Unit 9.1: Electrolysis and Faraday's Law

Electrolysis Overview

  • Definition: Electrolysis involves a non-spontaneous reaction that uses electrical energy to drive a redox reaction.

  • Key variable

    • Measured in coulombs (C);

    • Related to current ) and time through the equation:
      I = \frac{q}{t}

Relationship of Charge to Reactants and Products

  • Charge is related to the changes in concentrations of reactants and products.

  • Stoichiometry is used to analyze these relationships via dimensional analysis.

Applications of Electrolysis

  • Electroplating: Applying a coat of one metal over another;

    • Example: Silver plating on iron spoon.

      • At the anode, silver metal is oxidized to silver ions

      • At the cathode, silver ions are reduced back to solid silver on the surface of the iron spoon.

  • Water Splitting: Producing hydrogen and oxygen gases from water.

    • The reaction produces twice as much hydrogen at the cathode compared to oxygen at the anode.

  • Salt Electrolysis: Producing sodium metal and chlorine gas by electrolyzing molten sodium chloride at high temperatures.

    • Anode: Chloride ions are oxidized to chlorine gas.

    • Cathode: Sodium ions are reduced to liquid sodium.

Utilizing Dimensional Analysis in Calculations

  • Important for converting between different units in calculations involving electrolytic cells.

  • Use the following constants and units for conversion:

    • Current ($I$) in Amperes ($A$); 1 A = 1 C/s.

    • Time ($t$) in seconds (s).

    • Faraday's Constant: $F = 96,500 C/mol$.

Example Problem Walkthrough

  1. Calculating mass from charge: Given a current ($0.452 A$) and time ($1.5$ hours).

  2. Calculate charge ($q$):

    • Convert time to seconds: $1.5 ext{ hours} = 5400 s$.

    • Compute charge:
      q = I \times t = 0.452 A \times 5400 s = 2441 C

  3. Use Faraday's Constant:

    • To find moles of electrons:
      moles = \frac{q}{F} = \frac{2441 C}{96500 C/mol} = 0.0253 mol \text{ of electrons}

  4. Molar mass conversion:

    • For sodium ($Na$): Molar mass = $22.99 g/mol$.

      • Calculate mass of sodium produced:
        Mass = moles \times molar \ mass = 0.0253 \text{ mol} \times 22.99 \text{ g/mol} = 0.58 g

Calculating Mass of Chlorine Produced

  • Use charge and Faraday's constant for chlorine:

    • At the half-reaction, $Cl^-$ to $Cl_2(g)$ requires 2 mol of electrons for every mol of chlorine.

    • Molar mass of $Cl_2$ = $70.9 g/mol$.

    • Calculate moles of chlorine based on charge:
      Moles{Cl2} = \frac{0.0253 ext{ mol e}^-}{2} = 0.01265 ext{ mol Cl_2}

    • Calculate grams:
      Mass{Cl2} = moles \times molar \ mass = 0.01265 \text{ mol} \times 70.9 \text{ g/mol} = 0.90 g

Finding Time for Electroplating

  • Given mass of gold to be plated ($0.125 g$) with a current of $0.8 A$.

  1. Identify half-reaction for gold ($Au^{3+} + 3e^- \rightarrow Au$).

  2. Convert grams of gold to moles:

    • Molar mass of gold = $197 g/mol$.

  3. Equate mass to moles:
    Moles_{Au} = \frac{0.125 g}{197 g/mol} = 0.000635 \text{ mol Au}

  4. Calculate moles of electrons required:
    Moles_{e^-} = 3 \times 0.000635 = 0.001905 ext{ mol e}^-

  5. Convert moles to charge:

    • Calculate charge:
      q = moles \times F = 0.001905 \text{ mol e}^- \times 96,500 C/mol = 183.4 C

  6. Determine time using current:
    t = \frac{q}{I} = \frac{183.4 C}{0.8 A} = 229.25 s \approx 230 s

Multiple Choice Problem Summary

  • To electrolyze different ions ($Ag^+$, $Cu^{2+}$, $Fe^{3+}$, $Ti^{4+}$), the one requiring the greatest time corresponds to the ion needing the most electrons for reduction.

  • Titanium ($Ti^{4+}$) needs 4 electrons, thus requiring the longest time to plate out compared to others.

Conclusion

  • Review problems and practice calculating using the relationships between charge, mass, and time in electrolytic processes.

  • For practice, focus on questions 8, 9, and 11 to solidify understanding of electrolysis calculations.