Answer the following Questions.

Question1. How would you determine the standard electrode potential of the system Mg^{2+ | Mg?}

Solution :
The standard electrode potential of Mg²⁺ | Mg can be measured with respect to the standard hydrogen electrode, represented by Pt_(s), H_2(g) (1 atm) | H⁺_(aq)(1 M).

A cell, consisting of Mg | MgSO₄ (aq 1 M) as the anode and the standard hydrogen 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^ø  = E^øR  – E^øL

Here,E^øR for the standard hydrogen electrode is zero.

∴E^ø = 0 – E^øL

 = – E^øL

Question2. Can you store copper sulphate solutions in a zinc pot?

Solution :
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.

Question3. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Solution :
Substances that are stronger oxidising agents than ferrous ions can oxidise ferrous ions.

Fe²⁺ → Fe³⁺  + e^-1   ;   E^ø = −0.77 V

This implies that the substances having higher reduction potentials than +0.77 V can oxidise ferrous ions to ferric ions. Three substances that can do so are F₂, Cl₂, and O₂.

Question4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Solution :
For hydrogen electrode, , it is given that pH = 10

∴[H⁺] = 10⁻¹⁰ M

Now, using Nernst equation:

= −0.0591 log 10¹⁰

= −0.591 V

Question5. Calculate the emf of the cell in which the following reaction takes place:

Ni_(s) + 2Ag + (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s)
 

Given that E^øcell = 1.05 V

Solution :
Applying Nernst equation we have:

= 1.05 − 0.02955 log 4 × 10⁴

= 1.05 − 0.02955 (log 10000 + log 4)

= 1.05 − 0.02955 (4 + 0.6021)

= 0.914 V

Question6. The cell in which the following reactions occurs:

has E^øcell =  0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution :
Here, n = 2, E^øcell = 0.236 T = 298 K

We know that:

Question7. Why does the conductivity of a solution decrease with dilution?

Solution :
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.

8. Suggest a way to determine the Λ^∘m value of water.

Solution :
Applying Kohlrausch’s law of independent migration of ions, the  value of water can be determined as follows:

Λ°m(H₂O)  = Λ°m(HCl) + Λ°m(NaOH) – Λ°m(NaCl)

Hence, by knowing the Λ°m values of HCl, NaOH, and NaCl, the Λ°m value of water can be determined.

Question9. The molar conductivity of 0.025 mol L⁻¹ methanoic acid is 

46.1 S cm2 mol^−1.

Calculate its degree of dissociation and dissociation constant. Given λ ^°(H+)

= 349.6 S cm² mol⁻¹ and λ °(HCOO⁻) = 54.6 S cm² mol

Solution :
C = 0.025 mol L⁻¹

Now, degree of dissociation: 

Thus, dissociation constant: 

Question10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Solution :
I = 0.5 A

t = 2 hours = 2 × 60 × 60 s = 7200 s

Thus, Q = It 

= 0.5 A × 7200 s

= 3600 C

We know that 96487C = 6.023 X 10²³number of electrons. 

Then, 

Hence, 2.25 X 10²²number of electrons will flow through the wire.

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

Solution :
Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.

Question12. What is the quantity of electricity in coulombs needed to reduce 1 mol of

cr₂ O₇²? Consider the reaction:

Cr₂ O₇^2- + 14H+ + 6e^– → Cr³+ + 8H₂O

Solution :
The given reaction is as follows:

Cr₂ O₇^2- + 14H+ + 6e^– → Cr³+ + 8H₂O

Therefore, to reduce 1 mole of cr₂ O₇², the required quantity of electricity will be: 

=6 F

= 6 × 96487 C

= 578922 C

Question13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Solution :
A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO₂) as the cathode, and a 38% solution of sulphuric acid (H₂SO₄) as an electrolyte.

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

At anode: Pb_(s) + SO²-4(aq)  → PbSO₄(s) + 2e-

At cathode:

The overall cell reaction is given by,

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.

Question14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

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

Question15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Solution :
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, 

Fe_(s) → Fe²⁺ (aq) + 2e-

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 H₂CO_3, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.

The reaction corresponding at the cathode is given by,

The overall reaction is: 

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide(Fe₂O₃, x)H₂Oi.e., rust.

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

Question16. 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

Solution :
The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Al, Zn, Fe, Cu

Question17. Given the standard electrode potentials,

K^+/K = −2.93V, Ag⁺/Ag = 0.80V,

Hg²⁺/Hg = 0.79V

Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = − 0.74V

Arrange these metals in their increasing order of reducing power.

Solution :
The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K^+/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

Question18. Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place. Further show:

(i) Which of the electrode is negatively charged?

(ii) The carriers of the current in the cell.

(iii) Individual reaction at each electrode.

Solution :
The galvanic cell in which the given reaction takes place is depicted as:

(i) Zn electrode (anode) is negatively charged.

(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.

(iii) The reaction taking place at the anode is given by,

Zn_(s)  →   Zn²⁺(aq) + 2e-

The reaction taking place at the cathode is given by,

Ag⁺(aq)  + e^–  →  Ag(s)

Question19. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd

(ii) Fe^{2+(aq) + Ag+}(aq) → Fe³⁺(aq) + Ag(s)

Calculate the Δ_rG^θ and equilibrium constant of the reactions.

Solution :
(i) E^ø Cr³⁺ / Cr  = – 0.74 V

   Eø Cd²⁺ / Cd  = – 0.40 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

E^ø  = E^øR  – E^øL

     = -0.40 – (-0.74)

     = + 0.34 V

ΔrG^ø = -nFE^øcell

In the given equation, 

n = 6

F = 96487 C mol⁻¹

E^øcell = +0.34 V

Then, ΔrG^ø = −6 × 96487 C mol⁻¹ × 0.34 V

= −196833.48 CV mol⁻¹

= −196833.48 J mol⁻¹

= −196.83 kJ mol⁻¹

Again,

ΔrG^ø = −RT ln K

= 34.496

⇒  K = antilog (34.496)

= 3.13 × 10³⁴

(ii) E^ø Fe3+ / Fe²⁺  = – 0.77 V

   E^ø Ag⁺ / Ag  = – 0.80 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

E^ø  = E^øR  –  E^øL

       = 0.80 – 0.77

     = 0.03 L

Here, n = 1.

Then, ΔrG^ø = -nFE^ø

= −1 × 96487 C mol⁻¹ × 0.03 V

= −2894.61 J mol⁻¹

= −2.89 kJ mol⁻¹

Again, ΔrG^{ø =  – 2.303 RT In K}

= 0.5073

∴ K = antilog (0.5073)

= 3.2 (approximately)

Question20. Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s) | Mg²⁺(0.001M) || Cu²+(0.0001 M) | Cu(s)

(ii) Fe(s) | Fe²⁺(0.001M) || H+(1M)|H₂(g)(1bar) | Pt(s)

(iii) Sn(s) | Sn²⁺(0.050 M) || ^H+(0.020 M) | H₂(g) (1 bar) | Pt(s)

(iv) Pt(s) | Br₂(l) | Br⁻(0.010 M) || ^H+(0.030 M) | H₂(g) (1 bar) | Pt(s).

Solution :
(i) For the given reaction, the Nernst equation can be given as:

= 2.7 − 0.02955

= 2.67 V (approximately)

(ii) For the given reaction, the Nernst equation can be given as:

= 0.52865 V

= 0.53 V (approximately)

(iii) For the given reaction, the Nernst equation can be given as: 

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

(iv) For the given reaction, the Nernst equation can be given as:

Question21. In the button cells widely used in watches and other devices the following reaction takes place:

Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH−(aq)

Determine and for the reaction.

Solution :

∵ E^ø  = 1.104 V

We know that,

ΔrG^ø = -nFE^ø

= −2 × 96487 × 1.04

= −213043.296 J

= −213.04 kJ

Question22. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Solution :
Conductivity of a solution is defined 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 specific conductance. It is represented by the symbolκ. If ρ is resistivity, then we can write:

k = 1 / p

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., 

(Since a = 1, l = 1)

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 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte). 

A_m = kv

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 A_m  withfor strong and weak electrolytes is shown in the following plot:

Question23. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm⁻¹. Calculate its molar conductivity.

Solution :
Given, 

κ = 0.0248 S cm⁻¹

c = 0.20 M

⇒ Molar conductivity, A_m  = (k × 1000) / c

= 0.0248 × 1000 /0.20

= 124 Scm²mol⁻¹

Question24. 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 × 10⁻³ S cm^−1.

Solution :
Given, 

Conductivity, κ = 0.146 × 10⁻³ S cm⁻¹

Resistance, R = 1500 Ω

⇒ Cell constant = κ × R

= 0.146 × 10^{−3 × 1500}

= 0.219 cm⁻¹

Question25. 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

10² × κ/S m⁻¹ 1.237 11.85 23.15 55.53 106.74

Calculate λ_m for all concentrations and draw a plot between λ_m and c½. Find the value of λº_m.

Solution :
Given, 

κ = 1.237 × 10^{−2 S m−1,} c = 0.001 M

Then, κ = 1.237 × 10⁻⁴ S cm^−1, c^½ = 0.0316 M^1/2

= 123.7 S cm² mol⁻¹

Given, 

κ = 11.85 × 10⁻² S m⁻¹, c = 0.010M

Then, κ = 11.85 × 10⁻⁴ S cm^−1, c^½ = 0.1 M^1/2

= 118.5 S cm² mol⁻¹

Given, 

κ = 23.15 × 10^{−2 S m−1}, c = 0.020 M

Then, κ = 23.15 × 10⁻⁴ S cm⁻¹, c^1/2 = 0.1414 M^1/2

= 115.8 S cm² mol⁻¹

Given, 

κ = 55.53 × 10⁻² S m^−1, c = 0.050 M

Then, κ = 55.53 × 10⁻⁴ S cm⁻¹, c^1/2 = 0.2236 M^1/2

= 111.1 1 S cm² mol−¹

Given, 

κ = 106.74 × 10^{−2 S} m⁻¹, c = 0.100 M

Then, κ = 106.74 × 10⁻⁴ S cm⁻¹, c^1/2 = 0.3162 M^1/2

= 106.74 S cm² mol⁻¹

Now, we have the following data:

Since the line interrupts λm at 124.0 S cm² mol^−1, λº_m= 124.0 S cm² mol⁻¹.

Question26. Conductivity of 0.00241 M acetic acid is 7.896 × 10^{−5 S} cm⁻¹. Calculate its molar conductivity and if A_mº for acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?

Solution :
Given, κ = 7.896 × 10⁻⁵ S m⁻¹

c = 0.00241 mol L⁻¹

Then, molar conductivity, A_{m  =  k / c}

= 32.76S cm^{2 mol−1}

Again, A_mº = 390.5 S cm^{2 mol−1}

Now,

= 0.084

⇒ Dissociation constant, 

⇒ (0.00241 mol L^-1)(0.084)²  / (1-0.084)

= 1.86 × 10⁻⁵ mol L⁻¹

Question27. How much charge is required for the following reductions:

(i) 1 mol of Al³⁺ to Al.

(ii) 1 mol of Cu²⁺ to Cu.

(iii) 1 mol of MnO₄^–   to Mn^2+.

Solution :
(i)

Al³⁺  +  3e^–  → Al

⇒ Required charge = 3 F

= 3 × 96487 C

= 289461 C

(ii) Cu²⁺ +  2e^–  →  Cu

⇒ Required charge = 2 F

= 2 × 96487 C

= 192974 C

(iii) MnO₄^–  → Mn²⁺

i.e.,

Mn⁷⁺  + 5e^–  →  Mn²⁺

⇒ Required charge = 5 F

= 5 × 96487 C

= 482435 C

Question28. How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl₂.

(ii) 40.0 g of Al from molten Al₂O₃.

Solution :
(i) According to the question,

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calcium(2 × 20) /40

= 1 F

(ii) According to the question,

Electricity required to produce 27 g of Al = 3 F

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

= 4.44 F

Question29. How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H₂O to O₂.

(ii) 1 mol of FeO to Fe₂O₃.

Solution :
(i) According to the question,

H₂O  → H₂ + ½O₂.

Now, we can write:

O^2-  → ½O₂ + 2e^–

Electricity required for the oxidation of 1 mol of H₂O to O₂ = 2 F

= 2 × 96487 C

= 192974 C

(ii) According to the question,

Fe²⁺ → Fe³⁺ + e^-1

Electricity required for the oxidation of 1 mol of FeO to Fe₂O₃ = 1 F

= 96487 C

Question30. A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Solution :
Given, 

Current = 5A

Time = 20 × 60 = 1200 s

∵ Charge = current × time

= 5 × 1200

= 6000 C

According to the reaction,

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C  = (58.71 X 6000) / (2 X 96487) g

= 1.825 g

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

Question31. Three electrolytic cells A,B,C containing solutions of ZnSO₄, AgNO₃ 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 flow? What mass of copper and zinc were deposited?

Solution :
According to the reaction: 

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by = 96487 X 1.45 / 108 C

= 1295.43 C

Given, 

Current = 1.5 A

⇒ Time  = 1295.43 /1.5s

= 863.6 s

= 864 s

= 14.40 min

Again,

i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit  = (63.5×1295.43) / (2×96487) g

= 0.426 g of Cu

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

Therefore, 1295.43 C of charge will deposit

= (65.4×1295.43) / (2×96487) g

= 0.439 g of Zn

Question32. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe³⁺(aq) and I⁻(aq)

(ii) Ag⁺ (aq) and Cu(s)

(iii) Fe³⁺ (aq) ^{and Br−} (aq)

(iv) Ag(s) and Fe³⁺ (aq)

(v) Br₂ (aq) and Fe²⁺ (aq).

Solution :

Since Eº  for the overall reaction is positive, the reaction between Fe^3+₍aq) and I⁻(aq) is feasible.

Since Eº for the overall reaction is positive, the reaction between Ag⁺ _(aq) and Cu_(s) is feasible.

Since Eº for the overall reaction is negative, the reaction between Fe^3+_(aq) and Br⁻_(aq) is not feasible.

Since Eº for the overall reaction is negative, the reaction between Ag _(s) and Fe³⁺ _(aq) is not feasible.

Since Eº for the overall reaction is positive, the reaction between Br_2(aq) and Fe^2+_(aq) is feasible.

Question33. Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes.

(ii) An aqueous solution of AgNO₃with platinum electrodes.

(iii) A dilute solution of H₂SO₄with platinum electrodes.

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Solution :
(i) At cathode:

The following reduction reactions compete to take place at the cathode.

The reaction with a higher value of Eº takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode: 

The Ag anode is attacked by NO₃^–  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 Eº 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 by NO₃^–  ions. Therefore, OH⁻ or NO₃^–
ions can be oxidized at the anode. But OH⁻ ions having a lower discharge potential and get preference and decompose to liberate O₂.

OH^–  → OH + e^–

4OH^–  → 2H₂O +O₂

(iii) At the cathode, the following reduction reaction occurs to produce H₂ gas.

H+ (aq) + e-  → ½ H₂(g)

At the anode, the following processes are possible.

For dilute sulphuric acid, reaction (i) is preferred to produce O₂ 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  Eº 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 with a lower value of  Eº is preferred. But due to the over-potential of oxygen, Cl⁻ gets oxidized at the anode to produce Cl₂ gas.