NCERT Solutions Class 12 Chapter- 02 Electrochemistry Exercise 2.1

SimplyAcad has offered NCERT solutions of Electrochemistry Chapter 2 to help students get ready for the Chemistry board examination. The solutions will allow students to get a better understanding of the concepts related to the chapter.

The questions are prepared by the experts to test students on various grounds including knowledge of the basics of the chapter, its related functions and properties, etc. Therefore, these solutions will definitely benefit and make practice and revision their daily habit. The solutions provided here are in accordance with the latest 2024-25 guidelines. 

 

NCERT Solutions Class 12 Chapter- 02 Electrochemistry Exercise 2.1

Question 1. How would you determine the standard electrode potential of the system Mg2+ | Mg?  Electrochemistry Chapter 2

Solution :
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.

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

Eø  = EøR  – EøL

Here,EøR for the standard hydrogen electrode is zero.

∴Eø = 0 – EøL

= – EøL

Question 2. Can you store copper sulphate solutions in a zinc pot? Electrochemistry Chapter 2

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.

Question 3. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. Electrochemistry Chapter 2

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

Fe2+ → Fe3+  + e-1   ;   Eø = −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.

Question 4. Calculate the potential of a hydrogen electrode in contact with a solution whose pH is 10. \textbf

{Solution:} The concentration of \([H^+]\) ions can be determined using the pH: \[ \text{pH} = 10 \] \[ \text{We know, } \text{pH} = -\log_{10} [H^+] \] So, \[ [H^+] = 10^{-\text{pH}} \text{ M} \] \[ [H^+] = 10^{-10} \text{ M} \] The electrode reaction for the hydrogen electrode is: \[ \text{H}^+ + e^- \rightarrow \frac{1}{2} \text{H}_2 \] Using the Nernst equation, the potential \(E_{\text{cell}}\) of the hydrogen electrode can be calculated as: \[ E_{\text{cell}} = E^\circ_{\text{H}^+ / \frac{1}{2} \text{H}_2} – \frac{0.0591}{1} \log \frac{P_{\text{H}_2}}{[H^+]} \] Given: \[ E^\circ_{\text{H}^+ / \frac{1}{2} \text{H}_2} = 0 \text{ V} \quad (\text{standard hydrogen electrode potential}) \] \[ P_{\text{H}_2} = 1 \text{ bar} \] Number of electrons transferred (\(n\)) = 1 \[ E_{\text{cell}} = 0 – 0.0591 \times \log \left(\frac{1}{10^{-10}}\right) \] \[ E_{\text{cell}} = – 0.0591 \times 10 \] \[ E_{\text{cell}} = – 0.591 \text{ V} \] \textbf{Final answer:} The potential of the hydrogen electrode is \(-0.591 \text{ V}\). 

Question 5. 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 Eøcell = 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

Question 6. The cell in which the following reactions occurs:

has Eøcell =  0.236 V at 298 K.  

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. Electrochemistry Chapter 2

Solution :
Here, n = 2, Eøcell = 0.236 T = 298 K

We know that:

Question 7. Why does the conductivity of a solution decrease with dilution? Electrochemistry Chapter 2

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.

Question 8. Suggest a way to determine the Λ∘m value of water. Electrochemistry Chapter 2

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.

Question 9. 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:

Question 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? Electrochemistry Chapter 2

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.

Question 11. Suggest a list of metals that are extracted electrolytically. Electrochemistry Chapter 2

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

Question 12. What is the quantity of electricity in coulombs needed to reduce 1 mol of

cr2 O72? Consider the reaction:

Cr2 O72- + 14H+ + 6e– → Cr3+ + 8H2O

Solution :
The given reaction is as follows:

Cr2 O72- + 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

Question 13. Write the chemistry of recharging a lead storage battery, highlighting all the materials that are evolved during recharging. \textbf

{Solution:} In a lead storage battery, the anode is made of porous lead (\textbf{Pb}), and the cathode is made of lead dioxide (\textbf{PbO$_2$}). During the discharging process, the chemical reactions are as follows: At the anode: \[ \text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2e^- \] At the cathode: \[ \text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O} \] The overall discharging reaction is: \[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \] During the recharging process, the reactions are reversed. The sulfate ions (\textbf{SO$_4^{2-}$}) are released from the lead sulfate (\textbf{PbSO$_4$}) at both electrodes, and lead and lead dioxide are reformed: Overall recharging reaction: \[ 2\text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \text{Pb}(\text{s}) + \text{PbO}_2(\text{s}) + 4\text{H}^+ + 2\text{SO}_4^{2-} \] The materials involved during the recharging process include: – **Lead (Pb)** at the anode – **Lead dioxide (PbO$_2$)** at the cathode – **Hydrogen ions (H$^+$)** – **Sulfate ions (SO$_4^{2-}$)** These reactions ensure the battery’s electrodes are restored to their original state, ready for another discharge cycle. \end{document}

Question 14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells. Electrochemistry Chapter 2

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

Question 15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell. Electrochemistry Chapter 2

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 H2CO3, 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.

Question 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

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

Question 17. 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

Question 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.

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)

Question 19. Calculate the standard cell potentials of a galvanic cell in which the following reactions take place: (i) \[ 2\text{Cr (s)} + 3\text{Cd}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{Cd (s)} \] (ii) \[ \text{Fe}^{2+}(\text{aq}) + \text{Ag}^{+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag (s)} \] Calculate the $\Delta_rG^\circ$ and equilibrium constant for the reactions. \textbf

{Solution:} Given: \[ E^\circ_{\text{cell}}(\text{Cr}^{3+}/\text{Cr}) = -0.74 \text{ V} \] \[ E^\circ_{\text{cell}}(\text{Cd}^{2+}/\text{Cd}) = -0.40 \text{ V} \] (a) Calculation of standard cell potential ($E^\circ_{\text{cell}}$) \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = (-0.40) – (-0.74) = +0.34 \text{ V} \] (b) Calculation of $\Delta_rG^\circ$ \[ \Delta_rG^\circ = -nFE^\circ_{\text{cell}} \] \[ \Delta_rG^\circ = -(6 \text{ mol}) \times (96500 \text{ C mol}^{-1}) \times (0.34 \text{ V}) \] \[ \Delta_rG^\circ = -196860 \text{ C V} = -196860 \text{ J} = -196.86 \text{ kJ} \] (c) Calculation of equilibrium constant ($K_c$) \[ \Delta_rG^\circ = -2.303 \, RT \, \log K_c \] \[ \log K_c = \frac{\Delta_rG^\circ}{-2.303 \, RT} = \frac{-(-196860 \text{ J})}{2.303 \times 8.314 \text{ J K}^{-1} \times 298 \text{ K}} \] \[ \log K_c = 34.501 \] \[ K_c = \text{Antilog } (34.501) \approx 3.17 \times 10^{34} \] For the second reaction: Given: \[ E^\circ_{\text{cell}}(\text{Ag}^+/\text{Ag}) = 0.80 \text{ V} \] \[ E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77 \text{ V} \] (a) Calculation of standard cell potential ($E^\circ_{\text{cell}}$) \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = (0.80) – (0.77) \text{ V} = 0.03 \text{ V} \] (b) Calculation of $\Delta_rG^\circ$ \[ \Delta_rG^\circ = -nFE^\circ_{\text{cell}} \] \[ \Delta_rG^\circ = -(1 \text{ mol}) \times (96500 \text{ C mol}^{-1}) \times (0.03 \text{ V}) \] \[ \Delta_rG^\circ = -2.895 \text{ kJ} \] (c) Calculation of equilibrium constant ($K_c$) \[ \Delta_rG^\circ = -2.303 \, RT \, \log K_c \] \[ \log K_c = \frac{\Delta_rG^\circ}{-2.303 \, RT} = \frac{-(-2895 \text{ J})}{2.303 \times 8.314 \text{ J K}^{-1} \times 298 \text{ K}} \] \[ \log K_c = 0.5074 \] \[ K_c = \text{Antilog } (0.5074) \approx 3.22 \] Thus, the equilibrium constant $K_c$ for the second reaction is approximately 3.22. 

Question 20. {Question:} Write the Nernst equation and emf of the following cell at 298 K: \[ \text{B) Fe(s) | Fe}^{2+}(0.001 \text{ M}) \parallel \text{H}^{+}(1 \text{ M}) | \text{H}_2(\text{g}, 1 \text{ bar}) | \text{Pt(s)} \] \textbf

{Solution:} Given cell notation: \[ \text{Fe(s) | Fe}^{2+}(0.001 \text{ M}) \parallel \text{H}^{+}(1 \text{ M}) | \text{H}_2(\text{g}, 1 \text{ bar}) | \text{Pt(s)} \] According to the given cell, the Fe electrode acts as the anode and the hydrogen electrode acts as the cathode. \textbf{Anode Reaction:} The electrode at which oxidation takes place is called the anode. It is a negatively charged electrode. \[ \text{Fe(s)} \rightarrow \text{Fe}^{2+}(\text{aq}) + 2\text{e}^{-} \] \textbf{Cathode Reaction:} The electrode at which reduction takes place is called the cathode. It is a positively charged electrode. \[ 2\text{H}^{+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{H}_2(\text{g}) \] \textbf{Overall Cell Reaction:} \[ \text{Fe(s)} + 2\text{H}^{+}(\text{aq}) \rightarrow \text{Fe}^{2+}(\text{aq}) + \text{H}_2(\text{g}) \] \textbf{Standard Cell Potential ($E^\circ_{\text{cell}}$)} From the electrochemical series: – The standard reduction potential of $\text{H}^{+}|\text{H}_2(\text{g})$ is $E^\circ = 0.0$ V. – The standard reduction potential of $\text{Fe}^{2+}|\text{Fe(s)}$ is $E^\circ = -0.44$ V. Since the reduction potential of Fe is lower than that of hydrogen, Fe gets oxidized, and $\text{H}^{+}$ gets reduced. Thus, Fe acts as the anode and hydrogen as the cathode. The standard cell potential of the galvanic cell can be represented as: \[ \text{Fe(s) | Fe}^{2+}(0.001 \text{ M}) \parallel \text{H}^{+}(1 \text{ M}) | \text{H}_2(\text{g}, 1 \text{ bar}) | \text{Pt(s)} \] The standard cell potential is calculated as: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = E^\circ_{\text{H}^+/\text{H}_2(\text{g})} – E^\circ_{\text{Fe}^{2+}/\text{Fe(s)}} \] \[ E^\circ_{\text{cell}} = 0 – (-0.44) = 0.44 \text{ V} \] \textbf{Nernst Equation:} The Nernst equation for the galvanic cell is: \[ E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n} \log Q \] For the cell reaction: \[ \text{Fe(s)} + 2\text{H}^{+}(\text{aq}) \rightarrow \text{Fe}^{2+}(\text{aq}) + \text{H}_2(\text{g}) \] \[ E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n} \log \frac{[\text{Fe}^{2+}]P_{\text{H}_2}}{[\text{H}^{+}]^2} \] \textbf{EMF of the Cell:} Using the Nernst equation, the EMF of the cell can be calculated: \[ E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n} \log \frac{[\text{Fe}^{2+}]P_{\text{H}_2}}{[\text{H}^{+}]^2} \] Given values: – $E^\circ_{\text{cell}} = 0.44$ V – $[\text{Fe}^{2+}] = 0.001$ M – $[\text{H}^{+}] = 1$ M – $P_{\text{H}_2} = 1$ bar – $n = 2$ \[ E_{\text{cell}} = 0.44 – \frac{0.0591}{2} \log \left(\frac{0.001 \times 1}{1^2}\right) \] \[ E_{\text{cell}} = 0.44 – \frac{0.0591}{2} \log (10^{-3}) \] \[ E_{\text{cell}} = 0.44 + 3 \times 0.02955 \approx 0.53 \text{ V} \] \textbf{Final Answer:} (i) The Nernst equation for the given cell: \[ E_{\text{cell}} = E^\circ_{\text{cell}} \frac{0.0591}{n} \log \frac{[\text{Fe}^{2+}]P_{\text{H}_2}}{[\text{H}^{+}]^2} \] (ii) EMF of the cell is approximately $0.53$ V. 

Question 21. 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 :

∵ Eø  = 1.104 V

We know that,

ΔrGø = -nFEø

= −2 × 96487 × 1.04

= −213043.296 J

= −213.04 kJ

Question 22. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration. \textbf

{Solution:} \textbf{Specific Conductance/Conductivity ($\kappa$)} The inverse of resistivity ($\rho$) is called conductivity or specific conductance. It is defined as the reciprocal of specific resistance and is represented by the symbol $\kappa$. \[ \kappa = \frac{1}{\rho} \] Since, we know, \[ R = \rho \times \frac{l}{A} \] \[ \frac{1}{\rho} = \frac{1}{R} \times \frac{l}{A} \] \[ \therefore \kappa = G \times \text{Cell Constant} \] Where, \[ \text{Cell Constant} = \frac{l}{A} \] \[ G = \text{Conductance} \] \textbf{Specific Conductance} is defined as the conductance of a solution of 1 cm in length and 1 sq. cm area of cross-section. The SI unit of conductivity is $S \, m^{-1}$. The unit of cell constant is $cm^{-1}$ or $m^{-1}$. \textbf{Molar Conductivity ($\Lambda_m$)} Molar conductivity is defined as the conductance of all the ions produced by one mole of electrolyte present in a given volume of solution. Alternatively, it can be defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte. \[ \Lambda_m = \kappa \times V \] Where $V$ is the volume of solution containing 1 mole of electrolyte. If the concentration of solution is $M$ (mol $L^{-1}$), then: \[ \Lambda_m = \kappa \times \frac{1000}{M} \] Where $M$ is the molarity. The SI units of molar conductivity are $\Omega^{-1} \, cm^2 \, mol^{-1}$ or $S \, cm^2 \, mol^{-1}$. \textbf{Variation of Conductivity with Concentration} Conductivity always decreases with the 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 the decrease in concentration. \textbf{Variation of Molar Conductivity with Concentration} Molar conductivity increases with the decrease in concentration. This is because, with dilution, the degree of dissociation increases, and the total number of current-carrying ions increases. \textbf{Variation of Molar Conductivity of Strong Electrolytes} For strong electrolytes, molar conductivity decreases slowly with concentration as the ions are fully dissociated, and the increase in molar conductivity with dilution is due to the decrease in inter-ionic interactions. \textbf{Variation of Molar Conductivity of Weak Electrolytes} For weak electrolytes, molar conductivity increases more rapidly with dilution. This is because the weak electrolyte is not fully dissociated at higher concentrations, and with dilution, more ions are produced, increasing the molar conductivity. 

Question 23. 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, Am  = (k × 1000) / c

= 0.0248 × 1000 /0.20

= 124 Scm2mol−1

Question 24. 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

Question 25. 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 λm for all concentrations and draw a plot between λm and c½. Find the value of λºm.

Solution :
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

= 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

 

= 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

= 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

 

= 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

 

= 106.74 S cm2 mol−1

Now, we have the following data:

C1/2  / M1/2	0.0316	0.1	0.1414	0.2236	0.3162
λm (Scm2 mol-1)	123.7	118.5	115.8	111.1	106.74

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

Question 26. The conductivity of 0.00241 M acetic acid is $7.896 \times 10^{-5} \, \text{S} \, \text{cm}^{-1}$. Calculate its molar conductivity. If $\lambda_{m}^{0}$ for acetic acid is 390.5 $\text{S} \, \text{cm}^{2} \, \text{mol}^{-1}$, what is its dissociation constant? \textbf

{Solution:} \textbf{Given:} Conductivity, $K = 7.896 \times 10^{-5} \, \text{S} \, \text{cm}^{-1}$ \\ Concentration, $C = 0.00241 \, \text{mol} \, \text{L}^{-1}$ \textbf{Molar Conductivity} $\Lambda_m$: \[ \Lambda_m = \frac{K \times 1000}{C} = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} = 32.76 \, \text{S} \, \text{cm}^{2} \, \text{mol}^{-1} \] \textbf{Degree of Dissociation} $\alpha$: \[ \alpha = \frac{\Lambda_m}{\lambda_{m}^{0}} = \frac{32.76}{390.5} = 0.084 \] \textbf{Dissociation Constant} $K_a$: \[ K_a = \frac{C \alpha^2}{1 – \alpha} = \frac{0.00241 \times (0.084)^2}{1 – 0.084} = 1.86 \times 10^{-5} \, \text{mol} \, \text{L}^{-1} \] Thus, the dissociation constant of acetic acid is $1.86 \times 10^{-5} \, \text{mol} \, \text{L}^{-1}$.

Question 27. 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+.

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

Question 28. i) How much electricity in terms of Faraday is required to produce 20.0 g of Ca from molten \text{CaCl}_2? (ii) How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten \text{Al}_2\text{O}_3? \textbf

{Solution} \textbf{(i) Calculation for Calcium} The electrode reaction for the production of calcium is: \[ \text{Ca}^{2+} (\text{aq}) + 2e^- \rightarrow \text{Ca} (\text{s}) \] To produce 40 g of Ca from molten \text{CaCl}_2, the electricity required is 2 Faradays (2F), because the molar mass of calcium is 40 g/mol and 2 moles of electrons are needed per mole of Ca. To produce 20 g of Ca: \[ \text{Electricity required} = \frac{2F \times 20 \text{ g}}{40 \text{ g}} = 1F \] \textbf{(ii) Calculation for Aluminum} The electrode reaction for the production of aluminum is: \[ \text{Al}_2\text{O}_3 (\text{l}) + 6e^- \rightarrow 2\text{Al} (\text{s}) + 3\text{O}^{2-} \] To produce 54 g of Al (from 2 moles of Al atoms), the electricity required is 6 Faradays (6F), because the molar mass of aluminum is 27 g/mol. To produce 40 g of Al: \[ \text{Electricity required} = \frac{6F \times 40 \text{ g}}{54 \text{ g}} \approx 4.44F \] Thus, the electricity required to produce 20 g of Ca is 1 Faraday and for 40 g of Al is approximately 4.44 Faradays. \end{document}

Question 29. 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:

O2-  → 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

Question 30. 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.

Question 31. Three electrolytic cells \textbf{A}, \textbf{B}, and \textbf{C} containing solutions of ZnSO$_4$, AgNO$_3$, and CuSO$_4$, respectively, are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell \textbf{B}. How long did the current flow? What mass of copper and zinc were deposited? \textbf

{Solution} \textbf{Cell B (AgNO$_3$):} At the cathode: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \] 1 mole (108 g) of Ag is deposited by 96500 C. For 1.45 g of Ag, the charge required is: \[ Q = \frac{96500 \times 1.45}{108} = 1295.6 \text{ C} \] The current flow time \(t\) is given by: \[ Q = It \] \[ 1295.6 = 1.5 \times t \] \[ t = \frac{1295.6}{1.5} = 864 \text{ s} \] \textbf{Cell A (ZnSO$_4$):} At the cathode: \[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \] 2 moles of electrons (2 × 96500 C) deposit 1 mole (65.3 g) of zinc. For 1295.6 C of electricity, the mass of zinc deposited is: \[ \text{Mass of Zn} = \frac{65.3 \times 1295.6}{2 \times 96500} = 0.438 \text{ g} \] \textbf{Cell C (CuSO$_4$):} At the cathode: \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \] 2 moles of electrons (2 × 96500 C) deposit 1 mole (63.5 g) of copper. For 1295.6 C of electricity, the mass of copper deposited is: \[ \text{Mass of Cu} = \frac{63.5 \times 1295.6}{2 \times 96500} = 0.426 \text{ g} \] Thus, the time for which the current flowed is 864 seconds, and the masses of zinc and copper deposited are 0.438 g and 0.426 g, respectively.

Question 32. 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 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 Eº for the overall reaction is positive, the reaction between Br2(aq) and Fe2+(aq) is feasible.

Question 33. 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.