Electrochemistry

When electrons flow from a metal that loses them easily to one that gains them readily, electric current is produced—the idea behind a galvanic cell. Standard electrode potentials give the cell voltage, and the half-cell with the higher reduction potential is the cathode. Electrolysis forces a nonspontaneous reaction with an external power supply, and Faraday’s law predicts deposited mass. Here you change the emf to follow electron flow in a galvanic cell and compare it with electrolysis.

Electron Flow Generates Electricity

💡 Analogy: Water Wheel

①Water flowing from high to low turns a water wheel

②In electrochemistry, 'electrons' flow and do work (current)

③Electrons move from a metal that easily loses electrons (Zn) to one that easily gains them (Cu)

④This electron flow is 'current' — the principle of a battery

⑤Battery = galvanic cell; plating = electrolysis

Galvanic (Voltaic) Cell

EMF E° (V)1.1

🔍 Trace the Operation!

①At Zn: Zn → Zn²⁺ + 2e⁻ (oxidation, releases electrons)

②Electrons travel through external wire to Cu

③At Cu: Cu²⁺ + 2e⁻ → Cu (reduction, gains electrons)

④Salt bridge: ions move to maintain neutrality (without it, current stops)

⑤Electrons flow from anode (−) to cathode (+)

Standard Electrode Potentials

EMF Calculation

E°_cell = E°_cathode − E°_anode

Higher reduction potential is the cathode (+)

Selected Standard Reduction Potentials

ChartStandard Reduction Potentials (25°C)

Half Reaction	E° (V)
F₂ + 2e⁻ → 2F⁻	+2.87
Cu²⁺ + 2e⁻ → Cu	+0.34
2H⁺ + 2e⁻ → H₂ (SHE)	0.00
Zn²⁺ + 2e⁻ → Zn	-0.76
Li⁺ + e⁻ → Li	-3.04

💡 How to Read the Table

①Higher E° = easier to be reduced (gains electrons readily)

②Lower E° = easier to be oxidized

③Zn-Cu cell: E° = 0.34 − (−0.76) = 1.10 V

④SHE (standard hydrogen electrode) is the reference (E° = 0.00 V)

Electrolysis

Faraday's Law

m = MItnF

m: deposited mass, M: molar mass, I: current, t: time, n: # electrons, F: 96500 C/mol

Galvanic Cell vs Electrolysis

ChartComparison of the Two

Item	Galvanic Cell	Electrolysis
Energy conversion	chem→electric	electric→chem
Spontaneity	spontaneous (E°>0)	non-spontaneous (needs external power)
Anode	(−) negative	(+) positive
Cathode	(+) positive	(−) negative
Application	batteries, fuel cells	plating, electrorefining

Worked Examples

Example 1

Find the standard cell EMF E°cell of a Zn-Cu galvanic cell. (E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V)

E°cell = E°(cathode) − E°(anode). The higher reduction potential (Cu) is the cathode (+).

E°_cell = E°_cathode − E°_anode

Cu is the cathode, Zn the anode; substitute.

E°_cell = 0.34 − (−0.76) = 1.10 V

▸ 1.10 V

The higher reduction potential is the cathode (+). If E°cell > 0, the reaction is spontaneous → works as a galvanic cell.

Example 2

Electrolyzing CuSO₄ solution at 2 A for 5 min (300 s), what mass of Cu is deposited? (Cu molar mass 64, n = 2, F = 96500)

Use Faraday’s law m = MIt/nF. First the charge It = 2 × 300 = 600 C.

m = MItnF, It = 2 × 300 = 600 C

Substitute the values.

m = 64 × 6002 × 96500 ≈ 0.199 g

▸ about 0.20 g

Faraday’s law: deposited mass is proportional to charge (It) and inversely to electron count n. Cu²⁺ has n=2.

Summary

Key Formula

E°_cell = E°_cathode − E°_anode

E°cell > 0: spontaneous (galvanic cell)

CSAT-style

Which statement about galvanic cells and electrolysis is INCORRECT?

①A galvanic cell converts chemical energy to electrical energy

②In a galvanic cell, the anode is the (−) electrode

③Electrolysis is a non-spontaneous reaction needing an external source

④In electrolysis, the anode is the (−) electrode

⑤In both, oxidation occurs at the anode and reduction at the cathode

▸ ④ In electrolysis, the anode is the (−) electrode

In electrolysis, the electrode connected to the (+) terminal is the anode = (+) electrode.

The electrolysis anode is the (+) electrode, so ④ is incorrect (the galvanic anode is (−)).

🎯 Exam Points

①E°cell = E°cathode − E°anode — higher reduction potential = (+) electrode

②Galvanic: spontaneous, anode(−), cathode(+)

③Electrolysis: non-spontaneous, anode(+), cathode(−) ← careful!

④Faraday: m = MIt/nF (deposited mass)

⑤Oxidation = lose e⁻ (OIL), Reduction = gain e⁻ (RIG)

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