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# 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)
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.
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)
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.
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
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)
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.
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
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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