(Redirected from Faraday's law of electrolysis)

Faraday's laws of electrolysis are quantitative relationships based on the electrochemical research published by Michael Faraday in 1834.[1] They relate the amount of material produced at an electrode during an electrochemical reaction to the total charge passed or, equivalently, the average current and total time.

Mathematical form

Faraday's laws can be summarized by

${\displaystyle m\ =\ \left({Q \over F}\right)\left({M \over z}\right)}$

where:

• m is the mass of the substance liberated at an electrode in gms
• Q is the total electric charge passed through the substance in coulombs
• F = 96485.33289(59) C mol−1 is the Faraday constant
• M is the molar mass of the substance in grams per mol
• z is the valency number of ions of the substance (electrons transferred per ion).

Note that M/z is the same as the equivalent weight of the substance altered.

For Faraday's first law, M, F, and z are constants, so that the larger the value of Q the larger m will be.

For Faraday's second law, Q, F, and z are constants, so that the larger the value of M/z (equivalent weight) the larger m will be.

In the simple case of constant-current electrolysis, ${\displaystyle Q=It}$ leading to

${\displaystyle m\ =\ \left({It \over F}\right)\left({M \over z}\right)}$

and then to

${\displaystyle n\ =\ \left({It \over F}\right)\left({1 \over z}\right)}$

where:

• n is the amount of substance ("number of moles") liberated: n = m/M
• t is the total time the constant current was applied.

In the more complicated case of a variable electric current, the total charge Q is the electric current I(${\displaystyle \tau }$) integrated over time ${\displaystyle \tau }$:

${\displaystyle Q=\int _{0}^{t}I(\tau )\ d\tau }$

Here t is the total electrolysis time.[2]