electrochemistry

1. Electrochemistry
ECHMTB2-Lecture 5
BY: George Tsoeunyane
12/08/2022

2. Junction
Potentials
• Potential develops at any
point where there is charge
separation.
• Ions they move in the
presence of electric field.
• Differences in ion mobility
gives rise to liquid junction
potentials.
Ecell = Ecathode – Eanode + Ej
• Junction Potential can develop in between
solutions of different concentration

3. Electrical
double layer
• The surface of the metal
electrode is shown as having
excess of positive charge as
the consequence of an applied
positive potential.
• The double layer exists
between metal-solution
interface.
• The solution layer consists of
two parts
i. Tightly bound inner
layer(potential decreases
linearly with distance)
ii. Loosely bound outer
layer(decrease is
exponential).

4. Currents in Electrochemical Cells
• All electrochemical reactions takes place at the solution-electrode interfaces.
• Currents within the electrochemical cells are limited by (i) charge transfer resistance (ii) mass
transport resistance and ohmic solution resistance.
• Mode of transport (i) diffusion (ii) convection & (iii) migration.
• As the results of transportation of species, there Faradic and non-faradic currents within the
electrochemical cell.
• Faradic currents are due to transfer of electrons from the electrode by redox reactions.
• Non-Faradic currents are due to other processes rather than redox reactions.

5. Modes of Transport In The Electrochemical cell.
Types of mass transport
• Convection
• Migration
• Diffusion

6. Nernst Equation and its permutations
1. Meaningful cell emf can only be obtained at equilibrium.
2. The emf comprises of two electrode potentials.
3. The potential of one electrode can be defined in relation to reference electrode.
• These enables us to extract electroanalytical data from the cells by potentiometric approach.
• The amount of material in the analyte solution are related to the electrode potential by according to
Nernst equation.

7. Nernst equation
• Suppose we have a half-cell reaction
pP + qQ + …ne- → rR + sS + …
• The Nernst equation for the reaction would be
𝐸𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒 = 𝐸𝑜 +
𝑅𝑇
𝑛𝐹
ln
(𝑎𝑅)𝑟
. (𝑎𝑆)𝑠
…
(𝑎𝑝)𝑝. (𝑎𝑄)𝑞 …
Where; R – ideal gas law constant (8.316 J mol-1 K-1), T – absolute temperature (K), F – Faraday
constant, number of electrons involved in a process and a – activity of involved species (γ[X]).

8. Nernst equation
• Example: A sample of iron-containing ore is crushed and the powder extracted to form a clear
aqueous solution. A clean iron rod is immersed in the solution and the cell Fe | Fe2+ || SCE is
therefore made. The emf at equilibrium was measured as 0.714 V at 298 K and the SCE was
positive electrode which 0.242 V. What is the concentration of iron? (assume that all iron exist as
simple aquo ion in a +2 oxidation state and that the solution is “quite”.
• Solution:
The procedure is to adopt a two step process, initially determine the electrode potential EFe|Fe2+ from
the emf. Then concentration will be calculated using Nernst equation.

9. • Step 1
Emf = SCE - EFe|Fe2+
Therefore, EFe|Fe2+ = -0.472 V
• Step 2
Determining the concentration of [Fe2+] – it is actually activity of aquo Fe (a((Fe2+)).
EFe2+,Fe = Eo
Fe2+, Fe +
𝑅𝑇
𝑛𝐹
ln[
𝑎(𝐹𝑒2+)
𝑎(𝐹𝑒)
]
From standard reduction potential, Eo
Fe2+, Fe = -0.44 V
The unit
𝑅𝑇
𝐹
= (8.316 𝐽 𝑚𝑜𝑙−1
𝐾−1
× 298 𝐾)/(96486 J 𝑉−1
) = 0.0257 V
In the redox couple, Fe is metallic therefore its activity is unity.
Therefore ,
-0.472 V = -0.44 V +
0.0257 V
2
(ln(a(𝐹𝑒2+
))
ln(a(Fe2+) =
2 −0.44 𝑉+0.44 𝑉
0.0257 𝑉
Rearranging, we get
ln(a(𝐹𝑒2+
) = -2.490
a(𝐹𝑒2+
)= exp(-2.490) = 0.0829.

10. (i) Notice all three terms, E, Eo and RT/F have the same unit V.
(ii) When rearranging to make ln(aF2+) the subject of the equation, V units cancel out to leave the
logarithm value unitless.
(iii) Therefore the activity, a, is also unitless.
• Example 2:
i. Calculate the concentration of the iron at 298 K within the cell Fe | Fe2+ || SCE. The emf at
equilibrium was measured as 0.735 V.
ii. If the activity of Fe2+ is 10-5, calculate the new value of EFe2+, Fe and hence the new emf.

11. Types of redox electrodes
• The electrodes are grouped according to the type or the way measurements are made.
• Electrodes that under go redox reaction are called redox electrodes.
• There are commonly used electrodes such as copper, cobalt, silver, zinc nickel and other transition
metals.
• Other p-block metals such as tin, indium and lead can function as redox electrodes.
• The s-block metals do not make good redox electrodes because elemental metal is highly reactive –
forms the oxide layer that coat the metal
➢ Oxide layer leads to (i) poor reproducibility, (ii) poor electronic conductivity, therefore electrode
potentials of these metals are difficult to interpret.

12. Redox electrodes
• Redox electrodes are useful since, the metal M is a pure solid and have a lower valence ionic
analyte Mn+ in solution.
• Therefore, the activity of the metal M is always unity. The denominator of the logarithm term in
Nernst equation therefore becomes 1; the equation is reduced to
• EM(n+),M = Eo
M(n+),M +
𝑅𝑇
𝑛𝐹
ln(𝑎(𝑀𝑛+
)
• The equation allows a calibration curve to be drawn.
• The intercept of the curve would be standard electrode potential of the species under redox.
• If the Nernst equation is written in terms of log10 instead of natural logarithm, then the factor of
2.303 is introduced as follows

13. Redox electrodes
• EM(n+),M = Eo
M(n+),M + 2.303
𝑅𝑇
𝑛𝐹
𝑙𝑜𝑔10(𝑎(𝑀𝑛+)
• This equation will then give us concentration per decade.
At the temperature of 298 K
The term
2.303𝑅𝑇
𝑛𝐹
=
2.303 × 8.316
𝐽
𝑚𝑜𝑙.𝐾
×298𝐾
96486
𝐽
𝑚𝑜𝑙.𝑉
= (0.0591/n) = (59.1/n) mV
After performing unit cancellation.
• Therefore according to one electron couple M+,M, will change its electrode potential, EM+,M , by
59.1 mV per ten fold change in concentration.

14. Redox Electrodes
• Note that
59.1
𝑛
𝑚𝑉 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑁𝑒𝑟𝑛𝑠𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤ℎ𝑒𝑛 𝑙𝑜𝑔10 is used to the equation.
• The logarithm base 10 gives us number of decades of concentration.
• Then the Nernst equation reduces to
• EMn+,M = Eo
Mn+, M – [number of decades of dilution × 59.1
𝑚𝑉
𝑛
]
• The resultant equation is an approximate of Nernst equation.
• Example: A new way of extracting Nickle from its ore ore is been investigated. The first step is to crash
the rock powder, roast it, and then extract soluble Ni2+ species in aqueous solution. The activity of nickel
a(Ni2+) is monitored by potentiometric method, where a wire of pure nickel metal functions as an
electrode and is immersed in aliquot samples taken from the plant. The wire monitors the electrode
potential ENi2+,N . If Eo
Ni2+,Ni = -0.230 V, what is the ENi2+,N if a(Ni2+) = 10-6.

15. Limitations of Nernst equation