The document presents the Nernst equation for calculating the electromotive force (e.m.f) of galvanic cells at non-standard conditions, highlighting its relation to temperature, concentration, and redox reactions. It provides historical context on Walther Hermann Nernst, who developed the equation in 1887 and won the Nobel Prize in 1920. Additionally, the document includes examples and practice problems demonstrating the application of the Nernst equation and calculations of cell potentials and equilibrium constants.

1. To calculate the e.m.f of a Galvanic Cell
(Cell potential) at Non-standard
Conditions: Nernst Equation for a Cell
Reaction at Non-standard Conditions

2. Nernst Equation
Standard Conditions
C = 1 M
T = 250C (298 K)
Gases = 1 atm
E0
(anode)
E0
(cathode)
E0
cell = [E0
(cathode)] – [E0
(anode)]
The values of E0 can be directly obtained from the
electrochemical series

3. Non-standard
Conditions
C = other than 1 M
T = other than 250C
(298 K)
Gases = other than 1
atm
E (anode)
E (cathode)
E0
cell ???
Nernst Equation (contd…)
Nernst Equation
Nernst Equation correlates E to the T and C of the
electrode system

4. Brief History of Nernst
Full Name: Walther Hermann Nernst
Nationality: German
Profession: Chemist
Born: 25 June 1864
Died: 18 November 1941
Contributions: Known for his work in
Thermodynamics, Physical Chemistry,
Solid State Physics and
Electrochemistry. Developed Nernst
Equation in 1887
Recognition: Won 1920 Nobel Prize in
Chemistry for Nernst Heat Theorem that
paved way for “Third Law of
Thermodynamics”

5. • Nernst equation gives relation between
i. EM
n+/M and E0
M
n+/M
ii. Eredox and E0
redox
iii. Ecell and E0
cell

6. Nernst Equation (contd…)
Consider an electrode reaction (reduction):
Mn+(aq) + ne- ----------M(s)
The reduction potential of electrode system at
non-standard condition;
Where,
E = Reduction Potential (Non-standard conditions
E0 = Standard Reduction Potential
R = Gas Constant ( 8.314 JK-1mol-1
T = Temperature in K
F = One faraday = 96500 coulombs
n = No. of electrons exchanged

7. Nernst Equation (contd…)
Since electrode systems are set up at room temperature,
T can be taken as 298 K, reduction potential is given by
The Oxidation Potential of the electrode systems are
given by

8. Nernst Equation (contd…)
Calculate the electrode potential
of a given half cell (Fe2+ (0.1M)/Fe) at 250C.
Fe
Fe2+ (0.1M) According to Nernst Equation
Concentration of solid is always taken a
Fe2+ (0.1M) + 2e- --------- Fe(s)

9. Nernst Equation (contd…)
Given;
E0
Fe
2+
/Fe = -0.44 V (Electrochemical series)
n = 2 and, [Fe2+] = 0.1 M
= -0.44 – (0.0295) = -0.4695 V
Activity: Calculate the electrode potential of
Mg2+/Mg at 250C when the concentration of Mg2+
ions is 0.1M and [Mg 2+]= -2.38
Answer: -2.39V

10. Nernst Equation (contd…)
Nernst Equation for general Redox Reaction at 298 K;
aA + bB ------- cC + dD
e.m.f of a cell (Ecell) is given by
Calculate the e.m.f. the following cell at 298 K given
that, E0
Cr
3+
/Cr = -0.75V & E0
Fe
2+
/Fe = -0.44V
Cr(s) / Cr3+(0.1M) // Fe2+ (0.01M) / Fe(s)

11. Nernst Equation (contd…)
Given;
E0
Cr
3+
/Cr = -0.75V (RP) & E0
Fe
2+
/Fe = -0.44V (RP)
[Cr3+] = 0.1M & [Fe2+] = 0.01M
Cell Rep: Cr(s) / Cr3+(0.1M) // Fe2+ (0.01M) / Fe(s)
Cell Reactions
[Cr(s) ----------- Cr3+(0.1M) + 3e- ]
[Fe2+ (0.01M) + 2e- -------- Fe(s) ]
Anode (OHCR)
Cathode (RHCR)
X 2
X 3
2Cr(s) + 3Fe2+(0.01M) -------- 2Cr3+(0.1M) + 3Fe(s) NCR)
n = ??? n = 6
E0
cell = [E0
(cathode)] – [E0
(anode)] = [-0.44] – [-0.75] = 0.31 V

12. Nernst Equation (contd…)

13. Activity
1) Calculate the cell potential at 298 K for the cell;
Zn(s) / Zn2+ (0.1M) // Sn2+ (0.001M) / Sn (s)
2) Calculate the potential of the following cell at
298 K:
Sn4+(1.50M) + Zn(s) --- Sn2+(0.50 M) + Zn2+ (2.0M)
Answer:
1) 0.561V
2) 0.895
Nernst Equation (contd…)

14. Equilibrium Constant (Kc) of Redox Reaction
• Zn(s) + Cu2+(aq) ↔ Zn2+(aq) + Cu(s)
At “Equilibrium”, Ecell = 0
--------Eq (I)
----------------------Eq (II)

15. • Calculate the Kc for the cell reaction;
• Zn(s) + Cu2+(aq) ↔ Zn2+(aq) + Cu(s);
• E0
Zn
2+
/Zn = -0.76V & E0
Cu
2+
/Cu = +0.34 V
• Half cell Reactions
• n = 2
• Kc = antilog 37.29
• = 1.95 x 1037
Equilibrium Constant (Kc) of Redox Reaction
E0cell = +0.34V - -0.76V = +1.10V

16. Practice Problems
a) Calculate the Kc for the following reaction;
3Sn4+ + 2Cr -------- 3Sn2+ + 2Cr3+
(E0
Sn4+/Sn2+ = +0.15 V & E0
Cr3+/Cr = -0.71V)
Answer:
E0
Cell = 0.15+0.71 = 0.86V, n = 6
Kc = antilog [6 x 0.86]/[0.059] = [5.16]/[0.059] =
1 x 1090
Equilibrium Constant (Kc) of Redox Reaction

17. Relation between “Standard Free Energy
Change” (∆G0) and “Standard Cell Potential”
(E0
cell)
• Free Energy Change (∆G) measures the amount
of useful work that can be obtained.
• Electrical work is obtained by operating
electrochemical cell
• ∆G = Welectrical = -nFEcell
• ∆G0 = -nFE0
cell
• For spontaneous cell reaction ∆G must be
negative & Ecell must be positive

18. • For the cell; Mg(s)/Mg+2(aq)//Ag+(aq)/Ag(s)
calculate Kc at 250C and the Maximum Work that
can be obtained by the operating cell.
• (E0
Mg
2+/Mg = -2.38 V & E0
Ag+/Ag = +0.80V)
• Write cell reaction & find “n”
• Calculate E0
cell
• E0
cell = 3.18 V
• W(max) = ∆G0 = -nFE0
cell
• -(2 x 96500 x 3.18) = -611810 J
= antilog 107.8 = 6.26 x 10107

19. • Practice Problems
• For the reduction of silver ions with
copper metal the standard cell potential
was found to be +0.46V at 250C. Calculate
the work done in operating the cell.
• Ans: -89.0 kJ
• The value of standard free energy change in
the Daniel cell is -212.3 kJ at 298K. Calculate
the equilibrium constant for the reaction.
• Ans: 1.6 x 1037

20. Practice Problems
• T.B Q nos: 40 (pp:3.73),
• MCQ Q No: 36, 91