19.9-19.11: Electrolysis of Molten Salts-Stoichiometry of Elecrolysis

Electrolysis: the process of producing a chemical change in an electrolytic cell, which is an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.

Downs Cell: a commercial electrochemical cell used to obtain sodium metal by the electrolysis of molten sodium chloride.



The half-reactions are:
Na⁺ (l) + e^– → Na (l)
Cl^– (l) → (½)Cl₂ (g) + e^–
Where the reduction of Na⁺ to Na occurs at the cathode and the oxidation of Cl^– to Cl₂ occurs at the anode. No water is used.

CaCl₂ is added to reduce the melting point of NaCl from 801^oC to 580^oC. And the electrodes do not need to be separated because the non-spontaneous reaction does not want to proceed anyway. This is the only way to commercially produce Na(s).

The cell reaction is:
Na⁺ (l) + Cl^– (l) → Na (l) + (½)Cl₂ (g)

Overvoltage:

“The presence of water in a solution of sodium chloride must be examined in respect to its reduction and oxidation in both electrodes. Usually, water is electrolysed as mentioned in electrolysis of water yielding gaseous oxygen in the anode and gaseous hydrogen in the cathode. On the other hand, sodium chloride in water dissociates in Na+ and Cl– ions, cation, which is the positive ion, will be attracted to the cathode (+), thus reducing the sodium ion. The anion will then be attracted to the anode (–) oxidizing chloride ion.

The following half reactions describes the process mentioned:
1. Cathode: Na+(aq) + e– → Na(s) (E° = –2.71 V)
2. Anode: 2 Cl–(aq) → Cl2(g) + 2 e– (E° = +1.36 V)
3. Cathode: 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) (E° = –0.83 V)
4. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– → (E° = +1.23 V)
Reaction 1 is discarded as it has the most negative value on standard reduction potential thus making it less thermodynamically favorable in the process.
When comparing the reduction potentials in reactions 2 and 4, the reduction of chloride ion is favored. Thus, if the Cl– ion is favored for reduction, then the water reaction is favored for oxidation producing gaseous oxygen, however experiments show gaseous chlorine is produced and not oxygen.
Although the initial analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E°cell. This may be due to kinetic rather than thermodynamic considerations. In fact, it has been proven that the activation energy for the chloride ion is very low, hence favorable in kinetic terms. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame, the voltage of the external source has to be increased (hence, overvoltage.)
Finally, reaction 3 is favorable because it describes the proliferation of OH– ions thus letting a probable reduction of H+ ions less favorable an option.
The overall reaction for the process according to the analysis would be the following:
Anode (oxidation): 2 Cl–(aq) → Cl2(g) + 2 e–
Cathode (reduction): 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
Overall reaction: 2 H2O + 2 Cl–(aq) → H2(g) + Cl2(g) + 2 OH–(aq)
As the overall reaction indicates, the concentration of chloride ions is reduced in comparison to OH– ions (whose concentration increases). The reaction also shows the production of gaseous hydrogen, chlorine and aqueous sodium hydroxide.”

Source

Electroplating of Metals:



Copper (II) ion leaves the anode and plates out on the cathode. Unreactive substances (such as gold) collect as anode mud under the anode (very valuable).


Stoichiometry of Electrolysis:

Ampere: Denoted A, is the SI base unit for current. The coulomb ( C ) is the SI unit of electric charge and is equivalent to an ampere-second.

Coulombs = Amperes x Seconds
(Electric Charge = Electric Current x Time Lapse)
and
Amperes = Coulombs / Seconds

One faraday (96,485 Coulombs) is equivalent to the charge on one mole of electrons.

Example:

System: Cu²⁺ (aq) + 2e^– → Cu (s)
A constant current passes through this system for 6.50 hours and 6.75g Cu (s) was deposits. What was the current?

Amperes = Coulombs / Seconds

To get the electric current, convert Cu from grams to moles, use stoichiometry to determine moles of e^– to moles of Cu, then use moles of e^– to convert to Coulombs:
6.75g Cu x (1 mol Cu / 63.5g Cu) x (2 mol e^– / 1 mol Cu) x (96485 C / 1 mol e^–) = 20512.6 C

To get the time, convert 6.50h to seconds:
6.50h x (3600 s / 1 h) = 23400s

Divide Coulombs by Seconds to get Amperes:
Amperes = 20512.6 C / 23400s = 0.876 Amps