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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