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Page No 93: - Chapter 3 Electrochemistry Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 3.6:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6272/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7642a8a3.gif and https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6272/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_5b471a93.gif  for the reaction.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6272/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_3cbe101c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6272/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6272/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m322b2d38.gif = 1.104 V

We know that,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6272/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m1d70d9f9.gif

= −2 × 96487 × 1.04

= −213043.296 J

= −213.04 kJ

Question 3.7:

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

Answer:

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_5276534c.gif

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., https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_2f5adb50.gif

(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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m6ef48cac.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_2c7719e4.gif

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 https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m4318a684.gif withhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m4be7e146.gif for strong and weak electrolytes is shown in the following plot:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6279/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m7f124026.jpg

Question 3.8:

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

Answer:

Given,

κ = 0.0248 S cm−1

c = 0.20 M

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6281/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif  Molar conductivity, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6281/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_153127ed.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6281/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_1cad934.gif

= 124 Scm2mol−1

Question 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 conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1.

Answer:

Given,

Conductivity, κ = 0.146 × 10−3 S cm−1

Resistance, R = 1500 Ω

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6282/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif  Cell constant = κ × R

= 0.146 × 10−3 × 1500

= 0.219 cm−1

Question 3.10:

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

Concentration/M 0.001 0.010 0.020 0.050 0.100

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

Calculate https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m4318a684.gif for all concentrations and draw a plot between https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m5c1e4faa.gif and c½. Find the value ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_66981bef.gif .

Answer:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_718f2e5d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_5b09a7fe.gif

= 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_718f2e5d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_187f1ca3.gif

= 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_718f2e5d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m3b24da58.gif

= 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m42f207da.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_467cb6dc.gif

= 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_718f2e5d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m7fa187b3.gif

= 106.74 S cm2 mol−1

Now, we have the following data:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m14add555.gif

0.0316

0.1

0.1414

0.2236

0.3162

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m3757b090.gif

123.7

118.5

115.8

111.1

106.74

Since the line interruptshttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m4318a684.gif at 124.0 S cm2 mol−1https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_66981bef.gif = 124.0 S cm2 mol−1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6297/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_43d97fdd.jpg

Question 3.11:

Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_66981bef.gif for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

Answer:

Given, κ = 7.896 × 10−5 S m−1

c = 0.00241 mol L−1

Then, molar conductivity, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4c80ba4f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m1f081670.gif

= 32.76S cm2 mol−1

Again, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_66981bef.gif  = 390.5 S cm2 mol−1

Now, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_51386d51.gif

= 0.084

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif Dissociation constant, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m3fdaa82d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6300/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m53b0082c.gif

= 1.86 × 10−5 mol L−1

Question 3.12:

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 https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_47cc3b45.gif  to Mn2+.

Answer:

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m4edad211.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif  Required charge = 3 F

= 3 × 96487 C

= 289461 C

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m70d0504e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif Required charge = 2 F

= 2 × 96487 C

= 192974 C

(iii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_ma4d8667.gif

i.e.,https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7187dd05.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6301/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif  Required charge = 5 F

= 5 × 96487 C

= 482435 C

Question 3.13:

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,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6302/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_5f170bb2.gif

Electricity required to produce 40 g of calcium = 2 F

Therefore, electricity required to produce 20 g of calciumhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6302/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_130d7bd4.gif

= 1 F

(ii) According to the Question,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6302/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m29069595.gif

Electricity required to produce 27 g of Al = 3 F

Therefore, electricity required to produce 40 g of Al https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6302/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_2689df12.gif

= 4.44 F

Question 3.14:

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,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6304/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m1c39170.gif

Now, we can write:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6304/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m54435552.gif

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

= 2 × 96487 C

= 192974 C

(ii) According to the Question,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6304/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m2654bb76.gif

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

= 96487 C

Question 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

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6305/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif Charge = current × time

= 5 × 1200

= 6000 C

According to the reaction,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6305/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_78cff308.gif

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6305/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m65d6b40a.gif

= 1.825 g

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

Question 3.16:

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_2709dc3f.gif

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

Therefore, 1.45 g of Ag is deposited by =https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_3ce051ff.gif

= 1295.43 C

Given,

Current = 1.5 A

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4dd19828.gif Time https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m2aa8fc5a.gif

= 863.6 s

= 864 s

= 14.40 min

Again,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_1c8184f4.gif

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

Therefore, 1295.43 C of charge will deposit https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m41798b30.gif

= 0.426 g of Cu

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_68d48a7a.gif

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

Therefore, 1295.43 C of charge will deposit https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6307/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m306cdb09.gif

= 0.439 g of Zn

Question 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 Fe3+ (aq)

(v) Br(aq) and Fe2+ (aq).

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m5fe3d54b.gif

Sincehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a75cb5c.gif for the overall reaction is positive, the reaction between Fe3+(aq) and I(aq) is feasible.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_66c8498f.gif

Since https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a75cb5c.gif  for the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m5084326b.gif

Sincehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a75cb5c.gif  for the overall reaction is negative, the reaction between Fe3+(aq) and Br(aq) is not feasible.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_4b188a43.gif

Since https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a75cb5c.gif E for the overall reaction is negative, the reaction between Ag (s) and Fe3+(aq) is not feasible.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_m3b88045b.gif

Sincehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/17/262/6309/NCERT_17-11-08_Utpal_12_Chemistry_3_18_html_7a75cb5c.gif  for the overall reaction is positive, the reaction between Br2(aq) and Fe2+(aq) is feasible.

Also Read : Page-No-94:-Chapter-3-Electrochemistry-Exercise-Solutions-class-12-ncert-solutions-Chemistry

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