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Grade 11-12 (age 16-18)

high school Electrochemistry

Inside a battery, a metal that readily gives up electrons and ions that readily take them are kept apart with a wire between them, so the electrons can light a bulb as they travel along the wire. A device that draws current from a spontaneous redox reaction like this is a galvanic cell, and its voltage is calculated from the difference in standard reduction potentials of the two electrodes. The reverse, pushing electrons in with an external power supply to drive a reaction that would not happen on its own, is electrolysis, and the charge passed tells you how much substance is deposited. In the upper diagram, changing the anode metal changes the voltage; the lower one shows the electrolysis of copper sulfate solution for comparison with the galvanic cell.

Moving Electrons Make Electricity

💡 Think of a Water Wheel
①As water falling from high to low turns a water wheel, electrons flowing from the side that wants to give them up to the side that wants to take them do work
②Zinc gives up electrons easily and copper ions take them easily; joined by a wire, electrons flow from zinc toward copper
③The greater the difference in how strongly the two electrodes take electrons, the higher the voltage, just as a bigger drop gives water more force
④Dry cells and phone batteries are galvanic cells; electroplating and copper refining are electrolysis

The Galvanic Cell

Zn | Cu
🔍 What Happens Inside the Cell
①At the anode (−), the metal M gives up two electrons and dissolves as M²⁺: oxidation
②Those electrons travel through the wire to the copper cathode (+)
③At the cathode, Cu²⁺ from the solution takes the electrons and deposits as copper: reduction
④In the salt bridge, K⁺ moves toward the cathode, where Cu²⁺ is being used up, and NO₃⁻ moves toward the anode, where M²⁺ is building up, keeping both solutions electrically balanced; remove the bridge and the current soon stops
⑤With Mg as the anode the voltage is 2.71 V, with Pb only 0.47 V: the more readily a metal gives up electrons, the higher the voltage

Standard Electrode Potentials

Calculating the Cell Voltage
E°cell = E°cathode − E°anode
The electrode with the higher standard reduction potential is the cathode (+); the lower one is the anode (−)

Standard Reduction Potentials

Standard reduction potentials (25 °C)
Reduction half-reactionE° (V)
F₂ + 2e⁻ → 2F⁻+2.87
Cu²⁺ + 2e⁻ → Cu+0.34
2H⁺ + 2e⁻ → H₂ (SHE)0.00
Pb²⁺ + 2e⁻ → Pb−0.13
Fe²⁺ + 2e⁻ → Fe−0.44
Zn²⁺ + 2e⁻ → Zn−0.76
Mg²⁺ + 2e⁻ → Mg−2.37
Li⁺ + e⁻ → Li−3.04
💡 Reading the Potential Table
①A larger E° means easier reduction: the species takes electrons readily
②A smaller E° means the metal is oxidized more easily: it gives up electrons readily
③Pick two half-reactions: the upper one is the cathode, the lower one the anode, and the voltage is the difference between them
④For the Zn–Cu cell, 0.34 − (−0.76) = 1.10 V
⑤All values are relative to the standard hydrogen electrode (SHE), set at 0.00 V

Electrolysis

Faraday's Law
m = MItnF
m: mass deposited, M: molar mass, I: current, t: time, n: electrons per ion, F: 96500 C/mol (the charge on 1 mol of electrons)

Galvanic Cell vs Electrolysis

Comparing the Two Devices
FeatureGalvanic CellElectrolysis
Energy conversionchemical → electricalelectrical → chemical
Spontaneityspontaneous (E° > 0)nonspontaneous (needs a power supply)
Electrode where oxidation occurs(−) negative(+) positive
Electrode where reduction occurs(+) positive(−) negative
Usesdry cells, fuel cellselectroplating, copper refining

Worked Examples

Example 1
Find the standard voltage E°cell of the Zn–Cu galvanic cell. (E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V)
1
Cu, with the higher standard reduction potential, is the cathode (+); Zn, with the lower one, is the anode (−).
E°cell = E°cathode − E°anode
2
Substitute the two values.
E°cell = 0.34 − (−0.76) = 1.10 V
▸ 1.10 V
Since E° > 0, the reaction runs on its own and the cell works. This is the value shown in the upper diagram with Zn as the anode.
Example 2
A CuSO₄ solution is electrolyzed with a 2 A current for 5 minutes (300 s). What mass of Cu deposits on the negative electrode? (Molar mass of Cu 64 g/mol, F = 96500 C/mol)
1
The charge passed is current × time. Each Cu²⁺ needs 2 electrons to become Cu, so n = 2.
m = MItnF, It = 2 × 300 = 600 C
2
Substitute and calculate.
m = 64 × 6002 × 96500 ≈ 0.199 g
▸ about 0.20 g
600 C of charge is 600 ÷ 96500 ≈ 0.0062 mol of electrons, and half of that, about 0.0031 mol of Cu, is deposited. The mass deposited is proportional to the charge and inversely proportional to n.

Summary

Key Formula
E°cell = E°cathode − E°anode
If E° > 0, the reaction runs on its own and can be used as a cell
CSAT-style problem, adapted
Which statement about galvanic cells and electrolysis is NOT correct?
①A galvanic cell converts chemical energy into electrical energy
②In a galvanic cell the anode is the negative electrode
③Electrolysis is a nonspontaneous reaction that needs an external power supply
④In electrolysis the anode is the negative electrode
⑤In both devices oxidation occurs at the anode and reduction at the cathode
▸ ④ In electrolysis the anode is the negative electrode
1
In electrolysis, electrons are pulled out of the electrode joined to the supply's + terminal, so that electrode is the anode and it is positive.
2
So ④, "in electrolysis the anode is the negative electrode" is wrong; a negative anode belongs to a galvanic cell.
🎯 Exam Points
①E°cell = E°cathode − E°anode, and the electrode with the higher standard reduction potential is the positive one
②A galvanic cell runs a spontaneous reaction, with the anode negative and the cathode positive
③Electrolysis drives a nonspontaneous reaction: the electrode joined to the supply's + terminal is the anode and the one joined to − is the cathode, the opposite signs from a galvanic cell
④Faraday's law: m = MIt / nF
⑤In both devices oxidation happens at the anode and reduction at the cathode (OIL RIG: oxidation is loss, reduction is gain)
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