Electrochemistry is the study and quantification of the
controlled release of energy in redox systems.
Chemistry which generates electric circuits and potential.
Electrochemistry
—OIL RIG
Electrochemistry involves the study
of the transfer of electrons and
concerns mainly redox reactions.
Reduction is the gain of electrons
Oxidation is the loss of electrons
A process where the amount of charge during
the electrolysis process is specifically
controlled. This allows control over the amount
of a substance undergoing the reaction.
Quantitative Electrolysis
—First Law of Electrolysis by
Faraday
the amount of chemical reaction
that occurs at the electrodes is
proportional to the flow of
electrons (electricity passed
through the electrolyte).
the amount of electricity being
passed into the system has a
directly proportional output-mass
of the chemical deposition of the
electrolytic reaction, which is
quantifiable/measurable.
—Second Law of Electrolysis
by Faraday
the same amount of electricity
through
different electrolytes will
produce a different mass of
chemical deposition.
The mass of chemical deposition that
occurs will be different based on the
amount of electrons needed to be
discharged, even if the amount of
electricity passed through the system
remains the same for two
different electrolytes.
where F = Faraday's constant
L = Avogadro's constant
e = charge on the electron
n = number of moles (mol)
Q = Charge (in Coulombs)
The Faraday (F) constant, 9.65 x 104 C mol-1.(C =
Coulombs)
Q = Charge in Coulombs
I = Current in amperes or Coulombs/sec
t = time in seconds
Basic Formulate for Calculation
Problem Example
● An aqueous solution of copper (II) nitrate was electrolyzed using a
current of 1.5 A for 60 minutes at 25°C and 1 atm. Calculate
○ (a) the mass of copper metal deposited cathode
○ (b) the volume of gas evolved at the anode
Molar mass of Cu: 63.5 g/mol
Solution
● (a) The amount of electricity charge passed through the solution.
○ 𝑄 = 𝐼 𝑡
○ 𝑄 = 1.5 𝑎 × 60 60
○ 5.4 × 103
C
○ 96500 𝐶 = 1 𝐹
○ Convert it to Faraday’s constant: 5.6 × 10−2
𝐹
○ Half equation at the cathode: Cu 2+ (aq) + 2e- -> Cu (s)
○ Thus, 2 F of charge forms 1 mol Cu
○
1 𝑚𝑜𝑙 ×5.6 ×10−2𝐹
2 𝐹
𝐶𝑢 = 2.8 × 10−2
𝑚𝑜𝑙 𝐶𝑢
○ 𝑀𝑎𝑠𝑠 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 × 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
○ 2.8 × 10−2
𝑚𝑜𝑙 × 63.5 𝑔. 𝑚𝑜𝑙−1
○ 1.8 g