IB Chemistry R3.2 R3.2.15

Electrolysis Calculations

Faraday's laws: relating current, time, charge, and mass deposited.

Reactivity 3.2 HL Extension ⏱️ ~5 min revision

Key Equations

Step-by-Step Method

HL Extension

Faraday's Laws: 5-Step Calculation Method

Fundamental Formulas:

\[Q = I \times t, \quad n(\text{e}^-) = \frac{Q}{F}, \quad n(\text{product}) = \frac{Q}{zF} = \frac{I \times t}{zF}\]
  1. Convert time \(t\) to seconds (\(\text{s}\)).
  2. Calculate charge: \(Q = I \times t\) (in Coulombs, C).
  3. Find moles of electrons: \(n(\text{e}^-) = \frac{Q}{96500}\).
  4. Use half-equation stoichiometry (\(z\) = moles of electrons per mole of product) to find \(n(\text{product})\).
  5. Calculate mass (\(m = n \times M\)) or gas volume at STP (\(V = n \times 22.7\text{ dm}^3\)).

Worked Example

Worked Example

Copper Electrodeposition Calculation

Problem: Calculate mass of Cu deposited when \(2.50\text{ A}\) is passed through \(\text{CuSO}_4(\text{aq})\) for \(30.0\text{ min}\) (\(M(\text{Cu}) = 63.55\text{ g mol}^{-1}\)).

\[t = 30.0 \times 60 = 1800\text{ s}\] \[Q = 2.50 \times 1800 = 4500\text{ C}\] \[n(\text{Cu}) = \frac{4500}{2 \times 96500} = 0.0233\text{ mol}\] \[m(\text{Cu}) = 0.0233 \times 63.55 = \mathbf{1.48\text{ g}}\]
Examiner Trap

Common Faraday Calculation Pitfalls

  • Time unit: Always convert minutes/hours to seconds (\(1\text{ h} = 3600\text{ s}\)).
  • Number of electrons \(z\): Check ion charge (\(\text{Ag}^+ \implies z=1, \text{Cu}^{2+} \implies z=2, \text{Al}^{3+} \implies z=3\)).
  • Gas products: For \(\text{O}_2(\text{g})\) from \(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\), \(z = 4\).
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