A galvanic cell produces voltage equal to the potential of its positive electrode minus that of its negative electrode, so zinc against copper gives 1.10 V and lead dioxide against lead gives about 2.05 V. The cell’s internal resistance, found by comparing its open-circuit voltage with its voltage under a known current, decides how much of that voltage survives under load.
Part of Battery Building and Maintenance
The electrode reactions of a galvanic cell, how their potentials combine into a cell voltage, and how to measure internal resistance.
Each Electrode Has A Potential And The Cell Voltage Is Their Difference
An electrode in contact with its electrolyte has a characteristic potential, measured against a hydrogen reference defined as 0.00 V (Oxidation-Reduction). The electrode with the lower (more negative) potential gives up electrons and becomes the negative terminal of the cell; the one with the higher potential takes them and becomes the positive terminal. The cell voltage is the positive electrode’s potential minus the negative electrode’s.
At the negative electrode of a zinc cell the reaction is Zn → Zn²⁺ + 2e⁻, and at a copper positive electrode it is Cu²⁺ + 2e⁻ → Cu (Battery Fundamentals). With potentials of −0.76 V and +0.34 V the cell gives 0.34 − (−0.76) = 1.10 V, which is what the Daniell cell delivers.
Lead Has Two Potentials Because The Product Matters
The corpus quotes two different numbers for lead. The electrochemical-series table in Battery Fundamentals lists lead at −0.13 V. The Cell Voltage article uses −0.36 V together with +1.69 V for lead dioxide to get 1.69 − (−0.36) = 2.05 V for a lead-acid cell. The two are not in conflict. The explanation is general knowledge, not from the other articles: the first potential is for lead dissolving as a simple lead ion, the second for lead turning into lead sulfate, the insoluble product that actually forms in sulfuric acid. A lead-acid cell uses the second, and only the second reproduces the 2.0 to 2.1 V you measure.
Naming The Plates By Sign Avoids Confusion On Charge
On discharge the negative electrode is the anode. When the cell is recharged, current is forced through it the other way, and the textbook labels anode and cathode swap even though the plates stay where they are. In this tree the plates are always called positive and negative, by their sign during discharge, and nothing else.
What You Need
- A cell to test — any cell, from a single lead-acid cell to a Daniell cell
- A voltmeter — see Metering
- A known load resistor — one that draws a modest current, not a short
Method: Measure Open Voltage And Internal Resistance
Step 1: Open-circuit voltage. With nothing connected, read the voltage across the terminals. This is as close as you will get to the cell’s potential difference.
Step 2: Loaded voltage. Connect the load resistor, wait a few seconds and read the voltage again while measuring the current (or compute it from V ÷ R).
Step 3: Internal resistance. R_internal = (V_open − V_loaded) ÷ I. The Battery Fundamentals example: 1.1 V open, 0.9 V at 0.5 A, so R = (1.1 − 0.9) ÷ 0.5 = 0.4 Ω.
A rising internal resistance in a cell that was once low is the first measurable sign of a failing plate or separator (Maintenance and Reconditioning).
Worked Example: What A Lead-Acid Cell Gives At 10 A
Using the Cell Voltage article’s example cell with an internal resistance of 0.05 Ω, drawing 10 A: V = 2.05 − (10 × 0.05) = 1.55 V. If a short circuit were put across that cell, the current would be limited only by that resistance: I = 2.05 ÷ 0.05 = 41 A from one small cell. A large cell has a far lower internal resistance and so a far larger short-circuit current, which is why you never test with a short.
Common Mistakes
- Testing internal resistance with a dead short — the current is large, the leads heat and the spark can ignite hydrogen near a charged cell
- Mixing the two lead potentials — use −0.36 V with lead dioxide’s +1.69 V for lead-acid, and the −0.13 V series-table value for lead as a simple metal ion
Safety
Never short a charged lead-acid cell to measure it
A cell with an internal resistance of 0.05 Ω gives 41 A into a short by the arithmetic above, and a bigger cell gives much more; a tool or lead wire across the terminals can weld in place and the spark can ignite hydrogen in the cell headspace. Use a resistor of known value, and take rings and watches off first.
What’s Next
- Learn next: Lead-Acid Chemistry — the reactions of the plates in sulfuric acid
- Combine with: Cell Design — turning potentials into a built cell