Potato Battery & Electrochemical Circuit Lab

Science-fair galvanic cell physics, redox half-reactions & load-line simulation

1. Half-Cell Setup

1 cm (lower R_int)8 cm
1 cm (shallow)5 cm (more surface)
“You stuck me with that zinc nail and copper penny and prayed to the heavens above that I would become an alarm clock...” — Sarah's sentient potato
Topology:
Target Load:
Connected Science Fair Apparatus
Digital LCD Alarm Clock
Requires ≥ 1.20 V & 15 µA
12:00
ACTIVE
Open-Circuit Voc
2.20 V
2 × 1.10 V per cell
Internal Res. Rint
480 Ω
Electrolyte ohmic drop
Terminal Vterm
2.19 V
Under active load
Loop Current I
36.5 µA
P = 80.0 µW

2. Electrochemistry & V-I Graph

Anode Oxidation (-)
Zn(s) → Zn²⁺ + 2e⁻
E° = -0.76 V (gives up electrons)
Cathode Reduction (+)
2H⁺ + 2e⁻ → H₂(g) (on Cu)
E° = +0.34 V (receives electrons)
Net Cell Potential ($E_{\text{cell}}$): 1.10 V
Intersection (•) marks the closed-circuit operating point.
Electrochemical Note: Phosphoric acid in the raw potato provides hydronium ($H^+$) ions that are reduced to hydrogen gas at the copper penny, while zinc metal dissolves into the potato pulp.

Why the load changes a potato battery model

Read the explanation

This teaching model assigns zinc minus zero point seven six volts and copper plus zero point three four. Their difference is one point one volts per cell. With default russet resistivity forty five, spacing three point five and depth three, its resistance formula gives five hundred twenty five ohms per cell. Two in series give two point two volts and one thousand fifty ohms. These values are source fixtures, not measurements of an actual potato. Current equals open circuit voltage divided by internal resistance plus load resistance. For two cells and the modeled sixty thousand ohm clock, two point two divided by sixty one thousand fifty gives about thirty six microamps. Multiplying by the load resistance gives about two point one six volts at the load. Open circuit voltage and loaded voltage are different quantities. The moving bars partition the voltage between internal loss and the load. The source offers two branches of two series cells. Both branches keep the same two point two volt open circuit voltage. Their equal one thousand fifty ohm resistances in parallel become five hundred twenty five ohms. Into the source’s eight hundred fifty ohm LED approximation, current rises from about one point one six milliamps to one point six milliamps. Terminal voltage rises from zero point nine eight four to one point three six volts. Neither exceeds the model’s one point seven five volt LED threshold. This linear load is an educational approximation, not a measured diode curve.

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