NCERT Solutions Class 12th Chemistry Chapter – 3 Electrochemistry Question & Answer

NCERT Solutions Class 12th Chemistry Chapter – 3 Electrochemistry

NCERT Solutions Class 12th Chemistry Chapter – 3 Electrochemistry

?Chapter – 3?

✍Electrochemistry✍

?Question & Answer?

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

Answer: According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn, Fe, Cu.

Mg: Al: Zn: Fe: Cu

Q 2 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.

Answer: The reducing power increases with the lowering of reduction potential. In order of given standard electrode potential (increasing order) : K⁺/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag

Thus, in the order of reducing power, we can arrange the given metals as Ag< Hg < Cr < Mg < K

Q 3 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.

Answer: 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)

Q 4 Calculate the standard cell potentials of galvanic cell in which the followingreactions take place:
(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd
(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Calculate the ∆rGJ and equilibrium constant of the reactions.

Answer:  (i) E^ø _{Cr³⁺ / Cr}  = – 0.74 V

   E^ø _{Cd²⁺ / Cd}  = – 0.40 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

E^ø  = E^ø_R  – E^ø_L

     = -0.40 – (-0.74)

     = + 0.34 V

ΔrG^ø = –nFE^ø_cell

In the given equation,

n = 6

F = 96487 C mol ^{– 1}

E^ø_cell = +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^ø_R  – E^ø_L

       = 0.80 – 0.77

     = 0.03 L

Here, n = 1.

Then, ΔrG^ø = –nFE^ø_cell

= – 1 × 96487 C mol ^{– 1} × 0.03 V

= – 2894.61 J mol ^{– 1}

= – 2.89 kJ mol ^{– 1}

Again, ΔrG^ø = – 2.303 RT ln K

= 0.5073

K = antilog (0.5073)

= 3.2 (approximately)

Q 5 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)|Br–(0.010 M)|Br₂(l )||H⁺(0.030 M)| H₂(g) (1 bar)|Pt(s).

Answer: (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

Q 6 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.

Answer:

E^ø  = 1.104 V

We know that,

ΔrG^ø = –nFE^ø

= – 2 × 96487 × 1.04

= – 213043.296 J

= – 213.04 kJ

Q 7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer: Conductivity ($K$):
It is the conductance of unit cube of material. S.I unit is S/m. Common unit is S/cm.

The conductivity of an electrolytic solution always decreases with decrease in concentration that is on dilution. This is because with dilution, the degree of dissociation increases and the total number of current-carrying ions increases but the number of ions per unit volume decreases.

Molar conductivity $\Lambda$:
It is the ratio of the electrolytic conductivity k to the molar concentration C of the dissolved electrolyte.

Λ = k/C​

It is also defined as the conductance of a volume of solution containing 1 mole of dissolved electrolyte when placed between parallel electrodes 1 cm apart and large enough to contain between them all the solution.
The S.I unit of molar conductivity is Sm²/mol
The common unit of molar conductivity is Scm²/mol

The molar conductivity of strong and weak electrolytes increases with dilution (i.e, decrease in the concentration). This is because with dilution, the degree of dissociation increases and the total number of current-carrying ions increases.

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 with  √c  for strong and weak electrolytes is shown in the following plot:

Q  3.8: The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^–1. Calculate its molar conductivity

Answer: Given, k = 0.0248 S cm ^{– 1}

c = 0.20 M

Molar conductivity, A_m =  (k x 1000) / c

= 0.0248 x1000 / 0.20

124 Scm² mol ^{– 1}

Q 3.9: The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^–3 S cm^–1

Answer: Given,

Conductivity, k = 0.146 × 10⁻³ S cm−1

Resistance, R = 1500 Ω

Cell constant = k × R

= 0.146 × 10⁻³ × 1500

= 0.219 cm⁻¹

Q 10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0 Λ m.

Answer: The equation of plot is Λ_m ​= −51286C³ + 9416.C² − 602.5C+124.1

​

Q 3.11: Conductivity of 0.00241 M acetic acid is 7.896 × 10^–5 S cm^–1. Calculate its molar conductivity. If 0 Λ m for acetic acid is 390.5 S cm² mol^–1, what is its dissociation constant?

Answer: Given, K = 7.896 × 10 ^{– 5} S m ^{– 1}

c = 0.00241 mol L ^{– 1}

Then, molar conductivity, A_m = K/c

= 32.76S cm² mol ^{– 1}

Again, A_m^º = 390.5 S cm² 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}

Q 12 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 Mn²⁺?

Answer: (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

Q 13 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?

Answer: (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 = (3X40) / 27

= 4.44 F

Q 14 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?

Answer: (i) According to the question,

H₂O  → H₂ + ½ O₂

Now, we can write:

O^2- → ½ O₂ + 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

Q 3.15: 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?

Answer: 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.

Q 16 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?

Answer: 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

= 864/ 60 

= 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

Q 3.17: 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 Fe ³⁺ (aq)
(v) Br₂ (aq) and Fe²⁺ (aq).

Answer:

Since E^º for the overall reaction is positive, the reaction between Fe³⁺_(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 Fe³⁺_(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 for the overall reaction is positive, the reaction between Br_2(aq) and Fe²⁺_(aq) is feasible

Q  18 Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes.

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

The reaction with a higher value of  E^º 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 E^º 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 H_2(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 E^º 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 E^º is preferred. But due to the over-potential of oxygen, Cl ^– gets oxidized at the anode to produce Cl₂ gas.