NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Chemistry
Chapter : 3
Chapter Name : Electrochemistry

Q3.1 How would you determine the standard electrode potential of the system Mg 2+
|Mg?

Answer. The standard electrode potential Of Mg can be measured with respect to the standard hydrogen electrode, represented by
2+
|Mg

. A cell, consisting of Mg I MgSO ( aq 1M) as the anode and the standard hydrogen
∣ +
P t(s) , H2(g) (1atm) H (1M)
∣ (aq) 4

electrode as the cathode, is set up.

Then, the emf of the cell is measured and this measured emf is the standard electrode
potential of the magnesium electrode.
⊖ ⊖ ⊖
E = E − E
R l

Here, E for the standard hydrogen electrode is zero.
∘
R
θ ∘
∴ E = 0 − E
L

∘
= −E
l

Page : 70 , Block Name : Intext Questions

Q3.2 Can you store copper sulphate solutions in a zinc pot?

Answer. Zinc is more reactive than copper. Therefore, zinc can displace copper from its salt solution. If copper sulphate solution is stored in a zinc
pot, then zinc Will displace copper from the copper sulphate solution.
Zn + CuSO4 ⟶ ZnSO4 + Cu

Hence, copper sulphate solution cannot be stored in a zinc pot.

Page : 70 , Block Name : Intext Questions

Q3.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Answer. Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.
2∗ 3+ −1 2
Fe ⟶ Fe + e ;E = −0.77V

This implies that the substances having higher reduction potentials than
+0.77 V can oxidise ferozus ions to ferric ions. Three substances that can do so are F , 2

Cl , and O .
2 2

Page : 70 , Block Name : Intext Questions

Q3.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Page : 75 , Block Name : Intext Questions

Q3.5 Calculate the emf of the cell in which the following reaction takes place:

+ 2+
Ni(s) + 2Ag (0.002M) → Ni (0.160M) + 2Ag(s)

Given that
⊖
E = 1.05V
(cell )

Answer. ApplyingN ernstequationwehave :
2+
[Ni ]
∘ 0.0591
E( cell ) = E − log 2
(cell ) n +
[Ag ]

0.0591 (0.160)
= 1.05 − log
2 (0.002)2

0.16
= 1.05 − 0.02955 log
0.000004
4
=1.05 − 0.02955 log 4 × 10

=1.05 − 0.02955(log 10000 + log 4)

=1.05 − 0.02955(4 + 0.6021)

=0.914V

Page : 75 , Block Name : Intext Questions

Q3.6 The cell in which the following reaction occurs:
2Fe
3t
(aq) + 2I
−
(aq) → 2Fe
2∗
(aq) + I2 (s) has E
0

cell
= 0.236V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the
cell reaction.

Answer. Here, n =2 E ∘
cell
= 0.236V, T = 298K

We know that
∘ ∘
Δ, G = −nFEcell

= −2 × 96487 × 0.236

−1
= −45541.8643mol

−1
= −45.54k}mol
∘
Δ, G = −2.303RT log Kc
∘
Δ,G
⇒ log Ke = −
2.303RT
3
−45.54 × 10
= − =7.981
2.303 × 8.314 × 298

∴ Kc = Antilog (7.981)

7
= 9.57 × 10

Page : 75 , Block Name : Intext Questions

Q3.7 Why does the conductivity of a solution decrease with dilution?

Answer. The conductivity Of a solution iS the conductance Of ions present in a unit volume Of the solution. The number Of ions (responsible for
carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

Page : 85 , Block Name : Intext Questions

Q3.8 Suggest a way to determine the Λ° m value of water.

Answer. Applying Kohlrausch's law of independent migration of ions, the λ 0
m be determined as follows:

Page : 85 , Block Name : Intext Questions

Q3.9 The molar conductivity of 0.025 mol L −1
methanoic acid is 46.1 S cm mol
2 −1
. Calculate its degree of dissociation and dissociation constant.
Given λ (H )0 +

2 −1 0 2 −1
= 349.6Scm mol and λ (HCOO) = 54.6Scm mol

−1
C = 0.025molL

2 −1
Λm = 46.1Scm mol

0 + 2 −1
λ (H ) = 349.6Scm mol

0 − 2 −1
λ (HCOO ) = 54.6Scm mol
0 ∘ + 0 −
Λm (HCOOH) = λ (H ) + λ (HCOO )

= 349.6 + 54.6

2 −1
= 404.2Scm mol

Now, degree Of dissociation:
Λ= (HCOOH)
α =
∘
Λm (HCOOH)

46.1
=
404.2

= 0.114( approximately )

Thus, dissociation constant:
2
c ∝
K =
(1 − α)

−1 2
(0.025molL ) (0.114)

=
(1 − 0.114)

−4 −1
= 3.67 × 10 molL

Page : 85 , Block Name : Intext Questions

Q3.10 If a current of 0.5 ampere ows through a metallic wire for 2 hours, then how many electrons would ow through the wire?

I = 0.5A

t = 2 hours = 2 × 60 × 60s = 7200s

Thus, Q = I t

= 0.5A × 7200s

= 3600C

23
23
6.023 × 10 × 3600 22
We know that 96487C = 6.023 × 10 , number of electrons. Then, 3600C = number of electrons =2.25 × 10 num
96487

Hence, 2.25 × 10 number of electrons will ow through the wire.
22

Page : 88 , Block Name : Intext Questions

Q3.11 Suggest a list of metals that are extracted electrolytically.

Answer. Metals that are on the top of the reactivity senes such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted
electrolyticallv.

Page : 88 , Block Name : Intext Questions

Q3.12 Consider the reaction: Cr O What is the quantity of electricity in coulombs needed to reduce 1 mol of
2− + − 3+
2 + 14H + 6e → 2Cr + 7H2 O
7

?
2−
Cr2 O
7

The given reaction is as follows:

2 + − 3+
Cr2 O + 14H + 6e → 2Cr + 7H2 O
2−
Therefore, to reduce 1 mole of Cr2 O , the required quantity of electricity will be:
7

= 6F

= 6 × 96487C

= 578922C

Page : 88 , Block Name : Intext Questions

Q3.13 Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

A lead storage battery consists of a lead anode, a grid of lead packed with

lead oxide (PbO ) as the cathode, and a 38% solution of sulphuric acid
2

(H2 SO4 ) as an electrolyte.

When the battery is in use, the following cell reactions take place:

2− −
At anode: P b(s) + SO 4(aq) → P bSO4(s) + 2e

At cathode:PbO
2− + −
2(s)
+ SO + 4H + 2e ⟶ PbSO4(s) + 2H2 O(i)
4(aq) (aq)

The overall cell reaction is given by,
Pb(s) + PbO2(s) + 2H2 SO4(aq) ⟶ 2PbSO4(s) + 2H2 O(l)

When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, present at the anode and cathode is converted into and respectively.

Page : 92 , Block Name : Intext Questions

Q3.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Answer. Methane and methanol can be used as fuels in fuel cells.

Page : 92 , Block Name : Intext Questions

Q3.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Answer. In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That
spot behaves as the anode. The reaction at the anode is given by,
2+ −
Fe(,) ⟶ Fe + 2e
(aq)

Electrons released at the anodic spot move through the metallic object and go to another
spot of the object. There, in the presence of H ions, the electrons reduce oxygen. This spot behaves as the cathode. These H ions come either
+ +

from H CO which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the
2 3

atmosphere in water.
The reaction corresponding at the cathode is given by,
+ −
O2(g) + 4H + 4e ⟶ 2H2 O(j)
(aq)

The overall reaction is:
+ 2+
2Fe(s) + O2(g) + 4H ⟶ 2Fe + 2H2 O(i)
(aq) (α)

Also, ferrous ions are further oxidized bv atmospheric oxygen to ferric ions. These ferric ions combine With moisture, present in the surroundings, to
forrn hydrated ferric oxide
(Fe2 O3 , xH2 O)
i.e., rust.

Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.

Page : 92 , Block Name : Intext Questions

Q3.1 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

Answer. The following is the order in which the given metals displace each other from the solution of theil salts.
Mg, Al, Zn, Fe, Cu

Page : 93 , Block Name : Exercise

Q3.2 Given the standard electrode potentials,
+ +
K /K = −2.93V, Ag /Ag = 0.80V

2+
Hg /Hg = 0.79V
2+ 3+
Mg /Mg = −2.37V. Cr /Cr = −0.74V

Arrange these metals in their increasing order of reducing power.

Answer. The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of
+ 2+ 3+ 2+ +
K /K < Mg /Mg < Cr /Cr < Hg /Hg <Ag /Ag

Hence, the reducing power of the given metals increases in the following order:
Ag < Hg < Cr < Mg < K

Page : 93 , Block Name : Exercise

Q3.3 Depict the galvanic cell in which the reaction
(aq) + 2Ag(s)takes place. Further show:
+ 2+
Zn(s) + 2Ag (aq) → Zn

(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Answer. The galvanic cell in which the given reaction takes place is depicted as:
2+ +
∣ ∥ ∣Ag
Zn(s) Zn Ag
∣ (aq) ∥ (aq) ∣ (s)

(i) Zn electrode (anode) is negatively charged.
(ii) Ions are carriers Of current in the cell and in the external circuit, current Will ow from silver to Zinc.
(iii) The reaction taking place at the anode is given by,
2+ −
Zn(s) ⟶ Zn + 2c
(aq)

The reaction taking place at the cathode is given by,
+ −
Ag + e ⟶ Ag(s)
(aq)

Page : 93 , Block Name : Exercise

Q3.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) 2Cr(s) + 3Cd (aq) → 2Cr (aq) + 3Cd
2+ 3+

(ii) Fe 2+ ∗ 3+
(aq) + Ag (aq) → Fe (aq) + Ag(s)

Calculate the Δ G and equilibrium constant of the reactions.
r
θ

Page : 93 , Block Name : Exercise

Q3.5 Write the Nernst equation and emf of the following cells at 298 K:
(i)Mg(s) ∣∣Mg (0.001M) ∥
2+ 2+
∥Cu (0.0001M)|Cu(s)

(ii) Fe(s) ∣∣Fe
2+ +
(0.001M)llH (1M)∣
∣ H2 (g)(1bar)|Pt(s)

(iii) Sn(s) ∣∣sn 2+
(0.050M) ∥
∥H
+
(0.020M) |H2 (g)(1 bar )| Pt(s)

(iv) Pt(s) |Br (l)| Br (0.010M) ∥
− +
2 ∥H (0.030M) |H (g)(1bar)| Pt(s) 2

Answer. (i) For the given reaction, the Nernst equation can be given as:
2+
[Mg ]
∘ 0.0591
Ecell = E − log
cell n 2+
[Cu ]

0.0591 .001
= {0.34 − (−2.36)} − log
2 0001

0.0591
= 2.7 − log 10
2

=2.7 − 0.02955

=2.67V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:
2+
[Fe ]
∘ 0.0591
Ecell = E − log
cell n + 2
[H ]

0.0591 0.001
= {0 − (−0.44)} − log
2
2 1

= 0.44 − 0.02955(−3)

= 0.52865V

0.53V (approximately)

(iii) For the given reaction, the Nernst equation can be given as:
2+
[Sn ]
∘ 0.0591
Ecell = E − log 2
cell n +
[H ]

= 0.14 − 0.0295 × log 125

= 0.14 − 0.062

= 0.078V

= 0.08V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:
∘ 0.0591 1
Ecell = E − log
cell n − 2 + 2
[Br ] [H ]

0.0591 1
= (0 − 1.09) − log
2 2
2 (0.010) (0.030)

1
= −1.09 − 0.02955 × log
0.00000099

1
= −1.09 − 0.02955 × log
−3
9 × 10
7
= −1.09 − 0.02955 × log(1.11 × 10 )

7
= −1.09 − 0.02955 × log(1.11 × 10 )

= −1.09 − 0.208

= −1.298V

Page : 93 , Block Name : Exercise

Q3.6 In the button cells widely used in watches and other devices the following reaction takes place:
2+ −
Zn(s) + Ag O(s) + H2 O(I ) → Zn (aq) + 2Ag(s) + 2OH (aq)
2

Determine Δ, G and E ⊖ ⊖
for the reaction.

∘
∴ E = 1.104V

We know that,

∘ e
ΔrG = −nFE

= −2 × 96487 × 1.04

= −213043.296J

= −213.04kJ

Page : 93 , Block Name : Exercise

Q3.7 De ne conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration

Answer. Conductivity of a solution is de ned as the conductance of a solution of 1 cm in length and area Of cross-section 1 sq. cm. The inverse Of
resistivity is called conductivity or speci c conductance. tt is represented by the symbolK. If p is resistivity, then we can write:

1
κ =
ρ

The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum
electrodes with the unit area of
cross-section and at a distance of unit length.
I.e., G = κ = κ ⋅ 1 = K
a

l

Conductivity always decreases With a decrease in concentration, both for weak and
strong electrolytes. This is because the number of ions per unit volume that carry the
current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing t mole Of the electrolyte kept
between two electrodes With the
area Of cross-section A and distance Of unit length.
A
Λm = κ
l

NOW, / and A V (volume containing I mole Of the electrolyte),
∴ Λm = κV

Molar conductivity increases with a decrease in concentration. This is because the total
volume V of the solution containing one mole of the electrolyte increases on dilution.
The variation of Λ with √c for strong and weak electrolytes is shown in the following plot:
m

Page : 93 , Block Name : Exercise

Q3.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm . Calculate its molar conductivity
−1

Answer. Given
−1
κ = 0.0248Scm

c = 0.20M

κ×1000
∴ Molar conductivity, Λm =
c

0.0248×1000
=
0.2

2 −1
= 124Scm mol

Page : 93 , Block Name : Exercise

Q3.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl
solution at 298 K is 0.146 × 10 Scm . −3 −1

Given,

−3 −1
Conductivity, κ = 0.146 × 10 Scm

Resistance, R = 1500Ω

∴ Cell constant = κ × R

−3
= 0.146 × 10 × 1500

−1
= 0.219cm

Page : 93 , Block Name : Exercise

Q3.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Concentration/M 0.001 0.010 0.020 0.050 0.100

2 −1
10 × k/Sm 1.23711.8523.1555.53106.74

Calculate Λ 'for all concentrations and draw a plot between
m

Λm and c1/2. Find the value of \Lambda _ { m } ^ { 0 }

−2 −1
k = 1.237 × 10 Sm , c = 0.001M

−4 −1 1/2
Then, κ = 1.237 × 10 Scm , c1/2 = 0.0316M

κ
∴ Λm =
c

−4 −1 3
1.237×10 Scm 1000cm
= −1
×
0.001molL L

2 −1
= 123.7Scm mol

Given,

−2 −1
κ = 11.85 × 10 Sm , c = 0.010M

−4 −1 1/2
Then, κ = 11.85 × 10 Scm , c//2 = 0.1M
K
∴ Λm =
c

−4 −1 3
11.85×10 Scm 1000cm
= −1
×
0.010molL L

2 −1
= 118.5Scm mol

Given,

−2 −1
κ = 55.53 × 10 Sm , c = 0.050M

−4 −1 1/2 1/2
Then, k = 55.53 × 10 Scm ,c = 0.2236M
κ
∴ κ =
c

−4 −1 3
55.53×10 Scm 1000cm
= −1
×
0.050molL L

2 −1
= 111.11Scm mol

Given,

−2 −1
κ = 106.74 × 10 Sm , c = 0.100M

−4 −1 1/2 1/2
Then, κ = 106.74 × 10 Scm ,c = 0.3162M
κ
∴ Λm =
c

−4 −1 3
106.74×10 Scm 1000cm
= −1
×
0.100molL L

2 −1
= 106.74Scm mol

Now, we have following data:

Since the line interrupts Λ m at 124.0Scm mol
2 −1
, Λ
0

m=124.0
Scm mol
2 −1
.

Page : 94 , Block Name : Exercise

Q3.11 Conductivity Of 0.00241 M acetic acid is 7.896 x 10 . Calculate its molar conductivity and if Λ acetic is 390.5 Scm mol what is
−5 −1 0 2 −1
scm ,
m for

its dissociation constant ?

−5 −1
Given, κ = 7.896 × 10 Sm

−1
c = 0.00241molL
κ
Then, molar conductivity, Λm =
c

−5 −1 3
7.896×10 Scm 1000cm
= ×
−1
0.00241molL L

2 −1
= 32.76Scm mol
0 2 −1
Λm= = 390.5Scm mol
2 −1

Now, α =
Λm 32.76Scm mol
= −1
0 2
Am 390.5Scm mol

=0.084
2
cα
∴ Dissociation constant, Ka =
(1−α)

−1 2
(0.00241molL )(0.084)

=
(1−0.084)

−5 −1
= 1.86 × 10 molL

Page : 94 , Block Name : Exercise

Q3.12 How much charge is required for the following reductions:
(i) 1 mol of Al to Al.
3+

(ii) 1 mol of Cu to Cu 2+

(iii) 1 mol of MnO to Mn
− 2+

4

Answer. (i)
3+ −
(i) Al + 3e ⟶ Al

∴ Required charge = 3F

= 3 × 96487C

= 289461C
2+ −
(ii) Cu + 2e ⟶ Cu

∴ Required charge = 2F

= 2 × 96487c

= 192974C
− 2+
(iii) MnO ⟶ Mn
4

7+ − 2+
i.e., Mn + 5c ⟶ Mn

∴ Required charge = 5F

= 5 × 96487C

= 482435C

Page : 94 , Block Name : Exercise

Q3.13 How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl 2

(ii) 40.0 g of Al from molten Al O 2 3

Answer. (i) According to the question,
2∗ −1
Ca + 2e ⟶ Ca40g

Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium = = = 1F
2×20
F
40

(ii) According to the question,
3+ −
Al + 3e ⟶ Al

27 g

Electricity required to produce 27g of Al = 3F
3×40
Therefore, electricity required to produce 40 g of Al = F
27

= 4.44F

Page : 94 , Block Name : Exercise

Q3.14 How much electricity is required in coulomb for the oxidation Of
(i) 1mol of H O to O 2 2

(ii) 1mol of FeO to Fe O 2 3

Answer. (i) According to the question,
1
H2 O ⟶ H2 + O2
2

Now, we can write:

2− 1 −
O ⟶ O2 + 2e
2

Electricity required for the oxidation Of I mol Of H20 to 02
= 2 × 96487c

= 192974C

(ii) According to the question,
2∗ 3+ −1
Fe ⟶ Fe + e

Electricity required for the oxidation of 1 mol of Feo to Fe2 O3 = 1F

= 96487c

Page : 94 , Block Name : Exercise

Q3.15 A solution of is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the
cathode?

Given,

Current = 5A

Time = 20 × 60 = 1200s

∴ Charge = current × time

= 5 × 1200

= 6000C

According to the reaction,
2+ −
Ni + 2e ⟶Ni
(aq)

58.7g

Nickel deposited bv 2 x 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C =
58.71×6000
g
2×96487

= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.

Page : 94 , Block Name : Exercise

Q3.16 Three electrolytic cells A,B,C containing solutions Of znSO , AgNO 4 3
and CuSO4 , respectively are connected in series. A steady current of
1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current ow? What mass of copper
and zinc were deposited?

Answer. According to the reaction
+ −
Ag + e ⟶Ag
(aq) (s)

108g

i.e., 108 g of Ag is deposited by 96487 C.
96487×1.45
Therefore, 1.45 g of Ag is deposited by = C
108

= 1295.43c

Given

Current = 1.5A

1295.43
∴ s
1.5

= 863.6s

= 864s

= 14.40min

Again,
2+ −
Cu + 2e ⟶Cu(s)
(aq)

63.5g

2 × 96487C of charge deposit = 63.5g of Cu

Therefore, 1295.43 C Of charge Will deposit =
63.5×1295.43
g
2×96487

= 0.426 g of Cu
2+ −
Zn + 2e ⟶Zn(s)
(aq)

63.5g

i.e., 2 × 96487C of charge deposit = 65.4g of Zn

65.4×1295.43
Therefore, 1295.43c of charge will deposit = g
2×96487

= 0.439g of Zn

Page : 94 , Block Name : Exercise

Q3.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
3+ −
(i) Fe (aq) and I (aq)

+
(ii) Ag (aq) and Cu(s)

3+ −
(iii) Fe (aq) and Br (aq)

3+
(iv) Ag(s) and Fe (aq)

2+
(v) Br2 (aq) and Fe (aq)

Since E for the overall reaction is positive, the reaction between Br and is
o 2+
2(aq) and Fe (aq)

feasible.

Page : 94 , Block Name : Exercise

Q3.18 Predict the products Of electrolysis in each of the following:
(i) An aqueous solution Of AgNO with silver electrodes.
3

(ii) An aqueous solution of AgNO ,with platinum electrodes.
3

(iii) A dilute solution Of H SO With platinum electrodes.
2 4

(iv) An aqueous solution of CuCl with platinum electrodes.
2

Answer. (i) At cathode:
The following reduction reactions compete to take place at the cathode.

The reaction with a higher value of E takes place at the cathode. Therefore, deposition of saver will take place at the cathode.
∘

At anode:
The Ag anode is attacked by ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag
+

(ii) At cathode:
The following reduction reactions compete to take place at the cathode.

The reaction with a higher value of E' takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode :
Since Pt electrodes are inert, the anode is not attacked bv NO ions. Therefore, OH or NO can be oxidized at the anode. But OH¯ ions having a
− − −

3 3

lower discharge potential
and get preference and decompose to liberate O . 2
− −
OH ⟶ OH + e
−
4OH ⟶ 2H2 O + O2

(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.
+ − 1
H + e ⟶ H2(g)
(aq) 2

At the anode, the following pr«esses are possible.

For dilute sulphuric acid, reaction (i) is preferred to produce 02 gas. But for concentrated
sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.

The reaction with a higher value of E" takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
At anode:
The following oxidation reactions are possible at the anode.

At the anode, the reaction veith a lower value of E −0
is preferred. But due to the over- potential of oxygen, Cl- gets oxidized at the anode to produce
Cl gas.
2

Page : 94 , Block Name : Exercise