NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Chemistry Class 12

Created by: Mohit Rajpoot

Class 12 : NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The document NCERT Solutions - Electrochemistry Class 12 Notes | EduRev is a part of the Class 12 Course Chemistry Class 12.
All you need of Class 12 at this link: Class 12

NCERT QUESTION - Electrochemistry

Question 3.1: How would you determine the standard electrode potential of the system Mg2+  | Mg?

Ans:- The standard electrode potential of Mg2+  | Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), H2(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.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Here,NCERT Solutions - Electrochemistry Class 12 Notes | EduRev for the standard hydrogen electrode is zero.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Question 3.2: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans:- 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:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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., NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

(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.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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 NCERT Solutions - Electrochemistry Class 12 Notes | EduRevwithNCERT Solutions - Electrochemistry Class 12 Notes | EduRevfor strong and weak electrolytes is shown in the following plot:

Question 3.3: 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.

Ans:- (i) According to the question,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Now, we can write:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

= 2 × 96487 C

= 192974 C

(ii) According to the question,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

= 96487 C

Question 3.4:NCERT Solutions - Electrochemistry Class 12 Notes | EduRevCalculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Ans:- For hydrogen electrode,   NCERT Solutions - Electrochemistry Class 12 Notes | EduRev,  it is given that pH = 10

∴[H ] = 10−10 M

Now, using Nernst equation:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= −0.0591 log 1010

= −0.591 V

Question 3.5: Calculate the emf of the cell in which the following reaction takes place:

 NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Given that NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= 1.05 V

Ans:- Applying Nernst equation we have:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 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

Question 3.6: The cell in which the following reactions occurs:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev has NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= 0.236 V at 298 K.

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

Ans:- Here, n = 2, NCERT Solutions - Electrochemistry Class 12 Notes | EduRevT = 298 K

We know that:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= −2 × 96487 × 0.236

= −45541.864 J mol−1

= −45.54 kJ mol−1

Again, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev−2.303RT log Kc

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 7.981

Kc = Antilog (7.981)

= 9.57 × 107

Question 3.7: 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?

Ans:- Given,

Current = 5A

Time = 20 × 60 = 1200 s

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevCharge = current × time

= 5 × 1200  = 6000 C

According to the reaction,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C NCERT Solutions - Electrochemistry Class 12 Notes | EduRev = 1.825 g

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

Question 3.8: 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?

Ans:- According to the reaction:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

Therefore, 1.45 g of Ag is deposited by =NCERT Solutions - Electrochemistry Class 12 Notes | EduRev = 1295.43 C

Given,

Current = 1.5 A

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevTime NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 863.6 s = 864 s

= 14.40 min

Again,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

Therefore, 1295.43 C of charge will deposit NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 0.426 g of Cu

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

Therefore, 1295.43 C of charge will deposit NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 0.439 g of Zn

Question 3.9: Predict the products of electrolysis in each of the following:

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

(ii) An aqueous solution of AgNOwith platinum electrodes.

(iii) A dilute solution of H2SOwith platinum electrodes.

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

Ans:- (i) At cathode:

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The reaction with a higher value of NCERT Solutions - Electrochemistry Class 12 Notes | EduRevtakes place at the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:

The Ag anode is attacked by NCERT Solutions - Electrochemistry Class 12 Notes | EduRevions. 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.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The reaction with a higher value of NCERT Solutions - Electrochemistry Class 12 Notes | EduRevtakes 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 NCERT Solutions - Electrochemistry Class 12 Notes | EduRevions. Therefore, OH or NCERT Solutions - Electrochemistry Class 12 Notes | EduRevions can be oxidized at the anode. But OH ions having a lower discharge potential and get preference and decompose to liberate O2.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

At the anode, the following processes are possible.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

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.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The reaction with a higher value of NCERT Solutions - Electrochemistry Class 12 Notes | EduRevtakes 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.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

At the anode, the reaction with a lower value of NCERT Solutions - Electrochemistry Class 12 Notes | EduRevis preferred. But due to the over-potential of oxygen, Cl gets oxidized at the anode to produce Cl2 gas.

Question 3.10: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Ans:-  = 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 NCERT Solutions - Electrochemistry Class 12 Notes | EduRevnumber of electrons.

Then,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Hence, NCERT Solutions - Electrochemistry Class 12 Notes | EduRevnumber of electrons will flow through the wire.

Question 3.11: Suggest a list of metals that are extracted electrolytically.

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

Question 3.12: 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 NCERT Solutions - Electrochemistry Class 12 Notes | EduRevfor all concentrations and draw a plot between NCERT Solutions - Electrochemistry Class 12 Notes | EduRevand c½. Find the value ofNCERT Solutions - Electrochemistry Class 12 Notes | EduRev.

Ans:- Given,

κ = 1.237 × 10−2 S m−1, c = 0.001 M

Then, κ = 1.237 × 10−4 S cm−1, c½ = 0.0316 M1/2

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 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, c½ = 0.1 M1/2

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 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, c1/2 = 0.1414 M1/2

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 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, c1/2 = 0.2236 M1/2

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

 

= 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, c1/2 = 0.3162 M1/2

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRevNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 106.74 S cm2 mol−1

Now, we have the following data:

 

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Since the line interruptsNCERT Solutions - Electrochemistry Class 12 Notes | EduRevat 124.0 S cm2 mol−1, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= 124.0 S cm2 mol−1.

Question 3.13: Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Ans:- 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: NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

At cathode: NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The overall cell reaction is given by,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

When a battery is charged, the reverse of all these reactions takes place.

Hence, on charging, NCERT Solutions - Electrochemistry Class 12 Notes | EduRevpresent at the anode and cathode is converted into NCERT Solutions - Electrochemistry Class 12 Notes | EduRevand NCERT Solutions - Electrochemistry Class 12 Notes | EduRevrespectively.

Question 3.14: Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

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

Question 3.15: 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) Br(aq) and Fe2+ (aq).

Ans:-

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

SinceNCERT Solutions - Electrochemistry Class 12 Notes | EduRevfor the overall reaction is positive, the reaction between Fe3+ (aq) and I(aq) is feasible.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Since NCERT Solutions - Electrochemistry Class 12 Notes | EduRevfor the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

SinceNCERT Solutions - Electrochemistry Class 12 Notes | EduRev for the overall reaction is negative, the reaction between Fe3+ (aq) and Br(aq) is not feasible.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Since NCERT Solutions - Electrochemistry Class 12 Notes | EduRevE for the overall reaction is negative, the reaction between Ag (s) and Fe3+ (aq) is not feasible.

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

SinceNCERT Solutions - Electrochemistry Class 12 Notes | EduRev for the overall reaction is positive, the reaction between Br2(aq) and Fe2+ (aq) is feasible.

Question 3.16: 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

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

Mg, Al, Zn, Fe, Cu

Question 3.17: Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if NCERT Solutions - Electrochemistry Class 12 Notes | EduRevfor acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

Ans:- Given, κ = 7.896 × 10−5 S m−1

c = 0.00241 mol L−1

Then, molar conductivity, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 32.76S cm2 mol−1

Again, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= 390.5 S cm2 mol−1

Now, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 0.084

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevDissociation constant, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 1.86 × 10−5 mol L−1

Question 3.18: 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.

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

(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,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The reaction taking place at the cathode is given by,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Question 3.19: Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

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

(ii) Fe2+ (aq) Ag+ (aq) → Fe3+ (aq) + Ag(s)

Calculate the ΔrGθ and equilibrium constant of the reactions.

Ans:- (i) NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The galvanic cell of the given reaction is depicted as:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Now, the standard cell potential is

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

In the given equation,

n = 6

F = 96487 C mol−1

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= 0.34 V

Then, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= −6 × 96487 C mol−1 × 0.34 V

= −196833.48 CV mol−1

= −196833.48 J mol−1

= −196.83 kJ mol−1

Again,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev= −RT ln K

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 34.496

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevK = antilog (34.496)

= 3.13 × 1034

(ii)NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

The galvanic cell of the given reaction is depicted as:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Now, the standard cell potential is

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Here, n = 1.

Then, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= −1 × 96487 C mol−1 × 0.03 V

= −2894.61 J mol−1

= −2.89 kJ mol−1

Again, NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 0.5073

NCERT Solutions - Electrochemistry Class 12 Notes | EduRevK = antilog (0.5073)

= 3.2 (approximately)

Question 3.20: 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).

Ans:-

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 2.7 − 0.02955

= 2.67 V (approximately)

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 0.52865 V

= 0.53 V (approximately)

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

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 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:

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Question 3.21: 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.

Ans:- (i) According to the question,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calciumNCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 1 F

(ii) According to the question,

NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al NCERT Solutions - Electrochemistry Class 12 Notes | EduRev

= 4.44 F

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