Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 9 Electro Chemistry Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 9 Electro Chemistry

### 12th Chemistry Guide Electro Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
The number of electrons that have a total charge of 9650 coulombs is ………..
(a) 6.22 × 1023
(b) 6.022 × 1024
(c) 6.022 × 1022
(d) 6.022 × 10-34
Answer:
(c) 6.022 × 1022
Hint: IF = 96500 C = 1 mole of e = 6.023 × 1023 e
9650 C = $$\frac{6.22 \times 10^{23}}{96500} \times 9650$$ = 6.022 × 1022

Question 2.
Consider the following half cell reactions:
Mn2+ + 2e → Mn E° = – 1.18V
Mn2+ → Mn3+ + e E = – 1.51V
The E for the reaction 3Mn2+ → Mn+2Mn3+, and the possibility of the forward reaction are respectively.
(a) 2.69V and spontaneous
(b) – 2.69 and non spontaneous
(c) 0.33V and Spontaneous
(d) 4.18V and non spontaneous
Answer:
(b) – 2.69 and non spontaneous
Hint: Mn+ + 2e → Mn(E0red) = 1.18V
2[Mn2+ → Mn3+ + e] (E0ox) = – 1.51V
3Mn2++ → Mn3+ + 2Mn3+ + (E0cell) = ?
E0red = (E0ox) + (E0cell)
= – 1.51 – 1.18 and non spontaneous
= – 2.69 V
Since E° is – ve ∆G is +ve and the given forward cell reaction is non – spontaneous.

Question 3.
The button cell used in watches function as follows
Zn(s) + Ag2O(s) + H2O(1) $$\rightleftharpoons$$ 2Ag(s) + Zn2+(aq) + 2OH(aq) the half cell potentials are
Ag2O(s) + H2O(1) + 2e →2Ag(S) + 2OH(aq) E° = 034V. The cell potential will be
(a) 0.84V
(b) 1.34V
(c) 1.10V
(d) 0.42V
Answer:
(c) 1.10V
Hint: Anodic oxidation: (Reverse the given reaction)
(E0ox ) = 0.76V cathodic reduction
E0cell = (E0ox) + (E0red) = 0.76 + 0.34 = 1.1V

Question 4.
The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity of 5.76 × 10-3S cm-1at 298 K is ………….
(a) 2.88 S cm2 mo1-1
(b) 11.52 S cm2 mol-1
(c) 0.086 S cm2 mol-1
(d) 28.8 S cm2 mol-1
Answer:
(b) 11.52 S cm2 mol-1
Solution:

Question 5.

Calculate A0HOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25°C.
(a) 517.2
(b) 552.7
(c) 390.7
(d) 217.5
Answer:
(c) 390.7
Hint:

Question 6.
Faradays constant is defined as
(a) charge carried by I electron
(b) charge carried by one mole of electrons
(c) charge required to deposit one mole of substance
(d) charge carried by 6.22 × 1010 electrons
Answer:
(b) charge carried by one mole of electrons
Solution:
IF = 96500 C = 1 charge of mole of e = charge of 6.022 × 1023 e

Question 7.
How many faradays of electricity are required for the following reaction to occur
MnO4 → Mn2+
(a) 5F
(b) 3F
(C) IF
(d) 7F
Answer:
(a) 5F
Hint:
7MnO4 + 5e → Mn2+ + 4H2O
5 moles of electrons i.e., 5F charge is required.

Question 8.
A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g / mol and IF = 96500C).
(a) 4
(b) 2
(c) 8
(d) 6
Answer:
(b) 2
Solution:
m = ZIt
41mm 40sec = 2500 seconds
= $$\frac { 40 × 3.86 × 2500 }{ 2 x 96500 }$$
Z = $$\frac { m }{ n × 96500 }$$ = $$\frac { 40 }{ 2 × 96500 }$$
= 2g

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0.1 mol of chlorine gas using a current of 3A is ………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Answer:
(b) 107.2 minutes
Solution:
$$\frac { m }{ ZI }$$ (mass of 1 mole of Cl2 gas = 71)
t = $$\frac { m }{ ZI }$$ mass of 0.1 mole of Cl2 gas = 7.1 g mol-1)

Question 10.
The number of electrons delivered at the cathode during electrolysis by a current of 1 A in 60 seconds is (charge of electron = 1.6 × 10-19C)
(a) 6.22 × 1023
(b) 6.022 × 1020
(c) 3.75 × 1020
(d) 7.48 × 1023
Answer:
(c) 3.75 × 1020
Solution:
Q = It
= 1A × 60S
96500 C charge 6.022 × 1023 electrons
60 C charge = $$\frac{6.022 \times 10^{23}}{96500} \times 960$$
= 3.744 × 1020 electrons

Question 11.
Which of the following electrolytic solution has the least specific conductance?
(a) 2N
(b) 0.002N
(c) 0.02N
(d) 0.2N
Answer:
(b) 0.002N
Solution:
In general, specific conductance of an electrolyte decreases with dilution. SO, 0.002N solution has least specific conductance.

Question 12.
While charging lead storage battery
(a) PbSO4 on cathode is reduced to Pb
(b) PbSO4 on anode is oxidised to PbO4
(c) PbSO4 on anode is reduced to Pb
(d) PbSO4 on cathode is oxidised to Pb
Answer:
(c) PbSO4 on anode is reduced to Pb.
Solution:
Charging: anode: PbSO4(s) + 2e → Pb (s) + SO4-2 (aq)
Cathode: PbSO4(s) + 2H2O (1) → PbO2 (s) + SO4-2 (aq) + 2e

Question 13.
Among the following cells
I. Leclanche cell
II. Nickel – Cadmium cell
III. Lead storage battery
IV. Mercury cell
Primary cells are …………
(a) I and IV
(b) I and III
(c) III and IV
(d) II and III
Answer:
(a) I and IV

Question 14.
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
(a) Zinc is lighter than iron
(b) Zinc has lower melting point than iron
(c) Zinc has lower negative electrode potential than iron
(d) Zinc has higher negative electrode potential than iron
Answer:
(d) Zinc has higher negative electrode potential than iron
Solution:
E0Zn+|Zn = – 0.76V and E0Fe2+|Fe = 0.44V. Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

Question 15.
Assertion: pure iron when heated in dry air is converted with a layer Of rust.
Reason: Rust has the composition Fe3O4
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(d) both assertion and reason are false.
Solution:
Both are false

1. Dry air has no reaction with iron
2. Rust has the composition Fe2O3 x H2O

Question 16.
In H2 – O2 fuel cell the reaction occur at cathode is ……….
(a) O2(g) + 2H2O (l) + 4e → 4OH(aq)
(b) H+(aq) + OH(aq) → H2O (l)
(c) 2H2(g) + O2(g) → 2H2O (g)
(d) H+ + e → 1/2 H2
Answer:
(a) O2(g) + 2H2O (l) + 4e → 4OH(aq)
Solution:
(a) O2(g) + 2H2O (l) + 4e → 4OH(aq)

Question 17.
The equivalent conductance of M/36 solution of a weak monobasic acid is 6mho cm2 and at infinite dilution is 400 mho cm2. The dissociation constant of this acid is ………….
(a) 1.25 x 10-16
(b) 6.25 x 10 -6
(c) 1.25 x 10-4
(d) 6.25 x 10-5
Answer:
(b) 6.25 x 10 -6
Hint: α = $$\frac { 6 }{ 400 }$$
Ka = α2C = $$\frac { 6 }{ 400 }$$ x $$\frac { 6 }{ 400 }$$ x $$\frac { 1 }{ 36 }$$
= 6.25 x 10-6

Question 18.
A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (K = 1.25 x 10-3 S cm-1 ) in the cell and the measured resistance was 800Ω at 250 C . The cell constant is,
(a) 10-1 cm-1
(b) 10-1 cm-1
(c) 1 cm-1
(d) 5.7 x 10-12
Answer:
(c) 1 cm-1
Hint: R = p.$$\frac { 1 }{ A }$$
Cell constant = $$\frac { R }{ ρ }$$ = k.R $$(\frac { 1 }{ ρ } =k)$$ = 1.25 x 10-3 f-1cm-1 x 800Ω = 1cm-1

Question 19.
Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is 1.85 x 10-5 S m-1. Solubility product of the saltAB at 298K (Λ0m)AB = 14 x 10-3 S m2 mol-1.
(a) 5.7 x 10-2
(b) 1.32 x 1012
(c) 7.5 x 10-12
(d) 1.74 x 10-12
Answer:
(d) 1.74 x 10-12
Solution:

Question 20.
In the electrochemical cell: Zn|ZnSO4 (0.01M)||CuSO4 (1.0M)|Cu , the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that CuSO4 changed to 0.0 1M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2?
(a) E1 < E2
(b) E1 > E2
(c) E2 = 0↑E1
(d) E1 = E2
Answer:
(b) E1 > E2
Solution:

Question 21.
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

Then the species undergoing disproportional is …………..
(a) Br2
(b) BrO4
(c) BrO3
(d) HBrO
Answer:
(d) HBrO
Hint:

(Ecell)A = – 1.82 + 1.5 = – 0.32V
(Ecell)B = – 1.5 + 1.595 = + 0.095V
(Ecell)C = 1.595 + 1.0652 = – 0.529V
The species undergoing disproportionation is HBrO

Question 22.
For the cell reaction
2Fe3+(aq) + 2I(aq) → 2Fe2+ (aq) + I2(aq)
EC0cell = 0.24V at 298K. The standard Gibbs energy (∆ G0 ) of the cell reactions is …………
(a) – 46.32 KJ mol-1
(b) – 23.16 KJ mol-1
(c) 46.32 KJ mol-1
(d) 23.16 KJ mor-1
Answer:
(a) – 46.32 KJ mol-1
Solution:
n = 2; E0cell = 0.24V; ∆G = ?; F = 96500C
∆G0 = – nFE°
∆G0= – 2 × 96500 × 0.24
∆G0 = – 46320 J mol-1
∆G0 = – 46.32 KJ mol-1

Question 23.
A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?
(a) 31.75
(b) 15.8
(c) 7.5
(d) 63.5
Answer:
(b) 15.8
Solution:
m1 = 0.504 g m2 = ?
e1 = 1.008 e2 = 31.77
$$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{e}_{1}}{\mathrm{e}_{2}}$$
∴ $$\mathrm{m}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{e}_{1}} \cdot \mathrm{e}_{2}=\frac{0.504 \times 31.77}{1.008}=15.885 \mathrm{~g}$$

Question 24.
A gas X at 1 atm is bubble through a solution containing a mixture of 1MY and 1MZ-1 at 25°C . If the reduction potential of Z > Y> X, then
(a) Y will oxidize X and not Z
(b) Y will oxidize Z and not X
(c) Y will oxidize both X and Z
(d) Y will reduce both X and Z
Answer:
(a) Y will oxidize X and not Z
Solution:

Zoxidises Yand X
Y oxidises X and reduces Z (does not oxidise Z)
X nduces Y and Z

Question 25.
Cell equation: A2+ + 2B → A2+ + 2B
A2+ + 2e → AE° = + 0.34V and log10 K = 15.6 at 300K for cell reactions find E° for
B1 + e → B
(a) 0.80
(b) 1.26
(c) – 0.54
(d) – 10.94
Answer:
(a) 0.80
Solution:
A2+ + 2B → A2+ + 2B E°cell
Half reaction anode A → A2+ + 2e
ox = -.034V [Given : A2+ + 2e → A E° = +0.34V]
Cathode 2B+ + 2e → 2B E°red = ?]
log10K = 156; T = 300K; n = 2;
F = 96500C;
R 8.314 JK-1 mol-1
∆G = – 2.303 RT log K
:. -nFE° = – 2.303 RT log K

∴ E°red = E°cell – E°oxid
= 0.4643 – (-0.34)
= 0.4643 + 0.34
E° 0.8043 V = 0.80V

II. Short Answer Questions

Question 1.
Define anode and cathode
Answer:

1. Anode: The electrode at which the oxidation occur is called anode.
2. Cathode: The electrode at which the reduction occur is called cathode.

Question 2.
Why does conductivity of a solution decrease on dilution of the solution?
Answer:
Conductivity always decreases with decrease in concentration (on dilution of the solution) both for weak as much as for strong electrolytes. ¡t is because the number of ions per unit volume that carry the current is a solution decreases on dilution.

Question 3.
State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.
Answer:
Kohlrausch’s law:
It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Determination of the molar conductivity of weak electrolyte at infinite dilution.
It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally. However, the same can be calculated using Kohlraushs Law. For example, the molar conductance of CH3COOH, can be calculated using the experimentally determined molar conductivities of strong electrolytes HCI, NaCI and CH3COONa.
Λ°CH3COONa = λ°Na+ + λ°CH3COONa …….(1)
Λ°HCl = λ°H+ + λ°Cl ………………(2)
Λ°NaCl = λ°Na+ + λ°Cl …………….(3)
Equation (1) + Equation (2) – Equation (3) gives,
(Λ°CH3COONa) + (Λ°HCl) – (Λ°NaCl) = λ°H+ + λ°CH3COONa = Λ°CH3COONa

Question 4.
Describe the electrolysis of molten NaCI using inert electrodes.
Answer:
1. The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key.

2. The electrode which is attached to the negative end of the power supply is called the cathode and the one is which attached to the positive end is called the anode.

3. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time
pull the electrons from the anode.

Cell reactions:
Na+ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.
Cathode (reduction)
Na+(I) + eNa(1)
E0 = – 2.7 1V
Similarly, Cl ions are attracted towards anode where they losses their electrons and oxidised to chlorine gas. Anode (oxidation)
2Cl(1) Cl2(g) + 2e
E° = – 1 .36V
The overall reaction is,
2Na+(l) + 2Cl(l) → 2Na(l) + Cl2(g) (g)
E0 = 4.07 V

The negative E° value shows that the above reaction is a non spontaneous one. Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCI. In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.

Question 5.
State Faraday’s Laws of electrolysis.
Answer:
Faraday’s laws of electrolysis:
1. First law:
The mass of the substance (M) liberated at an electrode during electrolysis is directly proportional to the quantity of charge (Q) passed through the cell. M α Q

2. Second law:
When the same quantity of charge is passed through the solutions of different electrolytes, the amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents. M α Z

Question 6.
Describe the construction of Daniel cell. Write the cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell:
The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.

Reduction half cell:
A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

Joining the half cells :
The zinc and copper strips are externally connected using a wire through a switch (K) and a load (ex : volt meter)

The electrolytic solution in the cathodic and anodic compartment are connected using an inverted U tube containing agar-agar gel mixed with an inert electrolyte like Kcl, Na2SO4 etc.,
This acts as the salt bridge.
When the switch (K) closes the circuit, the electrons flow from zinc strip to copper strip.
This is due to the following redox reactions.

Anodic oxidation:
Zinc strip acts as the anode.
Here oxidation occurs.
Zinc is oxidised to Zn2+ ions and electrons.
Zn2+ ions enter the solution and electrons enter the zinc metal strip; then flow through the external wire and enter the copper strip.
Electrons are liberated at zinc electrode and hence it is negative.

Cathodic reduction:
Copper strip acts as the cathode.
Here reduction occurs. .
Cu2+ ions are reduced to copper metal.
Cu2+ ions in the solution accepts electrons flowing through the circuit from zinc to copper strip and gets deposited on the electrode as copper metal.
Here electrons are consumed and hence it is positive.

Salt bridge:

The electrolytes present in two half cells are connected using salt bridge. To maintain electrical neutrality Cl- ions (from KCl) move from the salt bridge.
Due to cathodic reduction, cathodic compartment contains more number of SO42- ions in the solution and becomes
negatively charged. To maintain electrical neutrality K+ ions (from KCl) move from the salt bridge.

Completion of circuit:

Electrons flow from negatively charged zinc anode into positively charged copper cathode through the external wire.
At the same time anions move towards anode and cations move towards cathode.
This completes the circuit.

Consumption of electrodes:

As the Daniel cell operates, the mass of zinc electrode gradually decreases, while the mass of copper electrode increases.
Hence the cell will function until the entire metallic zinc is converted into Zn2+ ions or the entire Cu2+ ions are converted into metallic copper.

Question 7.
Why is anode in galvanic cell considered to be negative and cathode positive electrode?
Answer:
A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.

The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it render ¡t effectively negative and the electron flow goes from it through the wire to the cathode.

Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This why the cathode is a positive electrode, because positive ions are reduced to metal atoms there.

Question 8.
The conductivity of a 0.01%M solution of a 1:1 weak electrolyte at 298K is 1.5 x 10-4 S cm-1.

1. molar conductivity of the solution
2. degree of dissociation and the dissociation constant of the weak electrolyte

Given that
λ0cation = 248.2 S cm2 mol-1
λ0anion = 51.8 S cm2 mol-1
Answer:
1. Molar conductivity
C = 0.01M
k = 1.5 × 10-4 S cm-1
(or)
K = 1.5 × 10-2 S m-1
$$\frac{\kappa \times 10^{-3}}{\mathrm{C}}$$ S m-1 mol-1 m3 = $$\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01}$$ S m2 mol-1
Λm = 1.5 × 10-3 s m-1

2. Degree of dissociation
α = $$\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\infty}^{\circ}}$$ (or) α = $$\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$$
= (248.2 + 51.8)S cm2 mol-1
= 300 S cm2 mol-1
Ka = $$\frac{\alpha^{2} C}{1-\alpha}$$
= $$\frac{(0.05)^{2}(0.01)}{1-0.05}$$
Ka = 2.63 × 10-5

Question 9.
Which of 0.1M HCl and 0.1 M KCl do you expect to have greater molar conductance and why?
Answer:
Compare to 0.1M HCl and 0.1 M KCl, 0.1M HCl has greater molar conductance.

1. Molar conductance of 0.1M HCl = 39.132 × 10-3 S m2 mol-1.
2. Molar conductance of 0.1 M KCl = 12.896 × 10-3 S m2 mol-1

Because, H+ ion in aqueous solution being smaller size than K+ ion and H+ ion have greater mobility than K ion. When mobility of the ion increases, conductivity of that ions also increases. Hence, 0. 1M HCI solution has greater molar conductance than 0.1 M KCI solution.

Question 10.
Arrange the following solutions in the decreasing order of specific conductance.

1. 0.01M KCI
2. 0.005M KCI
3. 0.1M KCI
4. 0.25 M KCI
5. 0.5 M KCI

Answer:
0.005M KCl > 0.01M KCI > 0.1M KCI > 0.25KCl > 0.5 KCI.
Specific conductance and concentration of the electrolyte. So if concentration decreases, specific conductance increases.

Question 11.
Why is AC current used instead of DC in measuring the electrolytic conductance?
Answer:
1. AC current to prevent electrolysis of the solution.

2. If we apply DC current to the cell the positive ions will be attracted to the negative plate and the negative ions to the positive plate. This will cause the composition of the electrolyte to change while measuring the equivalent conductance.

3. So DC current through the conductivity cell will lead to the electrolysis of the solution taken in the cell. To avoid such a electrolysis, we have to use AC current for measuring equivalent conductance.

Question 12.
0.1M NaCI solution is placed in two different cells having cell constant 0.5 and 0.25cm-1 respectively. Which of the two will have greater value of specific conductance.
Answer:
The specific conductance values are same. Because the reaction (cation) of cell constant does not change.

Question 13.
A current of 1 .608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu2+ after electrolysis assuming volume to be constant and the current efficiency is 100%.
Answer:
Given, I = I .608A
t = 50 min (or) 50 × 60 = 3000 S
V = 250 mL
C = 0.5M
η = 100%
The number of Faraday’s of electricity passed through the CuSO4 solution
Q = It
= Q = 1.608 × 3000
Q = 4824C
Number of Faraday’s of electricity = $$\frac { 4824C }{ 96500C }$$ = 0.5F
Electrolysis of CuSO4
Cu2+(aq) + 2e → Cu(s)
The above equation shows that 2F electricity will deposit 1 mole of Cu2+
0.5F electicity will deposit $$\frac { 1mol }{ 2F }$$ × 0.5F = 0.025 mol
Initial number of molar of Cu2+ in 250 ml of solution = $$\frac { 1mol }{ 250mL }$$ × 250mL = 0.125 mol
Number of molar of Cu2+ after electrolysis 0.125 – 0.025 = 0.1 mol
Concentration of Cu2+ = $$\frac { 0.1mol }{ 250mL }$$ × 1000 mL = 0.4 M

Question 14.
Can Fe3+ oxidises Bromide to bromine under standard conditions?
Given: $$\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^{\mathrm{o}}=0.771$$
$$\mathrm{E}_{\mathrm{Br}_{2} \mid \mathrm{Br}}^{\mathrm{O}}=1.09 \mathrm{~V}$$
Answer:
Required half cell reaction

E0cell = (E0ox) + (E0red) = – 1.09 + 0.771 = – 0.319V
We know that ∆G° = – nFE0cell
If E0cell is -ve; ∆G is +ve and the cell reaction is non-spontaneous.
Hence, Fe3+ cannot oxidise Bromide to Bromine.

Question 15.
Is it possible to store copper sulphate in an iron vessel for a long time?
Given:

Answer:
E0cell = (E0ox) + (E0red) = 0.44 V + 0.34V = 0.78V
These +ve E0cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 16.
Two metals M1 and M2 have reduction potential values of – xV and + yV respectively. Which will liberate H2 in H2SO4?
Answer:
Metals having negative reduction potential acts as powerful reducing agent. Since M1 has – xV, therefore M1 easily liberate H2 in H2SO4. Metals having higher oxidation potential will liberate H2 from H2SO4. Hence, the metal M1 having + xV, oxidation potential will liberate H2 from H2SO4.

Question 17.
Reduction potential of two metals M1 and M2 are
$$E_{\mathrm{M}_{1}^{2+}}^{0}=-2.3 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{M}_{1}^{2+}}^{0}=0.2 \mathrm{~V}$$
Predict which one Is better for coating the surface of iron.
Given:
$$\mathbf{E}_{\mathrm{Fe}^{2+} \mid \mathbf{F e}}^{0}=-\mathbf{0 . 4 4} \mathbf{V}$$
Answer:
Oxidation potential of M1 is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

Question 18.
Calculate the standard emf of the cell: Cd | Cd2+|| Cu2+ | Cu and determine the cell reaction. The standard reduction potentials of Cu2+ | Cu and Cd2+ | Cd are 0.34V and – 0.40 volts respectively. Predict the feasibility of the cell reaction.
Answer:

emf is +ve, so ∆G is (-)ve, the reaction is feasible.

Question 19.
In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidised at the anode and O2 at cathode. If 44.8 litre of H2 at 25°C and also pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu2+, how many grams of Cu deposited?
Answer:
Oxidation at anode:
2H2(g) + 4OH (aq) → 4H2O (1) + 4e
1 mole of hydrogen gas produces 2 moles of electrons at 25°C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
∴no. of moles of hydrogen gas produced = $$\frac{1 \mathrm{mole}}{22.4 \text { litres }}$$ x 44.8 litres = 2 moles of hydrogen
∴2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge. We know that Q = It
I = $$\frac { Q }{ t }$$ = $$\frac{4 \mathrm{F}}{10 \mathrm{mins}}$$ = $$\frac { 4×96500 }{ 10x60s }$$
I = 643.33 A
Electro deposition of copper
Cu2+(aq) + 2e → Cu(s)
2F charge is required to deposit
1 mole of copper i.e., 63.5 g
If the entire current produced in the fuel cell i.e., 4 F is utilised for electrolysis, then 2 x 63.5 i.e., 127.0 g copper will be deposited at cathode.

Question 20.
The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in the first cell. The amount of Cr deposited in the another cell? Given: molar mass of Nickel and chromium are 58.74 and 52gm-1 respectively.
Answer:

Question 21.
0.1M copper sulphate solution in which copper electrode is dipped at 25C. Calculate the electrode potential of copper.
Given : $$\left.\mathrm{E}_{\mathrm{Cu}^{2+}}^{\mathrm{o}}\right|_{\mathrm{Cu}}=0.34$$
Answer:
Given that
[Cu2+] = 0.1 M
E0Cu2+|Cu = 0.34
Ecell = ?
Cell reaction is Cu2+(aq) + 2e → Cu (s)
Ecell = E0 – $$\frac { 0.0591 }{ n }$$ log $$\frac{[\mathrm{Cu}]}{\left[\mathrm{Cu}^{2+}\right]}$$ = 0.34 – $$\frac { 0.0591 }{ 2 }$$ log $$\frac { 1 }{ 0.1 }$$
= 0.34 – 0.0296 = 0.31 V

Question 22.
For the cell Mg(s) | Mg2+(aq) || Ag+ (aq) | Ag (s), calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell. Given:
$$\mathrm{E}_{\mathrm{Mg}^{2+} \mid \mathrm{Mg}}^{\mathrm{o}}=-2.37 \mathrm{~V}$$ and $$\mathrm{E}_{\mathrm{Ag}^{2+} \mid \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V}$$
Answer:
Oxidation at anode
Mg → Mg2+ + 2e……………(1) (Eooxi = 2.37 V
Reduction at cathode
Ag+ + e → Ag……………(2) (Eored = 0.80 V
E0cell = (E0ox) + (E0red) = 2.37 + 0.80 = 3.17 V
Overall reaction
Mg + 2Ag+ → Mg2+ + 2Ag
∆G° = -nFE°
= – 2 × 96500 × 3.17
= – 6.118 × 105 J
We know that Wmax = ∆G°
Wmax = + 6.118 × 105 J
Relationship between ∆G° and Keq is,
∆G = – 2.303 RT logKeq
∆G = – 2.303 × 8.314 × 298 log Keq [25°C = 298 K]
log Keq = $$\frac{6.118 \times 10^{5}}{2.303 \times 8.314 \times 298}$$ = $$\frac{6.118 \times 10^{5}}{5705.84}$$
log Keq = 107.223
Keq = Antilog (107.223)
Kc = 1.5849

Question 23.
8.2 × 1012 litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 × 106 Cs-1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.
Answer:
Hydrolysis of water:
At anode: 2H2O → 4H+ + O2 + 4e …………..(1)
At cathode: 2H2O + 2e → H2 + 2OH
Overall reaction: 6H2O → 4H + 4OH +2H2 + O2
(or)
Equation (1) + (2) × 2
= 2H2O → 2H2 + O2
According to Faraday’s Law of electrolysis, to electrolyse two mole of Water
(36g ≃ 36 mL. of H2O), 4F charge is required alternatively, when 36 mL of water is electrolysed,
the charge generated = 4 × 96500 C.
When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to

Given that in 1 second, 2 × 106 C is generated therefore, the time required to generate
96500 × 1015 C is = $$\frac{1 \mathrm{S}}{2 \times 10^{6} \mathrm{C}}$$ × 96500 × 1015C = 48250 × 109 S
Number of year = $$\frac{48250 \times 10^{9}}{365 \times 24 \times 60 \times 60}$$ 1 year = 365 days
= 1.5299 × 106
= 365 × 24 hours
= 365 × 24 × 60 min
= 365 × 24 × 60 × 60 sec

Question 24.
Derive an expression for Nernst equation.
Answer:
Nernst equation is the one which relates the cell potential and the concentration of the species involved in an electrochemical reaction.
Let us consider an electrochemical cell for which the overall redox reaction is,
Answer:
xA + yB = lC + mD
The reaction quotient Q is,
$$\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}$$
We know that,
∆G = ∆G0 + RT ln Q
∆G = – nFEcell
∆G0 = -nFE0cell
equation (1) becomes
– nFEcell = -nFE0cell + RT ln Q
Subsitute the Q value in equation (2)
– nFEcell = – nFE0cell + RT ln $$\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}$$. ………..(3)
Divide the whole equation (3) by – nF
Ecell = E°cell – $$\frac { RT }{ nF }$$ ln $$\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$$
Ecell = E°cell – $$\frac { 2.303RT }{ nF }$$ log $$\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$$ ……………(4)
This is called the Nernst equation.
At 25°C (298 K) equation (4) becomes,
Ecell = E°cell – $$\frac { 2.303×8.314×298 }{ nx96500 }$$ log $$\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$$
Ecell = E°cell – $$\frac { 0.0591 }{ n }$$ log $$\left(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\right)$$

Question 25.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection. Al, Zn and Mg are used as sacrificial anodes.

Question 26.
Explain the function of H2 – O2 fuel cell.
Answer:
In this case, hydrogen act as a fuel and oxygen as an oxidant and the electrolyte is aqueous KOH maintained at 200°C and 20 – 40 atm. Porous graphite electrode containing Ni and NiO serves as the inert electrodes. Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively.
Oxidation occurs at the anode:
2H2(g)+ 4OH-(aq) → 4H2O(1) + 4e
Reduction occurs at the cathode O2(g) + 2 H2O(1) + 4e → 4 OH (aq)
The overall reaction is 2H2(g) + O2(g) → 2H2O(1)
The above reaction is the same as the hydrogen combustion reaction, however, they do not react directly ie., the oxidation and reduction reactions take place separately at the anode and cathode respectively like H2 – O2 fuel cell. Other fuel cell like propane – O2 and methane O2 have also been developed.

Question 27.
Ionic conductance at infinite dilution of Al3+ and SO42- are 189 and 160 mho cm2 equiv-1. Calculate the equivalent and molar conductance of the electrolyte Al2(SO4) at infinite dilution.

Solution:

III. Evaluate Yourself

Question 1.
Calculate the molar conductance of 0.01M aqueous KCI solution at 25°C . The specific conductance of KCl at 25°C is 14.114 x 10-2 Sm-1.
Answer:
Concentration of KCI solution = 0.01 M.
Specific conductance (K) = 14.114 × 10-2 S m-1
Molar conductance (Λm) = ?
Λm = $$\frac{\kappa \times 10^{-3}}{M}$$ = $$\frac{14.114 \times 10^{-2} \times 10^{-3}}{0.01}$$
S m-1 mol-1 m3
Λm = 14.114 × 10-5 × 102 = 14.114 × 10-3 Sm2 mol-1

Question 2.
The emf of the following cell at 25°C is equal to 0.34v. Calculate the reduction potential of copper electrode.
Pt(s) | H2(g,1atm) | H+ (aq,1M) || Cu2+(aq,1M) | Cu(s)
Answer:
SHE Value is zero
cell = E°R – E°L
= 0.34 – 0 = 0.34V
The reduction potential of copper electrode = 0.34V

Question 3.
Using the calculated emf value of zinc and copper electrode, calculate the emf of the following cell at 25°C.
Zn (s) | Zn2+ (aq, 1M) || Cu2+(aq, 1M) | Cu (s)
Answer:
cell = E°R – E°L
Ezn/zn2+ = 0.76V
ECu/Cu2+ = 0.76V
cell = 0.76 – (- 0.34V)
cell = 0.76 – (- 0.34)
cell = + 1.1 V

Question 4.
Write the overall redox reaction which takes place in the galvanic cell,
Pt(s) | Fe2+(aq),Fe2+(aq) || MnO4(aq), H+(aq), Mn2+(aq) || Pt(s)
Answer:
At Anode half cell – 5Fe2+(aq) → 5Fe3+(aq) + 5e
At cathode half cell – MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(1)
Overall redox reaction – 5Fe2+(aq) + MnO4(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(1)

Question 5.
The electrochemical cell reaction of the Daniel cell is
Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?
Answer:
Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
ln the case E°cell = 1.1V
Reaction quotient Q for the above reaction is, Q = $$\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$$
Ecell = E°cell – $$\frac { 0.0591 }{ n }$$ log $$\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}$$
If suppose concentration of Cu2+ is 1 .OM then the concentration of Zn2+ is 10M (why because,
ion concentration in the anode compartment increased by a 10 factor)
Ecell = 1.1 – $$\frac { 0.0591 }{ n }$$ log $$(\frac { 10 }{ 1 })$$
= 1.1 – 0.02955             ……………….(1)
= 1.070 V (cell voltage decreased)
Thus, the initial voltage is greater than E° because Q < 1. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode.

During this process, the Q = [Zn2+] [Cu2+] steadily increases and the cell voltage therefore steadily decreases. [Zn2+] will continue to increase in the anode compartment and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further leading to a further decrease in value.

Question 6.
A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g. calculate the equivalent mass of the metal.
Answer:
Given,
I = 0.15 amperes
t = 150 mins
= t = 150 × 6Osec
= t = 9000sec
Q = It
= Q = 0.15 × 9000 coulombs
= Q = 1350 coulombs
Hence, 135 coulombs of electricity deposit is equal to 0.783g of metal.
96500 coulombs of electricity, $$\frac { 0.783 × 96500 }{ 1350 }$$ = 55.97 gm of metal
Hence equivalent mass of the metal is 559.7

### 12th Chemistry Guide Electro Chemistry Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
When electric current is passed through an electrolytic solution, charge is carried by
a) electrons
b) ions
c) atoms
d) molecules
Answer:
b) ions

Question 2.
Ohm’s law
a) I = VR
b) I = $$\frac{R}{V}$$
c) I = $$\frac{V}{R}$$
d) V = $$\frac{I}{R}$$
Answer:
c) I = $$\frac{V}{R}$$

Question 3.
The cell constant of a conductivity cell with platinum electrodes at a distance of 0.5 cm and area of cross section 5 cm2 is
a) 10 cm-1
b) 0.1 cm-1
c) 2.5 cm-1
d) 0.25 cm-1
Answer:
b) 0.1 cm-1
Cell constant = $$\frac{l}{A}=\frac{0.5}{5}$$ = 0.1cm-1

Question 4.
Specific conductance (or) conductivity =
a) 1/ρ
b) $$\frac{1}{R} \frac{A}{l}$$
c) $$\mathrm{R} \frac{\mathrm{A}}{l}$$
d) both (a) and (b)
Answer:
d) both (a) & (b)

Question 5.
For 1 : 1 electrolyte like NaCl equivalent conductance is
a) less than molar conductance
b) greater than molar conductance
c) equal to molar conductance
d) zero
Answer:
c) equal to molar conductance

Question 6.
The relationship between molar conductance and equivalent conductance of 1M H2SO4 is
$$a) \Lambda_{\mathrm{m}}=\frac{\Lambda}{2} b) \Lambda_{\mathrm{m}}=2 \Lambda c) \Lambda_{\mathrm{m}}=\Lambda d) \frac{\Lambda_{\mathrm{m}}}{\Lambda}=0$$
Reason :

Question 7.
As concentration of the electrolyte decreases the specific conductance of the solution
a) decreases
b) increases
c) remains the same
d) becomes zero
Answer:
a) decreases

Question 8.
As concentration of the electorlyte decreases the molar conductance and equivalent conductance of the solution
a) decreases
b) increases
c) remains the same
d) becomes zero
Answer:
b) increases

Question 9.
In the measurement of conductivity of an electrolyte using wheatstone bridge arrangement which is correct?
a) PQ = RS
$$b) \frac{Q}{P}=\frac{R}{S} c) \frac{P}{Q}=\frac{R}{S} d) \frac{P}{Q}=\frac{S}{R}$$
Answer:
c) $$\frac{P}{Q}=\frac{R}{S}$$

Question 10.
Which among the following solution of NaCl has the maximum molar conductance value?
a) 10-4 M
b) 10-3 M
c) 10-2 M
d) 10-1 M
Answer:
a) 10-4 M
Reason : As dilution increases, molar conductance increases.

Question 11.
Molar conductance of an electrolyte approaches a limiting value in
a) highly concentrated solution
b) concentrated solution
c) very dilute solution
d) dilute solution
Answer:
c) very dilute solution

Question 12.
For a weak electrolyte there is a sudden increase in molar conductance as the concentration approaches
a) infinity
b) maximum value
c) zero
d) negative value
Answer:
c) zero

Question 13.
Limiting molar conductivity values of strong electrolytes can be determined by
a) Kohlrausch’s law
b) Faradays’ law
c) Nernst equation
d) extrapolating the straight line
Answer:
d) extrapolating the straight line

Question 14.
Limiting molar conductivity values of weak electrolytes can be determined by
a) Kohlrausch’s law
b) Faradays’ law
c) Nernst equation
d) extrapolating the straight line
Answer:
a) Kohlrausch’s law

Question 15.
Debye and Huckel derived an equation for the conductivity of strong electrolytes by assuming
a) partial dissociation
b) incomplete dissociation
c) complete dissociation
d) negligible dissociation
Answer:
c) complete dissociation

Question 16.
The basis for Kohlrausch’s law is
a) molar conductance
b) specific conductance
c) specific resistance
d) limiting molar conductance
Answer:
d) limiting molar conductance

Question 17.
Degree of dissociation of weak electrolytes can be calculated using the expression
$$a) [latex]\alpha=\frac{\mathrm{K}_{\mathrm{m}}}{\mathrm{K}_{\mathrm{m}}^{\mathrm{o}}} b) \quad \alpha=\frac{\mathrm{K}_{\mathrm{m}}^{\mathrm{O}}}{\mathrm{K}_{\mathrm{m}}} c) \alpha=\frac{\Lambda_{\mathrm{m}}^{\mathrm{o}}}{\Lambda_{\mathrm{m}}} d) \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{O}}}$$
Answer:
d) $$\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{O}}}$$

Question 18.
The device which converts chemical energy into electrical energy is known as
a) battery
b) Galvanic cell
c) Voltaic cell
d) all the above
Answer:
d) all the above

Question 19.
The device which converts electrical energy into chemical energy is known as
a) Galvanic cell
b) Voltaic cell
c) electrolytic cell
d) all the above
Answer:
c) electrolytic cell

Question 20.
In Daniel cell zinc undergoes
a) oxidation
b) reduction
c) hydrolysis
d) galvanisation
Answer:
a) oxidation

Question 21.
In Daniel cell copper ions undergo
a) oxidation
b) reduction
c) hydrolysis
d) galvanisation
Answer:
b) reduction

Question 22.
In Daniel cell the anode and cathode half cells are respectively
a) Cu | Cu2+, Zn | Zn2+
b) ZZn | Zn2+, Cu | Cu2+
c) Zn | Zn2+, Cu2+ | Cu
d) Cu2+ | Cu, Zn2+ | Zn
Answer:
c) Zn | Zn2+, Cu2+ | Cu

Question 23.
In Daniel cell, the number of electrons transfered in the redox reaction
Zn + CuSO4 → ZnSO4+ Cu is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 24.
Which of the following statement is correct with respect to electrolytic conductance? ULffi
a) Conductivity increases with the decrease ’ in Viscosity
b) Conductivity increases with increase in temperature
c) Molar conductance of a solution decreases with increase in dilution
d) Conductance decrease with increase in temperature.
Answer:
c) Molar conductance of a solution decreases with increase in dilution

Question 25.
In Daniel cell the Electronic solutions in the two half cells are connected using a
a) wire
c) copper strip
b) zinc strip
d) salt bridge
Answer:
d) salt bridge

Question 26.
A salt bridge is containing an inverted U tube
a) agar – agar gel with Kcl
b) agar – agar gel with Na2SO4
c) both (a) & (b)
d) none of the above
Answer:
c) both (a) & (b)

Question 27.
The general representation of a fuel cell is
a) Fuel / Electrode / Electrolyte / Electrode / Oxidant
b) Oxidant /Electrode / Electrolyte / Fuel
c) Fuel / Electrode / Electrolyte / Electrode / Reductant
d) Oxidant /Electrode / Electrolyte /Reductant
Answer:
a) Fuel / Electrode / Electrolyte / Electrode / Oxidant

Question 28.
Which of the following is correct?
a) 1 C = 1J × 1V
b) 1V = 1C × 1J
c) 1J = 1C × 1V
d) 1V = $$\frac{1 C}{1 \mathrm{~J}}$$

Question 29.
Ecell is equal to
a) (Eox)anode + (Ered)cathode
b) (Ered)cathode – (Ered)anode
c) (Ered)anode + (Ered)cathode
d) both (a) & (b)
Answer:
d) both (a) & (b)

Question 30.
The emf of standard hydrogen electrode is
assigned an arbitrary value of
a) zero
b) one
c) infinity
d) negative
Answer:
a) zero

Question 31.
Electrical charge carried by one electron is

Answer:
d) Both (a) & (b)

Question 32.
A cell reaction is spontaneous when AG° and ocell E are respectively
a) + ve and – ve
b) – ve and + ve
c) 1 and 0
d) 0 and 1
Answer:
b) – ve and + ve

Question 33.
Electrolysis is a
a) photochemical reaction
b) spontaneous reaction
c) non-spontaneous reaction
d) substitution reaction
Answer:
c) non-spontaneous reaction

Question 34.
If the atomic mass of an ion Mn+ is A, its electrochemical equivalent is
a) $$\frac{\mathrm{nA}}{\mathrm{F}}$$
b) $$\frac{\mathrm{F}}{\mathrm{nA}}$$
c) $$\frac{\mathrm{nF}}{\mathrm{A}}$$
d) $$\frac{A}{n F}$$
Answer:
d) $$\frac{A}{n F}$$
Reason Z = $$\frac{\text { Atomic mass }}{\text { Valency } \times \text { Faraday constant }}$$

Question 35.
When Q coulomb electric charge is passed through a series of aqueous solution of AgNO3, AlCl3 and CUSO4, the increasing order of mass of the metals liberated is
a) Ag > Cu > Al
b) AZ < Cu < Ag
c) Cu > Ag > Al
d) Ag < Cu < Al
Ans:
b) AZ < Cu < Ag
Reason: m a equivalent mass eAl < eCu < eAg \ mAl < mCu < mAg

Question 36.
Leclanche cell is a
a) primary battery
b) secondary battery
c) rechargeable battery
d) electrolytic cell
Answer:
a) primary battery

Question 37.
Lithium ion battery is a
a) primary battery
b) secondary battery
c) non – rechargeable battery
d) none of the above
Answer:
b) secondary battery

Question 38.
In cellular phones/ laptop Computers the battery used is
a) Lechlanche cell
b) Lead storage battery
c) Lithium – ion battery
d) H2 – O2 fuel cell
Answer:
c) Lithium – ion battery

Question 39.
Rusting of iron is
a) a hydrolysis reaction
b) an electrochemical redox reaction
c) an oxidation reaction
d) a reduction reaction
Answer:
b) an electrochemical redox reaction

Question 40.
The formula of rust is
a) PeO
b) Fe3O4
c) Fe (OH)3
d) Fe2O3. × H2O
Answer:
d) Fe2O3. × H2O

Question 41.
The most convenient method to protect the bottom of ship made of iron is
a) coating it with red lead oxide
b) white tin plating
c) connecting it with Mg block
d) connecting it with Pb block
Answer:
c) connecting it with Mg block
Reason: As Mg is more reactive metal than iron, it gets corroded in preference to iron. This is known as cathodic protection.

Question 42.
Standard reduction potential of three metals A, B and C are – 1.5 V, + 1 V and – 2 V respectively. The decreasing order of reducing power of these metals is
a) B > C > A
b) A > B > C
c) C > A > B
d) C > B > A
Answer:
c) C > A > B
– 2 V – 1.5 V + 1 V
Reason: lower the E° value higher is the reducing power.

Question 43.
Limiting molar conductivity of NH4OH (A?n)NH4OH is equal to

Answer:
d

Question 44.
A hydrogen gas electrode in made by dipping platinum wire in a solution of pH = 10 and by passing hydrogen gas around the platinum wire at one atmosphere pressure. The oxidation potential of the electrode would be
a) 0.59 V
b) 0.118 V
c) 1.18 V
d) 0.059 V
Answer:
a) 0.59 V

Reason: H2 —> 2H+ + 2e
Concentration 1 atm

Question 45.
A button cell used in watches functions as following:
Zn(s) + Ag2O + H2O(l) ⇌ 2Ag(s)+ Zn2+(aq) + 2OH(aq)
If half cell potentials are :
Zn2+(aq) + 2e(aq) → Zn(s) ; E° = -0.76 V
Ag2O + H2O(l) + 2e → 2Ag(s) + 2OH(aq) E° = – 0.34 V
The cell potential will be
a) 0.42 V
b) 0.84 V
c) 1.34 V
d)1.10V
Answer:
d) 1.10 V
Reason : Ecell – (Eox)anode + (Ered)cathode
= + 0.76 + 0.34 = + 1.10V

Question 46.
How many grams of cobalt metal will be deposited when a solution of Cobalt (II) chloride is electrolysed with a current of 10 amperes for 109 minutes? (1 Faraday = 96500 C; Atomic mass of (Co = 59 u)
a) 0.66
b) 4.0
c) 20.0
d) 40.0
Answer:
c) 20.0

Question 47.
Consider the half-cell reactions
Mn2+ + 2e → Mn; E° = – 1.18 V
Mn2+ → Mn3+ + e ; E° = – 1.51 V
The E° for the reaction 3Mn2+ → Mn + 2Mn3+ and the possibility of the forward reaction are respectively.
a) – 2.69 and not possible
b) – 4.18 V and possible
c) + 0.33 V and possible
d) + 2.69 V and not possible
Answer:
a) – 2.69 and not possible
Reason : E° = E°oxid + E°red
= – 1.51 +(-1.18) E°= – 2.69V
= 1.51 -1.18 ΔG°= -nFE°,
E° is negative, ΔG° is positive and hence the forward reaction is not possible.

Question 48.
The electrolyte used in Lechlanche cell is
a) Paste of KOH and ZnO
b) 38 % solution of H2SO4
c) Moist of paste NH4Cl and ZnCl2
d) Moist sodium hydroxide
Answer:
c) Moist of paste NH4Cl and ZnCl2

Question 49.
During the electrolysis of fused NaCl, which reaction occurs at anode.
a) chloride ions are oxidised
b) chloride ions are reduced
c) Sodium ions are oxidised
d) sodium ions are reduced
Answer:
a) chloride ions are oxidised

Question 50.
Amount of electricity that can deposit 108 gm of silver from AgNO3 solution is
a) 1 ampere
b) 1 coulomb
c) 1 Faraday
d) 1 Volt
Answer:
c) 1 Faraday
Reason: 1 g eq.mass of Ag (108 g) will be liberated by 1 Faraday

II. Match the following:

Question 1.

 A B i) Specific conductance a) Sm2 mol-1 ii) Conductance b) m-1 iii) Cell constant c) Sm2g equiv-1 iv) Molar conductance d) Sm-1 v) Equivalent conductance e) S

Answer:
d) Sm-1
e) S
b) m-1
a) Sm2 mol-1
c) Sm2g equiv-1

III. Pick out the correct statements

Question 1.
i) For strong electrolyte the plot ∧mVs√C is not a linear one.
ii) For a strong electrolyte, at high concentration the number of constituent ions in a given volume is high.
iii) At high concentration the ions experience a viscous drag due to greater solvation.
iv) At infinite dilution the ions are so close and the interaction between them becomes significant.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct Statement:
i) For strong electrolytes the plot ∧mVs√C is a straight line.
iv) At infinite dilution the ions are so apart and the interaction between them becomes insignificant. ‘

Question 2.
i) In Daniel cell electrons are liberated at : inc electrode and hence it is negative.
ii) Electrons flow from copper cathode to zinc anode.
iii) Through salt bridge ions can move into (or) out of the half cells.
iv) The zinc and copper strips are connected externally through salt bridge.

a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c) (i) & (iii)

Correct Statement:
(ii) Electrons flow from, zinc anode to copper cathode.
(iv) The zinc and copper strips are connected externally through a wire.

Question 3.
i) Electrolysis is carried out in an electrolytic cell by connecting it to an AC power supply.
ii) The device which is used to carry out the electrolysis is called the electrolytic cell.
iii) In the electrolytic cell cathode is + ve and anode is – ve.
iv) The electrochemical process occuring in the electrolytic cell and galvanic cell are the reverse of each other.
a) (i) & (ii) b) (ii) & (iii)
c) (iii) & (iv) d) (ii) & (iv)
Answer:
d) (ii) & (iv)

Correct Statement :
(i) Electrolysis is carried out in an electrolytic cell by connecting it to a DC power supply.
(iii) In the electrolytic cell cathode is – ve and anode is + ve.

Question 4.
i) The amount of substance deposited when 1 ampere current passed for 1 second is its equivalent mass.
ii) When same quantity of electric charge is passed through different solutions the masses liberated are inversely proportional to their equivalent masses.
iii) One gram equivalent mass of any substance can be liberated by passing one Faraday electricity.
iv) One Faraday of electric charge is carried by Avogadro number of electrons.

a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct Statement :
(i) The amount of substance deposited when 1 ampere current passed for 1 second is its electrochemical equivalent.
(ii) When same quantity of electric charge is passed through different solutions the masses liberated are directly proportional to their equivalent masses.

IV. Pick out the incorrect statements

Question 1.
i) At constant temperature, the current flowing through the cell (I) is directly proportional to the resistance of the cell.
ii) Resistance is the opposition that a cell offers to the flow of electric current through it.
iii) The conductivity of the electrolyte is measured using a conductivity cell.
iv) Resistance of an electrolytic solution is directly proportional to the cross sectional area and
inversely proportional to the length.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)

Correct Statement :
(i) At constant temperature, the current flowing through the cell (I) is directly proportional to the voltage across the cell (V).
(iv) Resistance of an electrolytic solution is directly proportional to the length and inversely proportional to the cross sectional area.

Question 2.
i) The conductance of lm3 electrolytic solution is called the specific conductance.
ii) Conductivity increases with increase in viscosity.
iii) Specific conductivity increases with increase in dilution.
iv) Conductivity of a cell can be measured using wheatstone bridge arrangement.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct Statement:
(ii) Conductivity increases with decrease in viscosity.
(iii) Specific conductivity decreases with increase in dilution.

Question 3.
i) Molar conductivity is due to the independent migration of cations in one direction and anions in the opposite direction.
ii) At infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance.
iii) It is possible to determine the molar conductance at infinite dilution for weak electrolytes experimentally.
iv) The solubility product of a sparingly soluble salt can not be determined using conductivity measurements.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct Statement:
(iii) It is impossible to determine the molar conductance at infinite dilution for weak electrolytes experimentally.
(iv) The solubility product of a sparingly soluble salt can be determined using conductivity measurements.

Question 4.
i) Corrosion of iron is called rusting.
ii) Coating of iron over zinc is called galvanisation.
iii) The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
iv) Greater the E° values in the spectrochemical series greater is the tendency of the species to donate electrons and undergo oxidation.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)

Correct Statement:
(ii) Coating of zinc over iron is called galvanisation.
(iv) Greater the E° values in the spectrochemical series greater is the tendency of the species to accept electrons and undergo reduction.

V. Assertion and reason

Question 1.
Assertion (A): If the temperature of the electrolytic solution increases, conductance also increases.
Reason (R): As temperature increases, the kinetic energy of the ions decreases and the attractive force between the ions increases.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong. Correct Reason R: As temperature increases, the kinetic energy of the ions increases and the attractive force between the ions decreases.

Question 2.
Assertion (A): In the wheatstone bridge arrangement AC power supply is used in the measurement of conductivity of an electrolyte in conductivity cell.
Reason (R): DC power supply will lead to the electrolysis of the solution taken in the cell.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A& R are correct, R explains A

Question 3.
Assertion (A): In lead acid storage battery, when, a potential greater than 2V is applied across electrodes the cell reactions of discharge process are reverse.
Reason (R): The electrochemical reactions which take place in galvanic cell may be reversed by applying a potential slightly greater than the emf generated by the cell.

a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
a) Both A & R are correct, R explains A

Question 4.
Assertion (A): Magnesium is used as a sacrificial anode to protect iron from rusting. Reason (R): The reduction potential of magnesium is higher than iron hence magnesium has lesser tendency to undergo corrosion in preference to iron.
a) Both A & R are correct, R explains A
b) Both A & R are correct, R does not explain A.
c) A is correct, but R is wrong.
d) A is wrong, but R is correct.
Answer:
c) A is correct, but R is wrong Correct Reason R: The reduction potential of magnesium is less than iron and hence magnesium has higher tendency to undergo corrosion in preference to iron.

VI. Two Mark Question

Question 1.
State Ohm’s law.
Answer:
At constant temperature, the current flowing through the cell (1) is directly proportional to the voltage across the cell (V).
ie) I ∝ V (or) I = V/R (or) V = IR

Question 2.
Define resistivity.
Answer:
Resistivity is defined as the resistance of an electrolyte confined between two electrodes having unit cross sectional area (1m2) and are separated by unit distance (1m).

Question 3.
Define specific conductance (or) conductivity.
Answer:
Specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimension, (lm3).

Question 4.
Define molar conductivity (∧m)
Answer:
Conductivity of a cell in which the electrodes are lm apart containing 1 mole of electrolyte in V m3 of electrolytic solution is called as molar conductivity.
Relation between specific conductance and molar conductance.
$$\Lambda_{\mathrm{m}}=\frac{\mathrm{K}\left(\mathrm{sm}^{-1}\right) \times 10^{3}}{\mathrm{M}} \mathrm{mot}^{-1} \mathrm{~m}^{3}$$

Question 5.
Define equivalent conductance (∧)
Answer:
Conductivity of a cell in which the electrodes are lm apart and containing 1 gram equivalent of electrolyte in Vm3 of electrolytic solution is called as equivalent conductance (∧)
$$\Lambda=\frac{\kappa \times 10^{-3}}{N} \mathrm{Sm}^{2} \mathrm{~g} \text { equiv }^{-1}$$
K = Specific conductance
N Normality

Question 6.
Define Electrochemical equivalent.
Answer:
Electrochemical equivalent is defined as the amount of substance deposited at the electrode by the passage of 1 coulomb charge.
Electrochemical equivalent
Z = $$\frac{\text { Equivalent mass }}{96500}$$

Question 7.
What is the role of salt bridge in Galvanic cell?
Answer:
To maintain the electrical neutrality in both the compartments, the non reactive anions Cl (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic compartment, at the same time some of the K+ ions move from the salt bridge into the cathodic compartment.

Question 8.
What are the applications of Kohlrausch’s law?
Answer:
It is used to calculate the molar conductance of a weak electrolyte at infinite dilution.
$$(\mathrm{eg}) \Lambda_{\mathrm{CH}_{\mathrm{COOH}}}^{\mathrm{o}}=\Lambda_{\mathrm{CH}_{2} \mathrm{COONa}}^{\mathrm{o}}+\Lambda_{\mathrm{HCl}}^{\mathrm{o}}-\Lambda_{\mathrm{NaCl}}^{\circ}$$
It is used to calculate the degree of dissociaton of weak electrolytes.
$$\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{o}}}$$
It is used to calculate the solubility product of sparingly soluble salts.
For 1:1 electrolyte
$$K_{S P}=\left(\frac{K \times 10^{-3}}{\Lambda_{m}^{o}}\right)^{2}$$

Question 9.
Define emf of a-cell.
Answer:

• The force that pushes the electrons away from the anode and pulls them towards the cathode is called the electromotive force (emf) or the cell potential.
• SI unit of emf is volt (V).

Question 10.
Define Joule.
Answer:
When there is one volt difference in electrical potential between the anode and cathode, the energy released for each coulomb of charge that moves between them is defined as one joule of energy.
1 J = 1 C × 1 V.

Question 11.
Write the factors affecting cell voltage.
Answer:

• Nature of the electrodes.
• Concentration of the electrolytes.
• Temperature at which the cell is operated.

Question 12.
What is the IUPAC definition of electrode potential (E)?
Answer:
Electrode potential is the electromotive force of a cell in which the electrode on the left is a standard hydrogen electrode and the electrode on the right is the electrode in question.

Question 13.
What is the IUPAC definition of standard electrode potential (E°)?
Answer:
Standard electrode potential (E°) is the value of the standard emf of a cell in which molecular hydrogen under standard pressure is oxidised to solvated protons at the left hand electrode.

Question 14.
Define electrolysis.
Answer:

• Electrolysis is a process in which the electrical energy is used to cause a non- spontaneous chemical reaction to occur.
• The energy is often used to decompose a compound into its elements.

Question 15.
What is galvanisation?
Answer:

• Coating of zinc over iron is called galvanisation.
• Galvanisation of iron prevents rusting.

Question 16.
How cathodic protection helps to protect the metal corrosion
Answer:

• Unlike galvanisation in this method the entire surface of iron is not covered with zinc.
• Metals like Mg or Zn which corrode more easily than iron can be used as a sacrificial anode.
• Iron material acts as a cathode.
• So iron is protected but Mg or Zn is corroded during rusting.
• This is known as cathodic protection.

Question 17.
How metals can be protected from corrosion?
Answer:

• Coating metal surface by paint.
• Galvanisation.
• Cathodic protection.
• Passivation.
• Alloy formation.

Question 18.
What is electrochemical series?
How is it useful to predict corrosion?
Answer:

• The series in which the standard aqueous electrode potential at 298 K for various metal – metal ion electrodes are arranged in the decreasing order of their standard reduction potential values is called electrochemical series.
• Standard reduction potential E° is a measure of the oxidising tendency of The species.
• Greater the E° value, greater is the tendency shown by the species to accept electrons and undergo reduction.
• Higher the E° value lower is the tendency to undergo corrosion..

VII.Three Mark Questions

Question 1.
The molar conductivity of a strong electrolyte and a weak electrolyte increases with dilution, why?
Answer:
Strong electrolytes:

• When dilution increases, the ions are far apart.
• Attractive force between the ions decreases.
• The interaction between the ions becomes insignificant.
• Hence molar conductivity increases and reaches a maximum value at infinite dilution.

Weak electrolytes:

• According to Ostwald’s dilution law, as dilution increases the dissociation of weak electrolyte increases.
• As dissociation increases the number of ions increases.
• Hence molar conductivity increases with dilution.

Question 2.
Write a note on Debye-Huckel and onsagar equation.
Answer:

• At infinite dilution, the interaction between the ions in the electrolyte solution is negligible.
• Except this condition, electrostatic interaction between the ions alters the properties of the solution from those expected from the free – ions – value.
• The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.
• They considered that each ion is surrounded by an ionic atmosphere of opposite sign.
• They derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociation.
• Later, this equation was further developed by onsagar.
• For a uni-univalent electrolyte the Debye, Huckel and onsagar equation is given as,

A and B are constants which depend only on the nature of the solvent and temperature.
D – dielectric constant of the medium, r) – viscosity of the medium.
T – temperature in Kelvin.

Question 3.
From the following data, prove that each constituent ion of the electrolyte makes a definite contribution towards the molar conductance irrespective of the nature of other ion with which it is associated.
Answer:

 Electrolyte ∧°m at 298 K KCl NaCl 149.86 126.45 KBr NaBr 151.92 128.51 KNO3 NaNO3 114.96 121.55

Solution:

• This value 23.41 for the difference between the (λ°m) values of K+ and Na+ is constant irrespective of the anion (Cl, Br or NO3) with which they are associated.
• This proves that at infinite dilution each constituent ion of the electrolyte makes a definite contribution towards the molar conductance of the electrolyte irrespective of the nature of the other ion with which it is Associated
• Thus Kohlraush’s law can be proved.
• Similarly we can conclude that

Question 4.
What is known as intercalation?
Answer:

• In lithium-ion battery, during discharge, the Li+ ions produced at the anode move towards cathode through the non-aqueous electrolyte.
• When a potential greater than the emf produced by the cell, is applied across the electrode, the cell reaction is reversed.
• Now the Li+ ions move from cathode to anode where they become embedded on the porous electrode.
• This is known as intercalation.

Question 5.
Write a note on standard Hydrogen Electrode (SHE).
Answer:

• Standard Hydrogen Electrode is used as a reference electrode.
• Its emf is assigned an arbitrary value of zero volt.
• It consists of a platinum electrode in contact with 1 M HC1 and 1 atm hydrogen gas.
• The hydrogen gas is bubbled through the solution at 25°C.
• SHE can act as a cathode as well as an anode. If SHE is used as a cathode, the reduction reaction is

2H+(aq,1M) + 2e → H2 (g, latm) E° = 0 volt
• If SHE is used as an anode, the oxidation reaction
is H2, (g, 1atm) →2H+(aq,1M) + 2e E°= 0 volt

Question 9.
Write a note on lithium – ion battery.
Answer:

• Lithium – ion battery is a secondary battery (rechargeable)
• Anode – Porous graphite
• Cathode – transition metal oxide such as Co02.
• Electrolyte – Lithium salt in an organic solvent.
• Oxidation at anode :
Li(s) → Li+(aq) + e ………………(1)
• Reduction at cathode :
Li++ CoO2(s) + e → LiCoO2(s) ……………..(2)
• Equation (1) + (2) gives the overall redox reaction.
Li(s) + CoO2(s) – LiCoO2(s) ………………. (3)
• Both electrodes allow Li+ ions to move in and out of their structures.
• Discharge process – Li+ ions produced at the anode move towards cathode through the non-aqueous electrolyte.
• Recharge process – A potential greater than
the emf produced by the cell is applied across the electrode, the cell reaction is reversed. .
• Li+ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
• Uses – cellular phones, laptop computers, digital cameras.

VIII. Five Mark Questions.

Question 1.
What are the factors affecting electrolytic conductance
Answer:

• As inter ionic attraction increases, conductance decreases.
• Solvent of high dielectric constant show high conductance in solution.
• As viscosity of the medium increases, conductance decreases.
• An temperature of the electrolytic solution increases, conductance increases because kinetic energy of ions increase and the attractive force between them decrease.
• As dilution increases specific conductance decreases as the number of ions for unit volume decreases.
• As dilution increases, molar conductance and equivalent conductance of a strong electrolyte increases, because inter ionic attractive force between the ions decreases with dilution.
• As dilution increases, molar conductance and equivalent conductance of a weak electrolyte increases, because degree of dissociation increases with dilution.

Question 2.
How is the conductivity of an electrolytic solution determined using a Wheatstone bridge?

Answer:

• Conductivity of an electrolytic solution is determined by using a Wheatstone bridge.
• Here one of the resistance is replaced by a conductivity cell containing the electrolytic solution of unknown conductivity.
• If DC power supply is used, it will lead to the electrolysis of the electrolytic solution.
• So, AC power supply is used to prevent electrolysis.
• A Wheatstone bridge is constituted using conductance and the curve is almost parallel to Am axis.
• This is because as the dilution increases dissociation of the weak electrolyte also increases.
• ∧°m values for strong electrolytes can be obtained by extrapolating the straight line.
• But this is not applicable for weak electrolytes, as the plot is not a linear one.
• ∧°m values of the weak electrolytes can be determined using Kohlrausch’s law.

Question 4.
Explain the thermodynamics of cell reactions.
Answer:
In a galvanic cell, the chemical energy is converted into electrical energy.
Electrical energy produced by the cell = Total charge of electrons x emf of the cell.
If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing agents in the overall cell reaction, the electrical energy produced by cell is Electrical energy = charge of ‘n’ mole of electrons x Ecell …………… (1)

Charge of 1 mole of electrons = 1 F
∴ Charge of n mole of electrons = n F
∴ Equation (1) becomes
Electrical energy = nFEcellE0u……………(2)

This energy is ueto do work
∴ Maximum work obtaJed from a galvanic cell is
(Wmax))cell nFEcell ………………….(3)
The (-). sign indicates work is done by the system on the surroundings.

According to ‘second aw of thermodynamics
Maximum work done by the system =
change -in-Gibbs free energyof the.’system
(Wcell = ∆G ………………..(4)
Combining’ (3) and (4)
∆G = -nFEcell ………………(5)

If Ecell is positive, ∆G is negative and the cefi reaction is spontaneous.

When all the cell components are m their
standard state, tbe above equation becomes
∆G° = -nFE°cell ……………..(6)
Also ∆G° = – RT In Keq ……………..(7)

Question 5.
Write about Lechlanche cell
Answer:

• Lechlanche cell is a primary battery (non – rechargeable)
• Anode : Zinc container
• Cathode : Graphite rod in contact with MnO2.
• Electrolyte : NH4Cl & ZnCl2 in water.
• EMF: 1.5 V

Oxidation at anode :

Equation (1) + (2) +(3) gives the overall redox reaction

Ammonia produced at the cathode combines with Zn2+ to form a complex ion [Zn (NH3)4]2+.
As the reaction proceeds, the concentration of NH4+ will decrease and the aqueous NH3 will increase which lead to the decrease in the emf of the cell.

Question 6.
Write about mercury button cell.
Answer:

• Mercury button cell is a primary battery (non – rechargeable)
• Anode : Zinc amalgamated with mercury.
• Cathode : HgO mixed with graphite.
• Electrolyte : Paste of KOH and ZnO.
• EMF :1.35 V
• Oxidation at anode :
• It has higher capacity and longer life.
• Uses : In pacemakers, electronic watches, cameras, etc.,

Question 7.
Write about lead storage battery.
Answer:

• Electrochemical reactions which take place in a galvanic cell may be reversed by applying a potential slightly greater than the emf . generated by the cell.
• This principle is used in secondary batteries to regenerate the original reactants.
• Lead storage battery is a secondary battery (reachargeable)
• Anode: Spongy lead.
• Cathode : Lead plate bearing PbO2.
• Electrolyte : 38 % by mass of H2SO4 with density 1.2 g / mL.
• EMF : emf of a single cell is 2 V. Usually six cells are combined in series to produce 12 volt.

Oxidation at anode:
$$\mathrm{Pb}_{(\mathrm{s})} \rightarrow \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}$$………..1
$$\mathrm{Pb}^{2+} ions combine with \mathrm{SO}_{4}^{2-} to form PbSO4 precipitate. \mathrm{Pb}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4}^{2} \underset{(\mathrm{aq})} \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}$$ ………………..(2)

Reduction at cathode;
$$\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{e} \rightarrow \mathrm{Pb}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{(1)}$$ …………(3)
Pb2+ ions combine with $$\mathrm{SO}_{4}^{2-}$$ from H2SO4 to form PbSO4 precipitate..
$$\mathrm{Pb}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}$$………………..(4)

Equation (1) + (2) + (3) + (4) gives the overall redox reaction.
$$\begin{array}{r} \mathrm{Pb}_{(s)}+\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{SO}_{4(\mathrm{aq})}^{2-} \rightarrow \\ 2 \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)} \end{array}$$
Emf of the cell depends on the concentration of H2SO4
As the cell reaction uses $$\mathrm{SO}_{4}{ }^{2}$$ ions, the concentration of H2SO4 decreases.
When the cell potential decreases to about 1.8 V, the cell has to be recharged.

Recharge of the cell:
A potential greater than 2 V is applied across the electrodes, the cell reactions of the discharge process are reversed.
During recharge, the role of anode and cathode is reversed and H2S04 is regenerated.
Oxidation occurs at the cathode (now act as anode)

\begin{aligned} \mathrm{PbSO}_{4(s)}+& 2 \mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow \\ & \mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e} \end{aligned}
Reduction occurs at the anode (now act as cathode)
$$\mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}_{(s)}+\mathrm{SO}_{4}^{2-}(\mathrm{aq})$$
Overall redox reaction is
$$\begin{array}{c} 2 \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_{2} \mathrm{O}_{(1)} \rightarrow \\ \mathrm{Pb}_{(\mathrm{s})}+\mathrm{PbO}_{2(\mathrm{~s})}+4 \mathrm{H}_{(\mathrm{aq})}+2 \mathrm{SO}_{4}^{2-} \end{array}$$

Thus the overall cell reaction is exactly the reverse of the redox reaction which takes place while discharging. .
Uses : In automobiles, trains, inverters etc.,

Question 8.
Explain the electrochemical mechanism of corrosion.
Answer:

• The redox process which causes deterioration of metal by oxygen and moisture is called corrosion.
• Rusting or corrosion of iron is an electrochemical process.
• Rusting requires both oxygen and water.
• Since it is an electrochemical redox process, it requires an anode and cathode in different places on the iron.
• Iron surface and a droplet of water on the surface form a tiny galvanic cell as shown in the figure.
• The region enclosed by water is exposed to low amount of oxygen and it acts as anode.
• The remaining area has high amount of oxygen and it acts as cathode.
• Based on the oxygen content, an electrochemical cell is formed.

Corrosion occurs at the anode ie, the region covered by water.
Oxidation at anode : Iron dissolves in the anode region.
$$2 \mathrm{Fe}_{(\mathrm{s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+4 \mathrm{e}^{-} \quad \mathrm{E}^{\circ}=0.44 \mathrm{~V}$$ ……………………. (1)
Electrons move through the iron metal from the anode to the cathode area.
At the cathode oxygen dissolved in water is reduced to water.

Reduction at cathode:
Atmospheric carbon dioxide and water react to give carbonic acid which furnishes the H+ions for reduction.
O2(g) + 4HM++(aq) + 4e → 2H2O(l)
E° = 1.23V ……………..(2)
Electrical circuit is completed by the migration of ions through water droplet.
The overall redox reaction is, (1) + (2)
$$2 \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}$$
E° = 0.44 + 1.23 = 1.67 V
The positive emf value shows that the reaction is spontaneous.
Fe2+ ions are further oxidised to Fe3+, which on further reaction with oxygen form rust.

VIII. Additional problems:

Problems based on conductivity :

Question 1.
A conductivity cell has two platinum electrodes separated by a distance 1.5cm and the cross sectional area of each electrode is 4.5sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15Ω. Find the specific conductance of the solution.
Solution:
l = 1.5 cm = 1.5 × 10-2 m
A = 4.5 cm2 = 4.5 × (10-4)m2
R = 15Ω

$$K=\frac{1}{R}\left(\frac{\ell}{A}\right)$$
$$K=\frac{1}{15 \Omega} \times \frac{1.5 \times 10^{-2} \mathrm{~m}}{4.5 \times\left(10^{-4}\right) \mathrm{m}^{2}}$$
= 2.22 Sm-1

Question 2.
0.04 N solution of a weak add has a specific conductance 4.23 × 10-4 Scm-1. The degree of dissociation of acid at this dilution is 0.0612. Calculate the equivalent conductance of weak acid at infinite dilution.
Solution:

Problems based on Kohlrausch’s Law

Question 1.
Equivalent conductances of NaCl, HCl and CH3COONa at infinite dilution are 126.45,426.16 and 91 S cm2g equiv-1. Calculate the equivalent conductance of CH3COOH at infinite dilution.
Solution:

= 91 + 426.16 – 126.45
= 517.16 -126.45
∧°CH3COOH = 390.71Sm2gequ-1

Question 2.
The equivalent conductance of M/36 solution of a Weak monobasic acid is 6 mho cm2 equivalent and infinite dilution is 400 mho cm2 equivalent-1. Calculate the dissociation constant of this acid.
Solution:

Problems based on Faraday’s law

Question 1.
If 50 milli ampere of current is passed through copper coulometer for 60 minutes, calculate the amount of copper deposited.
Solution:
I = 50 mA = 50 × 10-3 A;
t = 60 min = 60 × 60 = 3600 seconds

Question 2.
0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minuets. What is the electro chemical equivalent of copper?
Solution:
m = 0.1978 g; I = 0.2 A;
t = 50 mins = 50 × 60 = 3000 sec
Z = ?
m = ZIt
∴ $$Z=\frac{m}{I t}$$
$$=\frac{0.1978}{0.2 \times 3000}$$
Z = 3.297 × 10-4gC-1

Question 3.
What current strength in amperes will be required to liberate 10 g of iodine from potassium iodide solution in one hour?
Solution:
l = ?m = 10g;
t = 1 hr = 60 × 60 = 3600 sec

Question 4.
An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weight of silver and iodine will be liberated while 1.25 g copper is being deposited?
Solution:

Question 5.
On passing a current of 1 ampere for 16 min 5 sec through 1 litre solution of CuCl2 all copper of the solution was deposited at cathode. Calculate the normality of CuCl2 solution.
Solution:
1 = 1 ampere
t = 16 min 5 sec = 16 * 60 + 5 = 965 sec
N = ? m = ZIt

Problems based on emf

Question 1.
The reaction $$\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Co}^{2+} \rightleftharpoons \mathrm{Co}_{(\mathrm{s})}+\mathrm{Zn}^{2+}$$ occurs in a cell Compute the standard emf of the cell.
Solution:
Given that
$$E^{0} z_{n} / Z n^{2+}$$ = +0.76 V and $$E_{\mathrm{Co} / \mathrm{Co}}^{2+}$$ = +0.28 V
Zn(s) + Co2+ Co(s) +Zn2+
Oxidation potential of Co ion is change into reduction potential
+0.28 V = -0.28 V
E° Cell = E° oxidation (2n/Zn2+) + E° reduction(Co2+/Co)
E° cell = + 0.76 + (-0.28)
E°cell = +0.48 V
This reaction is feasible.

Question 2.
What is the E°el, of the reaction
$$\mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Sn}_{(\mathrm{aq})}^{2+} \rightleftharpoons \mathrm{Cu}_{(s)}+\mathrm{Sn}_{(\mathrm{aq})}^{4+}$$ at 25°C, If the equilibrium constant for the reaction is 1 × 106.
Solution :
Keq= 1 × 106 T = 25° C + 273 = 298 K
R = 8.314 JK-1 mol-1
Cu2+ + 2e → Cu
Sn2+ → Sn4+ + 2e

Question 3.
Cu+(aq) | is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction.
$$2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(s)}$$
Calculate E° for the above reaction
$${ }^{\mathrm{E}^{\mathrm{O}}} \mathrm{Cu}^{2+} \mid \mathrm{Cu} = 0.34$$ and $$\mathrm{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}^{+}}^{\mathrm{o}}=0.15 \mathrm{~V}$$
Solution:

Equation (1) = Equation (2) + Equation (3)
ΔG°1 = ΔG°2 + ΔG°3
– 0.68 F= 0.15F—1E°3F
-0.68 = -0.15 – E°3
3 = 0.68 — 0.15 = 0.53 V
$$2 \mathrm{Cu}_{(\mathrm{aq})}^{+} \rightleftharpoons \mathrm{Cu}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}$$
Half reactions are
Cu+ → Cu2+ + e
(Oxidation at anode)
E = 0.15V ……………….(4)
(Reduction at cathode)
Cu+ + e→ Cu
red =E°3= 0.53V ………………..(5)
Overall reaction is
2Cu+ → Cu2+ + Cu
:.E° = (E°oxid)Anode + (E°red)cathode
= – 0.15 + 0.53
cell = 0.38V

Question 4.
For the redox reaction
$$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightleftharpoons \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}_{(\mathrm{s})}$$
if E°cell = 1.1V calculate Ecell of the reaction
Solution:
E°ell = 1.1V; n = 2; Ecell=?
[Cu2+] = 0.1 M; [Zn2+] = 1M

Question 5.
Calculate the emf of the cell Zn | Zi2+ (0.001 M | Ag+ (0.1 M) | Ag. The standard potential of Ag+ | Ag half cell is + 0.80 V and Zn | Zn2+ is + 0.76 V.
Anode : Zn → Zn2+ + 2e E° = + 0.76 V
Cathode : 2Ag+ + 2e → 2Ag E° = + 0.80 V
Cell reaction : Zn + 2Ag+ ⇌ Zn2+ + 2Ag
E° = + 0.76 + 0.80 = + 1.56 V

Ecell = 1.56 – 0.02955 log 10-1
Ecell = 1.56 + 0.02955 = 1.58955 V