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:
    $Moles<em>{Cl</em>2} = \frac{0.0253 ext{ mol e}^-}{2} = 0.01265 ext{ mol Cl_2}$

  • Calculate grams:
    $Mass<em>{Cl</em>2} = moles \times molar \ mass = 0.01265 \text{ mol} \times 70.9 \text{ g/mol} = 0.90 g$

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.