ELECTROCHEMISTRY Important Formula For 1 year B.E students and also using for the first year PG student using this Formulas we can able to Find the Emf of the solution Like galvanized corrosion as well in physics we dont know to represent the chemical Cell

1. Electrochemistry

2. 2 v 1.2
Learning Objective
The students will be able to solve the numerical
problems related to electrochemistry

3. 3 v 1.2
• R =  x l/a
• l/a ( cell constant) = R (resistance) x 1/  (specific conductance)
• m = (1000 )/ c, where m = molar conductance,
= specific conductance, c = concentration
• equ = (1000 )/N, where equ = equivalent conductance,
N = normality
• Ecell = E0
cell -0.0591/n log [P]/[R]
•  G = -nFE
• Log Kequ = nE0/0.0591
Important Formulas

4. 4 v 1.2
• Ecell = 0.0591/n log C2/C1 (C2> C1), for concentration cell
• pH = Ecell / 0.0592, if glass (indicator)
• electrode and reference electrode, both are hydrogen
electrode
• Ecell = Eo
cal + 0.0591 pH
pH = (Ecell - 0.2422) / 0.0591
• pH = (Ecell + EG
0 –Ecal) /0.0591
Important Formulas

5. 5 v 1.2
• Calculate the valency of mercurous ions with the help of the following cell:
Hg | 0.001N mercurous nitrate in 0.1 N HNO3| 0.01N mercurous nitrate in 0.1 N HNO3| Hg
when the emf observed at 18°C is 0.0209 V.
• Ecell = 0.0591/n log C2/C1
0.029 = 0.0591/n log 0.01/0.001 = 0.0591/n x 1
n=0.0591/0.029 = 2
Hence valency of mercurous ion is 2 and its formula Hg2
2+
Solved problems

6. 6 v 1.2
Solved problems
• The potential of a hydrogen gas electrode in a solution of an acid of
unknown strength is measured as 0.1 V at 298 K against normal
hydrogen electrode. Calculate the pH of the solution.
• Hence Ecell = 0.1 V
• Ecell = Eo
SHE - [ Eo
H+/H2(unkown) + 0.0592 log [H+] ]
• 0.1 = 0 - [ 0+ 0.0592 log [H+] ]
• 0.1 = - 0.0592 log [H+]
• 0.1 = - ( - 0.0592 pH )
• pH = Ecell / 0.0592 = 0.1/ 0.0592
• pH = 1.69

7. 7 v 1.2
Solved problems
• A 0.01 N KCl solution shows a resistance 225 ohms in a conductivity
cell. The specific conductance of 0.01 N KCl solution at the temperature
of the experiment is 0.00141 mho/cm. If 0.02 N solution of an acid
shows a resistance of 80 ohms in the same cell, find the specific and
equivalent conductance of the acid.
Cell constant = sp conductance x resistance
= 0.00141 mho/cm x 225 ohm = 0.3173 cm-1
Therefore,
sp conductance of 0.02 N acid solution = cell const/ resistance of acid soln,
= 0.3173 cm-1/ 80 ohm
= 3.966 x 10-3 mho cm-1
equ of acid soln = 1000 x 3.966 x 10-3/ 0.02

8. 8 v 1.2
Solved Problems
• Write the cell construction, reaction and calculate the emf of the cell
when Ni and Al are coupled using the data Eo
Ni2+/Ni= -0.23V and
[Ni2+] = 0.02M; Eo
Al3+/Al = -1.66V and [Al3+] = 0.01M
Al|Al3+
(0.01M) || Ni2+
(0.02M)|Ni,
At the anode: 2Al 2Al3+ + 6e- ; At the cathode: 3Ni2+ + 6 e- 3 Ni
Net cell reaction: 2Al + 3Ni2+ 2Al3+ + 3 Ni
E cell = E0
Cell + (0.0592/n) log [Reactant]/[Product]
E0
Cell = E0c- E0a = Eo
Ni2+/Ni - Eo
Al3+/Al = -0.23 – (-1.66) = 1.43V
E cell= 1.43 + (0.0591/6) log [Ni2+]3/ [Al3+]2
= 1.43 + 0.009867 log [0.02]3/[0.01]2
=1.4192 V

9. 9 v 1.2
Solved Problems
• Calculate the equilibrium constant of the reaction:
Fe2+ (aq) +Ag + (aq) Fe3+ (aq) +Ag (s) at 25º C, if E0
Fe3+/Fe2+ = 0.77 V
and E0
Ag+/Ag = 0.8 V
E0
cell = 0.8 – (-0.77) V = 1.57 V
log Keq = (n E0
cell)/ 0.0591 = (1 x 1.57)/0.0591 = 26.52
Hence, Keq = antilog 26.52 = 3.31 x 1026

10. 10 v 1.2
Problems to solve
1. An iron rod is placed is a 0.5 M FeSO4 solution. Calculate reduction
potential assuming 90% dissociation of FeSO4. Given E0
Fe2+/Fe= -0.44 V
2. Given standard reduction potentials for Cd and Ag are -0.4 & 0.78 V
respectively. Write the cell representation, net reaction & 0.78 V.
Calculate the EMF of the cell.
3. Calculate the EMF of the following cell :
Zn/Zn2+
(0.1 M) || Cu2+
(1.75 M)/Cu.
Given E0
Zn2+/Zn = -0.76 V and E0
Cu2+/Cu =0.34 V
4. Construct as many cells as feasible by combining the following
electrodes Zn, Cu, Ag in contact with its 1 M solution of Zn2+, Cu+, Cu2+,
Ag 2+. Given E0
Zn2+/Zn = - 0.76 V; E0
Cu2+/Cu = 0.34 V, E0
Cu+/Cu = 0.52 V;
E0
Ag+/Ag = 0.8 V

11. 11 v 1.2
Problems to solve
5. For the cell Zn (s) | Zn2+ (0.01 M) || Cr3+ (0.01 M) | Cr (s). Calculate
i) E0
cell & Ecell at 298 K. ii) Equilibrium constant K for the cell reaction.
Reduction potential of Zn & Cr are -0.76 V and -0.74V respectively.
6. What is the concentration of Ni2+ in the cell at 250C , if the emf is
0.601 V?
Ni (s) | Ni2+ (c= ? ) || Cu2+ (0.75 M) | Cu (s)
Given, E0
Ni2+/Ni = 0.25 V, and E0
Cu2+/Cu = 0.34 V
7. The resistance of 0.098 N KCl was found to be 214.4 ohms at 298 K.
Calculate the equivalent conductance of the solution at the same
temperature if the cell constant is 0.878 cm-1.