Numerical problems on Electrochemistry

Fore BE First year Students (VTU)

1. Dr. Swastika Das
Professor of Chemistry
BLDEACET, Vijayapur.
Numerical Problems on
Electrochemistry- Module -1.

2. Numerical Problems :

⚫ 1. Calculate the EMF of the following silver concentration cell at
298 K.
Ag/ Ag
+
(0.01 M) // Ag
+
( 0.5 M)/ Ag.
C2 = 0.5 M ; C1 = 0.01 M
E
cell = 0.0591/n log ( C2/ C1) = 0.1004 V.

3. 2. Calculate the EMF of the following copper concentration cell at 298
K.

Cu/ Cu
+2
(0.072M) // Cu
+2
( 0.32 M)/ Cu.

Anode : Cu -------> Cu
+2
(0.072M) + 2e-
Cathode: Cu
+2
( 0.32 M) + 2e- -------> Cu


Net Reaction : Cu
+2
( 0.32 M) -------> Cu
+2
(0.072M)

E
cell = 0.0591/n log ( C2/ C1) = 0.0591/2 log ( 0.32/ 0.072)

E
cell = 0.019 V

4. ⚫ 3.The emf of the following concentration cell at 25
0
C is 0.8
Volts. Calculate C1.
⚫ Ag/AgNO3 (C1 M) // AgNO3(0.2M)/Ag.

⚫ Ecell = 0.0591/n log (C2/C1)
⚫ 0.8 = 0.0591/1 log ( 0.2/C1)
⚫ 0.8/ 0.0591 = log (0.2 / C1)
⚫ 13.54 = log (0.2 / C1)
⚫ Antilog (13.54) = 0.2 / C1
⚫ 3.467 x 10
-13
= 0.2 / C1
⚫ C1 = 0.2/ 3.467 x 10
13
= 5.76 x 10
-15
M.

5. 4. The emf of the following concentration cell at 25
0
C is
0.024 Volts. Calculate x.

⚫ Au/Au
+3
(0.03M) // Au
+3
(xM)/Au

Ecell = 0.0591/n log C2/C1
0.024= 0.0591/3 log (x/0.03)
0.024 x 3 /0.0591 = log (x/0.03)
1.2182 = log (x/0.03)
Antilog ( 1.2182) = x/0.03
16.527= x/0.03
x= 0.495M

6. Nernst equation for a cell:
At 298 K / 25
0
C ,
Ecell = E
0
cell - 0.0591/n{log [species oxidised ]/[species reduced ]}

⚫ E
0
cell = E
0
cathode - E
0
anode
⚫ n= number of electrons transferred

7. Nernst equation: Numerical Problems:
1. An electrochemical cell consists of an Fe electrode dipped in
0.1M FeSO4 and Ag electrode dipped in 0.05M AgNO3
solution. Write cell representation, cell reaction and calculate
EMF of the cell at 298 K given that the standard reduction
potentials of Fe and Ag electrodes are -0.44 V and +0.80 V
respectively.
Representation : Fe/ (0.1M) FeSO4 // (0.05M) AgNO3/ Ag
Cell Reaction : Fe + 2Ag
+
------ Fe
2+
+ 2 Ag
⚫ Ecell = E
0
cell - 0.0591/ n {log [Fe
2+
]/ [Ag
+
]
2
}
⚫ = 1.24 – (0.0295) {log [0.1]/ [0.05]
2
}
⚫ = 1.24 - (0.0295) {log 40} = 1.24-0.047 = 1.193 Volts

8. 2. Calculate emf of following cell at 298 K.
Ni(s)/ Ni
2+
(0.01 M) // Cu
2+
(0.5M)/Cu (s)
Given: Std EP of Ni= -0.25 V and Std EP of Cu= 0.34V.
Ni + Cu
2+
------ Ni
2+
+ Cu
reduced oxidized
Ecell = E
0
cell - 0.0591/ n {log [Ni
2+
]/ [Cu
2+
]}
E
0
cell = 0.34 – (- 0.25) = ).34+ 0.25 = 0.59V
Ecell = 0.59 - 0.0591/ 2 {log [0.01 ]/ [0.5] }
=0.59- (0.0295) log (0.02) = 0.59- (0.0295)x (-1.698)
= 0.59 +0.05 = 0.64 V.
Ecell = 0.64 V

9. ⚫ 3. Calculate emf of following cell at 298 K.
⚫ Fe(s)/ Fe
2+
(0.02 M) // Cd
2+
(1.0M)/Cd (s)

⚫ Given: Std EP of Fe= -0.44 V and Std EP of Cd= -0.40V.

⚫ Ecell = E
0
cell - 0.0591/ n {log [Fe
2+
]/ [Cd
2+
]}
⚫ = (-0.40 + 0.44) - 0.0591/2{log [0.02 ]/ [1.0]}
⚫ = 0.04 – (0.0295) log (0.02)
⚫ Ecell = 0.04+ 0.050 = 0.09 Volts.

10. cell

4. Calculate emf of following cell at 298 K.
Cd(s)/ Cd
2+
(0.25M) // Ag
+
(0.5M)/Ag (s)
Given: Std EP of Cd= -0.40 V and Std EP of Ag= 0.80V.
Cd + 2Ag
+
---- Cd
2+
+ 2Ag
2 moles of Ag+ and 1 mole of Cd involved in reaction.
Ecell = E
0
- 0.0591/ n {log [Cd
2+
]/ [Ag
+
]
2
}
E
0
cell = 0.80 + 0.40 = 1.2 V
Ecell = 1.2 - 0.0591/ 2 {log [0.25]/ [0.5]
2
}

Ecell = 1.2 - 0.0591/ 2 {log [0.25 ]/ [0.25]}
Ecell = 1.2 V

11. ⚫ 5. Consider the electrochemical cell as given:
Zn(s)/ Zn
2+
(0.005M) // Ag
+
(0.1M)/Ag (s).
The standard reduction potentials of Zn and Ag are
-0.76 V and +0.80V respectively. Write the redox electrode
reactions with their respective electrode potentials, net cell
reaction and calculate EMF of the cell at 24
0
C.
Anode : Zn------- Zn
+2
+2e-
Cathode: 2Ag
+
+ 2e- ------ 2Ag



⚫ Net reaction : Zn+ 2Ag
+
------ Zn
+2
+ 2Ag

12. ⚫ Ecell = E
0
cell - 2.303 RT / nF{log [Zn
2+
]/ [Ag
+
]
2
}

= (0.80+0.76) - 2.303 x8.314x297/2x96500{log [0.005 ]/ [0.1]
2
}
= 1.56 – {5686.701/ 193000 [log (0.5)] }
= 1.56 – {0.0294 x (-0.301)}
= 1.56 + 0.008849
Ecell = 1.5688 Volts.

13. 6. Compute the cell potential at 22.3
0
C of the Ag
+
/ Ag
couple with respect to Cu
+2
/Cu if the concentration
of the Cu ion and Ag ion


are 4.2x10
-6
M and 1.3x10
-3
M respectively. E
0
of Ag and Cu are 0.8V and 0.34V
respectively.

Ecell = E
0
cell - 2.303 RT / nF{log [Cu
2+
]/ [Ag
+
]
2
}
= 0.46 - 2.303 x8.314x295.3/2x96500{log [4.2x10
-6
]/
[1.3x10
-3
]
2
}
= 0.46 – {(5654.151/193000) log (3.23)}
= 0.46 – 0.0149 = 0.4451 V.

14. Cu /Cu Cu /Cu
Cu /Cu

7. Calculate the standard electrode potential E
0
Cu+2/Cu
if its electrode potential E, at 25
0
C is 0.296V when the
concentration of Cu
+2
ions is 0.015M.
E
+2
= E
0 +2
+ 0.0591/n log [Cu
+2
]
0.296 = E
0 +2
+ 0.0591/2 log [0.015]
E
0
Cu
+2
/Cu = 0.296 - 0.0591/2 log [0.015]
= 0.296 + 0.0540
E
0
Cu
+2
/Cu =0.35 V.

15. 9. Calculate the reduction electrode potential of copper when it is in
contact with 0.5 M copper sulphate solution at 298 K. The E
0
value of
copper is 0.34V.

E
Cu
+2
/Cu = E
0
Cu
+2
/Cu + 0.0591/n log [Cu
+2
]
= 0.34 + 0.0591/2 log [0.5]
=0.34 + (0.0295) ( -0.30)= 0.34 - 0.0088

E
Cu
+2
/Cu = 0.3312 V.

16. 10. The standard reduction potentials of copper and zinc are
0.34V and -0.76 V respectively. If the cell develop an EMF of
1.10 V, find the free energy change of overall cell reaction.
Zn + Cu
+2
------ Zn
+2
+ Cu.
n=2 mol electrons, E
0
= 1.10 V and
F= 96500 coulomb/mol
ΔG = -nFE = -2 x 96500 x 1.1
= -212.3 kJ/mol.
The cell reaction is spontaneous because of decrease in free energy.

17. Home Tasks:
1. A cell is constructed by coupling a zinc electrode dipped in
0.5 M ZnSO4 and a nickel elecorde dipped in 0.05 M
NiSO4. Write the cell representation and cell reaction.
Calculate the EMF of the cell , given the standard electrode
potential of Zn and Ni are -0.76 V and -0.25 V respectively.
[ Ans: 0.48 V]
2. A cell is formed by coupling a Nickel electrode dipped in
0.01 M Ni
2+
solution and a lead electorde dipped in 0.5 M
Pb
2+
solution. Write the cell representation and cell
reaction. Calculate the EMF of the cell , given the standard
electrode potential of Ni and Pb are -0.25 V and -0.13V
respectively. [ Ans: 0.17V]