°cell = 1.10 V, measured 1.01 V.
|1.01 − 1.10| ÷ 1.10 × 100% = 8.2%.
4Why must each metal be placed in a solution of its own ions rather than in any solution?
So each half-cell has its own half-reaction and the oxidizing agent cannot touch the reducing agent directly; otherwise electrons transfer in the beaker instead of through the wire.
5Why should a new salt bridge be used for each cell?
A reused bridge carries ions from the previous cell, contaminating the new half-cells and changing the voltage.
6Four cells of 1.5 V each are connected in series. What is the total voltage?
7Give one specific reason why a measured Mg–Cu cell potential is far below 2.71 V.
Magnesium's oxide coating hinders the reaction at the magnesium surface (also: non-standard concentrations, meter current, internal resistance).
More practice: warm-up
1When the multimeter reading is positive, the red lead is on which electrode?
The cathode (positive electrode).
2Why are metals sanded before the lab?
To remove oxide coatings.
3Do voltages add in series or in parallel?
4What is the E°cell of a Zn–Cu cell?
5Name one safety precaution for this lab.
Wear goggles (and gloves); collect metal-ion solutions in a waste container; read the SDS.
More practice: on your own
1A student measures a Zn–Cu cell at 1.06 V and a Cu–Ag cell at 0.43 V. Predict what a Zn–Ag cell should measure, and give the E°cell for comparison.
Measured values add approximately: 1.06 + 0.43 = about 1.49 V. E°cell = 0.80 − (−0.76) = 1.56 V.
2Explain why using 0.10 mol/L solutions in both half-cells of a Zn–Cu cell still gives a voltage close to 1.10 V.
When both ion concentrations are equal and both ions have the same charge (2+), the effects of the lower concentration on the two half-cells largely cancel, so the cell potential stays close to E°cell. (Only the ratio [Zn²⁺]/[Cu²⁺] matters for this cell.)
3A Zn–Cu cell's reading slowly drops from 1.04 V to 0.98 V over 10 minutes while connected to a small motor. Give two explanations.
The reaction changes the concentrations ([Cu²⁺] down, [Zn²⁺] up), lowering the potential; and current through the internal resistance (especially a paper salt bridge drying out) reduces the measured voltage.
4Design a data table for this investigation, including units and the number of decimal places.
Columns: Cell | Anode | Cathode | E°cell (V, 2 d.p.) | Measured E (V, 2 d.p.) | % difference (2 sig. figs.) | Temperature (°C). One row per cell.
5Use significant figures correctly: E°cell = 2.71 V, measured 1.95 V. Find the percent difference.
|1.95 − 2.71| ÷ 2.71 × 100% = 0.76 ÷ 2.71 × 100% = 28.04...% ≈ 28%.
Challenge
1A student builds Zn | Zn²⁺ || Cu²⁺ | Cu but uses 1.0 mol/L Zn²⁺ and only 0.010 mol/L Cu²⁺. Predict qualitatively whether the reading will be above or below 1.10 V, and explain using Le Châtelier's principle.
Below 1.10 V. With less Cu²⁺ (a reactant) the forward reaction Zn + Cu²⁺ → Zn²⁺ + Cu is less favoured, so the push on electrons is smaller. (The Nernst equation you will meet at university predicts about 1.04 V.)
2Lemon cells have an internal resistance of several thousand ohms. Use Ohm's law (from SPH3U) to estimate the largest current one lemon cell (0.9 V) could supply through a short circuit if its internal resistance is 3000 Ω. Why can't lemons charge a phone?
I = V/R = 0.9 V ÷ 3000 Ω = 3 × 10⁻⁴ A (0.3 mA). A phone needs on the order of 1 A, thousands of times more, so lemon batteries can't supply enough current, even when many are connected.
3Plan an extension: how would you find E° for an unknown metal X using only your copper half-cell and a voltmeter? State what you would measure and how you would calculate E°(X).
Build X | Xⁿ⁺ (1.0 mol/L) || Cu²⁺ (1.0 mol/L) | Cu. Measure E and note which electrode is the cathode. If Cu is the cathode, E°(X) = 0.34 − E; if X is the cathode, E°(X) = 0.34 + E. Repeat three times with fresh bridges and average. Report the result with an uncertainty, and discuss oxide layers and concentrations.
Quiz
1. A digital multimeter connected to a galvanic cell shows a positive reading. The red lead is on:
2. Which is the best explanation for a measured Mg–Cu cell reading of 1.9 V instead of 2.71 V?
3. E°cell is 0.46 V and the measured value is 0.41 V. What is the percent difference?
4. Three cells of 1.2 V each are connected in series. The total voltage is:
5. Why should each half-cell get a fresh salt bridge when testing many pairs?
6. A lemon battery lights an LED only faintly, even with the correct voltage. The best explanation is:
Big idea (F2.4, F2.5): real galvanic cells give voltages close to, but usually below, their calculated E°cell values. Measuring, then explaining each difference with a specific cause, is the core of scientific investigation.
How to help: Build the lemon battery together. Ask your teen to predict the chain voltage before measuring and to explain why it is a little lower than predicted. Ask: "Which way does current flow through the LED, and why does the longer leg go to copper?"
Common difficulties:
- Treating a negative meter reading as an error rather than information (it just means the leads are reversed).
- Writing vague error sources ("human error"). Encourage specific ones: oxide layers, concentrations, temperature, internal resistance.
- Confusing voltage and current: more cells in series raise voltage, not the current a weak cell can supply.
Safety: a grown-up should cut the lemons. LEDs are small parts that young children could swallow. Throw the lemons away afterwards.