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CHM 1046
General Chemistry II
Dr. Michael Blaber

Electrochemistry

Quantitative Aspects of Electrolysis

Reduction half-reactions provide information about the stoichiometric requirements for the production of the elemental form of the metal. In particular, the number of electrons needed per molar basis of metal reduced

Na⁺ + e^- ® Na
Cu²⁺ + 2e^- ® Cu
Al³⁺ + 3e^- ® Al

From the above information, 1 mole of electrons will allow the production (i.e. reduction) of 1 mole of sodium metal, 0.5 mole of copper metal, and 0.333 mole of aluminum metal

For any half-reaction the amount of substance reduced is proportional to the number of electrons that pass into the cell

We can quantitate the number of electrons by keeping track of the charge that passes into a cell

Charge is measured in Coulombs

1 mole of electrons has a charge of 96,500 Coulombs (C). This number of Coulombic charge is equal to 1 Faraday (F)

Charge of 1 mole of electrons = 96,500 C = 1F

The word, current, refers to the rate of flow of electricity

One ampere (amp) is equal to a rate of flow of charge of 1 Coulomb every second

amps = Coulombs/second

and

Coulombs = amps * seconds

In summary…

Amps = charge / unit of time

Charge µ the number of electrons

The number of electrons µ the number of moles of a compound reduced

Calculate the mass of copper produced in 1.5 hours by the electrolysis of molten CuCl₂ if the electrical current is 12.3 Amps

The charge in Couloumbs = amps * seconds

C = 12.3 amps * 1.5 hours * 60 min/hour * 60 sec/min

C = 6.64 x 10⁴

Therefore, 6.64 x 10⁴ Coulombs total pass into the reduction cell (cathode) during the reaction. How many moles of electrons is this?

6.64 x 10⁴ C * (1 mole e^-/ 96,500C) = 0.688 moles e^-

How many moles of Cu can be reduced with 0.688 moles of electrons?

Cu²⁺ + 2e^- ® Cu
(it takes 2 moles of e^- to reduce 1 mole of copper ion)

0.688 moles e^- * (1 mole Cu / 2 mole e^-) = 0.344 moles Cu

Finally, how many grams of Copper are there per mole?

0.344 moles Cu * (63.5 g/mole) = 21.8 grams

Therefore, 21.8 grams of Copper will be produced

Electrical Work

Relationship between E and DG:

A positive value for E (electromotive force) is associated with a spontaneous redox reaction
Likewise, a negative value for DG is associated with a spontaneous reaction or process

+E or -DG means a spontaneous reaction

The relationship between DG and E:

DG = -nFE

Useful work

For any spontaneous process, DG is a measure of the maximum useful work (w_max)that can be done by the process.
Therefore, -nFE also is a measure of the maximum useful work that can be done by a redox reaction:

w_max = -nFE

w_max = ~~moles~~ * (~~Coulombs~~/~~moles~~) * (Joules/~~Coulomb~~) = Joules

Note: work done by the system on the surroundings is indicated by a negative sign for w

Non-spontaneous redox reactions

Electrolysis describes the application of an external EMF to drive a non-spontaneous process
For a non-spontaneous process, DG is positive and E will be negative

To force the redox reaction to proceed, the external potential must be larger in magnitude than the EMF of the cell (i.e. E_cell)

E_external > - E_cell

When an external potential is applied, the surroundings do work on the system. Therefore, we expect the sign of the w term to be positive

w_max = +nFE_external

Units of power

A watt is a unit of energy expenditure, i.e. the amount of energy consumed or produced per unit of time. It has units of Joules/second:

1 watt = 1 Joule/second

and

1 Joule = 1 watt * second

How does the Power Company charge you for your electric usage? It charges you for the watt*hours that you consume, or kilowatt*hours

1 k watt hour = (1000 watt)*(1 hour) (360 sec/hour) ((1 J/s)/watt) = 3.6 x 10⁶ Joules

Calculate the number of kilowatt hours of electricity required to produce 8.75 x 10² kg of Zinc by electrolysis of Zn²⁺ if the applied EMF is 3.75 Volts

How many moles of Zinc is 8.75 x 10² kg?

8.75 x 10² kg * 1000g/kg * 1 mole/65.4g = 1.34 x 10⁴ moles

How many moles of electrons would account for this many moles of Zinc?

Zn²⁺ + 2e^- ® Zn

Thus, 2 moles of electrons are required to reduce 1 mole of zinc

(2 moles e^- / 1 mole Zn) * 1.34 x 10⁴ moles Zn = 2.68 x 10⁴ moles e^-

How many Coulombs is this?

(96,500 C / mole e^-) * 2.68 x 10⁴ moles e^- = 2.59 x 10⁹ C

Recall that 1 Volt is the potential difference required to impart 1 J of energy to a charge of 1 Coulomb. In other words:

1 V = 1Joule / 1 Coulomb

1 Joule = 1 volt * 1 Coulomb

In the present case, the applied EMF is 3.75 Volts. In other words, we have 2.59 x 10⁹ C of charge with an applied potential of 3.75 V. The energy associated with this amount of charge having this potential is:

3.75 V * 2.59 x 10⁹ C * (1 J/V*C) = 9.67 x 10⁹ J

The conversion of J into KWh is given above:

9.67 x 10⁹ J * (1 KWh / 3.6 x 10⁶ J) = 2.69 x 10³ KWh

2000 Dr. Michael Blaber