Solution of Nernst equation and its applications Test 02 - Electrochemistry (Class 12, CBSE) Class 12 Notes | EduRev

Class 12 : Solution of Nernst equation and its applications Test 02 - Electrochemistry (Class 12, CBSE) Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-02 
CLASS - XII CHEMISTRY (Electrochemistry) 
[ANSWERS] 
Topic: -Nernst equation and its applications 
1. E
0
cell  =E
0
cathode – E
0
Anode
=E
0
Ag +/Ag – E
0
Cu
2+
/Cu 
=0.80V – (+0.34V) 
 =+0.46V 
2. From the reduction potential values, it is indicated that Nickel (more negative value)
is more reactive than copper and will, then displace copper from CuSO 4
 Ni (s) + Cu
2+
(aq) ----Ni
2
+ (aq) +Cu(s). 
3. Factors affecting electrode potential values are –
a) Concentration of electrolyte
b) Temperature.
4. Daniel cell:
Zn(s) / Zn
2
 +(aq) //Cu
2
+(aq)/Cu(s) 
Nernst equation – at 298 K 
E cell = (E
0
Cu
2
+/Cu – E
0
Zn
2
+/Zn) - 
2
2
0.059 [ ]
 log
2 [ ]
Zn
cu
+
+
5. (i) Standard electrode potential and equilibrium constant
E
0
cell = 
c
2.303
 log k
RT
nF
Where E
0
cell = standard electrode potential of cell 
R = Gas constant 
T = temperature in Kelvin 
n = no .of electrons. 
F = Faraday’s constant and 
Kc = Equilibrium constant 
(ii) Standard electrode potential and Gibbs free energy change- 
? G
0
  = - n F E
0
 cell
Where   ? G
0
 = Change in Gibbs’ free energy
n          =  No. of electrons 
Page 2


CBSE TEST PAPER-02 
CLASS - XII CHEMISTRY (Electrochemistry) 
[ANSWERS] 
Topic: -Nernst equation and its applications 
1. E
0
cell  =E
0
cathode – E
0
Anode
=E
0
Ag +/Ag – E
0
Cu
2+
/Cu 
=0.80V – (+0.34V) 
 =+0.46V 
2. From the reduction potential values, it is indicated that Nickel (more negative value)
is more reactive than copper and will, then displace copper from CuSO 4
 Ni (s) + Cu
2+
(aq) ----Ni
2
+ (aq) +Cu(s). 
3. Factors affecting electrode potential values are –
a) Concentration of electrolyte
b) Temperature.
4. Daniel cell:
Zn(s) / Zn
2
 +(aq) //Cu
2
+(aq)/Cu(s) 
Nernst equation – at 298 K 
E cell = (E
0
Cu
2
+/Cu – E
0
Zn
2
+/Zn) - 
2
2
0.059 [ ]
 log
2 [ ]
Zn
cu
+
+
5. (i) Standard electrode potential and equilibrium constant
E
0
cell = 
c
2.303
 log k
RT
nF
Where E
0
cell = standard electrode potential of cell 
R = Gas constant 
T = temperature in Kelvin 
n = no .of electrons. 
F = Faraday’s constant and 
Kc = Equilibrium constant 
(ii) Standard electrode potential and Gibbs free energy change- 
? G
0
  = - n F E
0
 cell
Where   ? G
0
 = Change in Gibbs’ free energy
n          =  No. of electrons 
F           =  Faraday’s Constant 
E
0
 cell  =  Standard electrode Potential of cell. 
6. Fe
3+
 + 3e
- 
    Fe
According to Nernst Equation –
E Fe 
3+
/ Fe  =  E
0 
Fe
3+
/Fe –
0.059
n
 log 
3
1
[ ] Fe
+
=   0.771 V - 
0.059 1
 log 
3 0.1
=   0.771 V - 0.0197 V 
=   +0.7513V 
7. P
H  
= - Log  [H
+
]
The cell reaction is –
H
+
  + e-      
1
2
H 2 (g)
According to Nearest Equation 
E  =  E
0
 - 
0.059 1
 log 
[ ] n H
+
0.03V  =  0 + 
0.059
1
(- 
1
log 
[ ] H
+
)
= 0  +  0.059 P
H
PH = 
0.03
0.059
V
  =  5.07 V 
8. The cell reaction is
2 Cr   +  3 Fe
2+
  6e
-
 ? 2 Cr
3+
  + 3Fe
Nernst Equation – 
E cell = ( E
0
 Fe
2
+/ Fe- E
0
cr
3+
 /cr ) - 
2
3
3
2
0.059
 log 
6
Cr
Fe
+
+
? ?
? ?
? ?
? ?
=(-0.44v – (-0.74v) - 
( )
( )
2
3
0.10
0.059
 log 
6
0.01
= 0.3V - 
4
0.059
 log 10
6
= 0.3V – 0.0394V 
= +0.2606 V 
Page 3


CBSE TEST PAPER-02 
CLASS - XII CHEMISTRY (Electrochemistry) 
[ANSWERS] 
Topic: -Nernst equation and its applications 
1. E
0
cell  =E
0
cathode – E
0
Anode
=E
0
Ag +/Ag – E
0
Cu
2+
/Cu 
=0.80V – (+0.34V) 
 =+0.46V 
2. From the reduction potential values, it is indicated that Nickel (more negative value)
is more reactive than copper and will, then displace copper from CuSO 4
 Ni (s) + Cu
2+
(aq) ----Ni
2
+ (aq) +Cu(s). 
3. Factors affecting electrode potential values are –
a) Concentration of electrolyte
b) Temperature.
4. Daniel cell:
Zn(s) / Zn
2
 +(aq) //Cu
2
+(aq)/Cu(s) 
Nernst equation – at 298 K 
E cell = (E
0
Cu
2
+/Cu – E
0
Zn
2
+/Zn) - 
2
2
0.059 [ ]
 log
2 [ ]
Zn
cu
+
+
5. (i) Standard electrode potential and equilibrium constant
E
0
cell = 
c
2.303
 log k
RT
nF
Where E
0
cell = standard electrode potential of cell 
R = Gas constant 
T = temperature in Kelvin 
n = no .of electrons. 
F = Faraday’s constant and 
Kc = Equilibrium constant 
(ii) Standard electrode potential and Gibbs free energy change- 
? G
0
  = - n F E
0
 cell
Where   ? G
0
 = Change in Gibbs’ free energy
n          =  No. of electrons 
F           =  Faraday’s Constant 
E
0
 cell  =  Standard electrode Potential of cell. 
6. Fe
3+
 + 3e
- 
    Fe
According to Nernst Equation –
E Fe 
3+
/ Fe  =  E
0 
Fe
3+
/Fe –
0.059
n
 log 
3
1
[ ] Fe
+
=   0.771 V - 
0.059 1
 log 
3 0.1
=   0.771 V - 0.0197 V 
=   +0.7513V 
7. P
H  
= - Log  [H
+
]
The cell reaction is –
H
+
  + e-      
1
2
H 2 (g)
According to Nearest Equation 
E  =  E
0
 - 
0.059 1
 log 
[ ] n H
+
0.03V  =  0 + 
0.059
1
(- 
1
log 
[ ] H
+
)
= 0  +  0.059 P
H
PH = 
0.03
0.059
V
  =  5.07 V 
8. The cell reaction is
2 Cr   +  3 Fe
2+
  6e
-
 ? 2 Cr
3+
  + 3Fe
Nernst Equation – 
E cell = ( E
0
 Fe
2
+/ Fe- E
0
cr
3+
 /cr ) - 
2
3
3
2
0.059
 log 
6
Cr
Fe
+
+
? ?
? ?
? ?
? ?
=(-0.44v – (-0.74v) - 
( )
( )
2
3
0.10
0.059
 log 
6
0.01
= 0.3V - 
4
0.059
 log 10
6
= 0.3V – 0.0394V 
= +0.2606 V 
9. The half cell reactions are
Anode : Zn(s) ? Zn
2
 + (aq) +2e
-
Cathode : Cd
2
+ (aq) + 2e
-
 ? Cd (s)
Nernst Equation
E cell = (E
0
Cathode  -  E
0
anode) - 
2
2
0.059
 log 
Zn
n Cd
+
+
? ?
? ?
? ?
? ?
= (- 0.403 – (-0.763) - 
0.059 0.0004
 log 
2 0.2
 = 0.36V  - 0.0798V = 0.4398V 
0 0
cell
 =  - n F E G ?
= -2mol × 96500 C/mol × 0.4398V
=  -8488  J mol
-1
 
10. From the reaction, n =2
E
0
cell = E
0
cu
2
 +/cu  - E
0
Zn
2
+/Zn 
 = + 0.34v  - (-0.76v) = 1.10V 
E
0
cell = 
c
2.303
 log k
RT
nF
At 298k , E
0
cell ×
c
 log k
0.059
n
Log k c = E
0
cell ×
0.059
n
= 1.10 ×
2
 = 37.29
0.059
K c  = Antilog 37.29 
 = 1.95 ×10
37
11. [Ag
+
] = 5.3 ×10
-9 
M
12. E
0
cell = 0.03V
0
 = -2895J G ?
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