Electrode Kinetics: Exchange Current and Overpotential · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Why real electrodes need extra voltage

Science · Chemistry · ages 21-22
Name ______________________   Date ____________

  1. Which expression defines an electrode's overpotential?

    • Cell voltage minus the solution-resistance drop
    • Actual electrode potential minus its equilibrium potential
    • Equilibrium electrode potential minus the measured electrode potential

  2. What is exchange current density, j0?

    • The magnitude of either opposing current density at equilibrium
    • The sum of the opposing current-density magnitudes
    • The maximum net current density that reactant delivery can sustain

  3. At a matched hydrogen-evolution current density under comparable conditions, which surface usually needs the smaller activation overpotential magnitude?

    • Mercury, because slower surface steps need less driving force
    • Both, because equal equilibrium potentials imply equal kinetic rates
    • Platinum, because its hydrogen reaction steps are faster

  4. A suitable anode in concentrated aqueous brine produces mainly chlorine. Besides high chloride concentration, which kinetic factor helps explain this?

    • Oxygen evolution is faster on the selected anode
    • Oxygen evolution is slower on the selected anode
    • Oxygen evolution has no kinetic barrier on the selected anode

  5. A dissolved reactant gives 2 mA at 0.10 V, 7 mA at 0.20 V and 7 mA at 0.30 V under steady conditions. Reactant delivery sets the final current. What type of region is this final part?

    • A mass-transfer-limited region
    • A charge-transfer-controlled region
    • An equilibrium region with zero net current

  6. A shallow current-potential slope near equilibrium can still occur in a charge-transfer-controlled region.

    Circle one:   True   False

  7. For otherwise matched electrodes near equilibrium under charge-transfer control, a larger j0 requires a larger overpotential magnitude to sustain the same small current density.

    Circle one:   True   False

  8. At the operating potential, oxidation paths A and B carry partial currents of 3 mA and 8 mA. Each transfers the same number of electrons per product molecule. Which product forms faster?

    • Product A, because a smaller partial current means faster product formation
    • Both products form at equal rates
    • Product B, because its partial current is larger
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Why real electrodes need extra voltage W1-mt__z7Y780pyo-s1

  1. Actual electrode potential minus its equilibrium potential · Overpotential is eta = E - Eeq, with both potentials referring to the same electrode conditions and reference.
  2. The magnitude of either opposing current density at equilibrium · At equilibrium, oxidation and reduction current densities cancel. The magnitude of either opposing current density is j0.
  3. Platinum, because its hydrogen reaction steps are faster · Platinum usually supports faster hydrogen-evolution kinetics than mercury, so it needs less activation overpotential at a matched current density.
  4. Oxygen evolution is slower on the selected anode · Slow oxygen-evolution kinetics help chloride oxidation compete on the selected anode. Sodium reduction is a separate cathode question.
  5. A mass-transfer-limited region · The final current is limited by reactant delivery. More electrode driving force cannot increase that delivery under the stated conditions.
  6. True · Charge-transfer control does not require a steep slope. The slope depends on exchange current density and other kinetic parameters.
  7. False · A larger j0 gives more current at the same small overpotential. It therefore needs a smaller overpotential magnitude for the fixed target current density.
  8. Product B, because its partial current is larger · For equal electron counts per product molecule, formation rate is proportional to partial-current magnitude. Both oxidation currents are positive, so B forms faster because 8 mA exceeds 3 mA.
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