Electrochemistry

Master galvanic and electrolytic cells, cell potential, Nernst equation, and Faraday's laws for AP Chemistry.

# Electrochemistry

AP Chemistry Unit 9 connects redox chemistry to electrical energy: galvanic cells, electrolytic cells, cell potentials, the Nernst equation, and Faraday's laws.


1. Galvanic (Voltaic) Cells

Spontaneous redox → electrical energy.

  • Anode: oxidation (−), electrons leave
  • Cathode: reduction (+), electrons arrive
  • Salt bridge: maintains neutrality
  • Electron flow: anode → wire → cathode

Ecell=EcathodeEanodeE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Positive EcellE^\circ_{cell} → spontaneous.


2. Standard Reduction Potentials

  • More positive EE^\circ → stronger oxidising agent (better at being reduced)
  • More negative EE^\circ → stronger reducing agent (more easily oxidised)
  • SHE: E=0.00E^\circ = 0.00 V (reference)

3. Thermodynamics and Electrochemistry

ΔG=nFEcell\Delta G^\circ = -nFE^\circ_{cell}

where nn = moles of electrons, F=96485F = 96485 C/mol

Ecell=RTnFlnKorlnK=nFERTE^\circ_{cell} = \frac{RT}{nF}\ln K \quad \text{or} \quad \ln K = \frac{nFE^\circ}{RT}


4. Nernst Equation

Ecell=EcellRTnFlnQE_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q

At 298 K: Ecell=Ecell0.0592nlogQE_{cell} = E^\circ_{cell} - \frac{0.0592}{n}\log Q


5. Electrolytic Cells

Non-spontaneous redox driven by external voltage.

  • Anode still oxidation, cathode still reduction
  • External power source reverses natural electron flow
  • Examples: electroplating, electrolysis of water, aluminium extraction

Faraday's Laws

n=ItnForm=ItMnFn = \frac{It}{nF} \quad \text{or} \quad m = \frac{ItM}{nF}

where II = current (A), tt = time (s), nn = electrons per ion


6. Practice Questions

    1. Calculate EcellE^\circ_{cell} for Zn/Cu: EE^\circ(Cu²⁺/Cu) = +0.34 V, EE^\circ(Zn²⁺/Zn) = −0.76 V.
    1. Calculate ΔG\Delta G^\circ for this cell.
    1. Use the Nernst equation to find EcellE_{cell} when [Cu²⁺] = 0.010 M and [Zn²⁺] = 1.0 M.
    1. How long does it take to deposit 1.00 g of Cu from CuSO₄ using 2.00 A?
    1. Is a cell with E=0.50E^\circ = -0.50 V spontaneous? What would you need to make it work?

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Summary

  • Ecell=EcathodeEanodeE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}; positive = spontaneous
  • ΔG=nFE\Delta G^\circ = -nFE^\circ
  • Nernst: E=E(0.0592/n)logQE = E^\circ - (0.0592/n)\log Q
  • Electrolysis: m=ItM/(nF)m = ItM/(nF)
  • Galvanic: spontaneous; electrolytic: non-spontaneous (needs external voltage)

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