Electrochemistry and Electrolysis: Fundamentals, Stoichiometry, and the Chlor-Alkali Process

Electric Charge Measurements: Electric charge is measured in Coulombs ( $C$ ).
Time Units and Standards:
- The official SI unit for time is the second ( $s$ ).
- Definition of a second: It is based on the decay of radioactive isotopes. This standard is referred to as the atomic clock.
- Traditional units of time used in calculations include minutes, hours, and days.
Defining the Coulomb:
- A Coulomb is defined mathematically as Joules over Volts ( $C = J/V$ ).
The Faraday Constant ( $F$ ):
- Value: $9.65 \times 10^4$ .
- Units: It can be expressed as Volt-moles ( $V\,mol^{-1}$ ) or Coulomb-moles ( $C\,mol^{-1}$ ).
- The choice of unit depends on what is most salient to the specific calculation or system being analyzed.

Modeling Current:
- Calculations often assume a constant current.
- In real-world electronics, current is not always constant due to brownouts, electrical surges, and other fluctuations.
- Designing electronics to handle these fluctuations is complex, and modeling non-constant current is mathematically difficult, hence the assumption of constancy in introductory models.
Stoichiometry of Electroplating:
- Electroplating problems are essentially extended stoichiometry problems.
- General logical path for calculations:
  1. Determine the mass of the metal deposited (e.g., Copper).
  2. Relate the mass to moles of the substance.
  3. Relate moles of the substance to moles of electrons ( $e^-$ ) transferred.
  4. Convert moles of electrons to Coulombs using Faraday's constant.
  5. Convert Coulombs to Amperage ( $A$ ) using time.
Electrons per Mole: To find the relationship between the substance and electrons, one must examine the specific half-cell where electroplating occurs. There is no need to look at the overall balanced equation if the half-reaction electron transfer is known.

Scenario details:
- Substance: Copper ( $Cu$ ).
- Mass deposited: $404\,mg$ .
- Duration: $5\,hours$ .
Step-by-step Calculation:
1. Mass conversion: $404\,mg = 0.404\,g$ .
2. Molar Mass of Copper: Using the periodic table, Copper is approximately $63.6\,g/mol$ .
3. Electron Transfer: The half-reaction involves $2$ moles of electrons per $1$ mole of Copper ( $Cu^{2+} + 2e^- \rightarrow Cu$ ).
4. Faraday's Constant: $9.65 \times 10^4\,C/mol\,e^-$ .
5. Calculate Total Charge (Coulombs): $0.404\,g\,Cu \times \frac{1\,mol\,Cu}{63.6\,g\,Cu} \times \frac{2\,mol\,e^-}{1\,mol\,Cu} \times \frac{9.65 \times 10^4\,C}{1\,mol\,e^-} = 1.226 \times 10^3\,C$
6. Convert Time to Seconds: $5\,hours \times 3600\,s/hour = 18000\,s$ (Note: Mentioned in transcript as $1.802 \times 10^4\,s$ or roughly $18000\,s$ ; transcript specifically reads "fifteen thousand" and "one point eight zero, two seconds").
7. Calculate Current ( $I = Q/t$ ): $\frac{1.226 \times 10^3\,C}{18000\,s} = 0.0681\,A$
Calibration Note: Most people are not naturally "calibrated" to visualize amperage numbers in daily life unless they work extensively with electronics; therefore, following strict dimensional analysis is crucial for accuracy.

Symbol Clarification:
- $C$ (by itself): Stands for Coulombs (unit of charge).
- $A$ : Stands for Amperage (unit of current).
- $I$ : Often used in physics and chemistry as the symbol for Current.
Methodology Notes:
- The instructor notes that while they agree with the values used in textbook factors, they do not always agree with the specific mechanics or "murky factors" textbooks use for setup, preferring clear dimensional analysis.

Halogens as Oxidizing Agents:
- Chlorine ( $Cl_2$ ): Famous for producing bleach.
- Bromine ( $Br_2$ ): Used as an oxidizing agent in hot tubs to keep them clean.
- Fluorine ( $F_2$ ): Extremely strong oxidizing agent.
- Public Perception of Fluoride: There is often needless fear regarding fluoride in drinking water due to confusion between the dangers of elemental Fluorine ( $F_2$ ) and the benefits of the fluoride ion ( $F^-$ ) for humanity.
- Iodine ( $I_2$ ): Used as a topical antiseptic when dissolved in alcohol.
  - The characteristic purple mark left on the skin is used by medical professionals to ensure the area is clean and ready for a vaccination shot.
Source: Iodine is commonly extracted from seawater.

Scenario details:
- Current: $8.52\,mA$ ( $8.52 \times 10^{-3}\,A$ ).
- Time: $10\,minutes$ .
- Reaction focus: Iodine production at the cathode (though traditionally an oxidation at the anode, the transcript specifies cathode collection for this setup).
Step-by-step Calculation:
1. Calculate Charge ( $Q = I \times t$ ): $8.52 \times 10^{-3}\,C/s \times 600\,s = 5.112\,C$
2. Stoichiometry Transition:
  - Coulombs to Moles of Electrons: Use Faraday's Constant ( $9.65 \times 10^4\,C/mol\,e^-$ ).
  - Moles of Electrons to Moles of Iodine ( $I_2$ ): $1$ mole of $I_2$ produced for every $2$ moles of electrons.
  - Moles of Iodine to Grams: Molar mass of diatomic Iodine ( $I_2$ ) is $254\,g/mol$ .
3. Final Setup: $5.112\,C \times \frac{1\,mol\,e^-}{9.65 \times 10^4\,C} \times \frac{1\,mol\,I_2}{2\,mol\,e^-} \times \frac{254\,g\,I_2}{1\,mol\,I_2} = 6.73 \times 10^{-3}\,g$
4. Result: $6.73\,mg$ of $I_2$ .

The Chlor-Alkali Process: This is the major industrial process involving the electrolysis of seawater.
Primary Products:
- Bleach.
- Baking soda and baking powder.
- Sodium carbonate.
- Sodium hydroxide.
- Halogens (like Iodine and Chlorine).
Medicinal Processing of Iodine:
1. Iodine is extracted via electrolysis.
2. It is dissolved in ethanol (forming a tincture).
3. It is subsequently mixed with water and other stabilization products for retail/medical use.

Dialogue on Electron Ratios:
- Student/Discussion Point: Confusion regarding the $2:1$ ratio of electrons to copper.
- Response: The ratio comes directly from the half-reaction (Copper requires 2 electrons). One must be careful to include both the electron-to-metal ratio and the Faraday constant ( $9.65 \times 10^4\,C/mol\,e^-$ ) in the sequence of dimensional analysis.
Units check (Milliamps):
- Student/Discussion Point: Calculation error by a factor of 1000.
- Response: The error was due to not converting milliamps ( $mA$ ) to amps ( $A$ ) correctly initially. $8.52\,mA$ must be treated as $8.52 \times 10^{-3}\,C/s$ .