Electrolysis Stoichiometry

Stoichiometry of an Electrolytic Cell

The extent of chemical change that occurs in an electrolytic cell is stoichiometrically related to the number of moles of electrons that pass through the cell. From the perspective of the voltage source and circuit outside the electrodes, the flow of electrons is generally described in terms of electrical current using the SI units of coulombs and amperes. It takes 96,485 coulombs to constitute a mole of electrons, a unit known as the faraday (F).

This relation was first formulated by Michael Faraday in 1832, in the form of two laws of electrolysis:

1. The weights of substances formed at an electrode during electrolysis are directly proportional to the quantity of electricity that passes through the electrolyte.
2. The weights of different substances formed by the passage of the same quantity of electricity are proportional to the equivalent weight of each substance.

The equivalent weight of a substance is defined as the molar mass divided by the number of electrons required to oxidize or reduce each unit of the substance. Thus, one mole of V³⁺ corresponds to three equivalents of this species, and will require three faradays of charge to deposit it as metallic vanadium ($V^{3+} + 3e^- \rightarrow V$).

Most stoichiometric problems involving electrolysis can be solved without explicit use of Faraday's laws. The "chemistry" in these problems is usually very elementary; the major difficulties usually stem from unfamiliarity with the basic electrical units:

• current (in amperes) is the rate of charge transport: 1 amp = 1 $\frac {Coulombs}{second}$.
• power (in watts) is the rate of energy production or consumption: 1 w = 1 $\frac {Joule}{second}$.
• 1 kilowatt-hour = 3600 J.

Example

A metallic object to be plated with copper is placed in a solution of CuSO₄. What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?

To solve, set up a dimensional analysis equation:

$1.5\ hours \times \frac {3600\ seconds}{1\ hour} \times \frac {.22\ Coulombs}{second} \times \frac {1\ mole\ e^-}{96485\ Coulombs} \times \frac {1\ mole\ Cu^{2+}}{2\ mole\ e^-} \times \frac {63.54\ grams\ Cu}{1\ mole\ Cu} =$

The answer is 0.39 g Cu.

What if this question were asked in a different fashion? How would you go about solving it?

How many amps would it take to deposit 0.39 grams of Cu in 1.5 hours?

Again, use dimensional analysis relationships to solve:

$0.39\ g\ Cu \times \frac {1\ mole\ Cu}{63.54\ g\ Cu} \times \frac{2\ moles\ e^-}{1\ mole\ Cu^{2+}} \times \frac {96485\ Coulombs}{1\ mole\ e^-} = 1184\ Coulombs$

1.5 hours is the equivalent of 5400 seconds:

$\frac{1184\ Coulombs}{5400\ seconds} = 0.22\ Amps$