Electrochemistry Class 12 NCERT solutions – Chemistry Chapter 2

SimplyAcad has offered Electrochemistry Class 12 NCERT solutions 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 benefit and make practice and revise their daily habit. The solutions provided here are following the latest 2024-25 guidelines. 

Electrochemistry Class 12 NCERT solutions Chapter 2 Questions 1 to 10

Question 1: How would you determine the standard electrode potential of the system Mg²⁺ | Mg?

Solution:
The standard electrode potential of the Mg²⁺ | Mg system can be measured with respect to the standard hydrogen electrode (SHE), which is represented as Pt(s), H₂(g) (1 atm) | H⁺(aq) (1 M). To measure this potential:

1. Set up a galvanic cell where the anode is Mg | MgSO₄ (1 M) and the cathode is the standard hydrogen electrode.
2. Measure the emf of the cell.

The emf measured is the standard electrode potential of the magnesium electrode, denoted as

$E^\circ$

E∘. The relationship is given by:

$E^\circ = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}$

∘​ Since

$E^\circ_{\text{cathode}}$

∘​ for SHE is 0 V:

$E^\circ = 0 – E^\circ_{\text{anode}} = -E^\circ_{\text{anode}}$

​ Thus, the emf measured is equal to the negative of the standard electrode potential of Mg.

---

Question 2: Can you store copper sulfate solutions in a zinc pot?

Solution:
No, you cannot store copper sulfate solution in a zinc pot. Zinc is more reactive than copper, and it can displace copper from its sulfate solution. The reaction is as follows:

$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$

Zn+CuSO4​→ZnSO4​+Cu This reaction shows that zinc will react with copper sulfate, leading to the deposition of copper and the formation of zinc sulfate, which would corrode the zinc pot.

---

Question 3: Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.

Solution:
Ferrous ions (

$\text{Fe}^{2+}$

) can be oxidized to ferric ions (

$\text{Fe}^{3+}$

) by substances that have higher reduction potentials than the

$\text{Fe}^{3+}/\text{Fe}^{2+}$

couple (0.77 V). Three substances that can oxidize ferrous ions are:

1. Fluorine (
   $$F2​) with

   $E^\circ = +2.87 \, \text{V}$ 

2. Chlorine (Cl2​) with
   $E^\circ = +1.36 \, \text{V}$ 
3. Oxygen (O
   $$2​
   

   $E^\circ = +1.23 \, \text{V}$ 

These substances have higher standard electrode potentials and can oxidize ferrous ions under suitable conditions.

---

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

Solution:
Given that the pH of the solution is 10, we can calculate the hydrogen ion concentration (

$[\text{H}^+]$

using:

$\text{pH} = -\log[\text{H}^+]$ $[\text{H}^+] = 10^{-10} \, \text{M}$

The Nernst equation for the hydrogen electrode is given by:

$E_{\text{cell}} = E^\circ – \frac{0.0591}{n} \log \frac{[\text{H}_2]}{[\text{H}^+]}$

 

For a hydrogen electrode,

$E^\circ = 0 \, \text{V}$

and

$[{\text{H}}_2]$

=1 atm. Since

$n = 1$

, we have:

$E_{\text{cell}} = 0 – 0.0591 \log \frac{1}{10^{-10}}$ $E_{\text{cell}} = -0.0591 \times (-10) = 0.591 \, \text{V}$

 

Therefore, the potential of the hydrogen electrode is

$E_{\text{cell}} = 0.591 \, \text{V}$

 

---

Question 5: Calculate the emf of the cell in which the following reaction takes place:

 

$\text{Ni(s)} + 2\text{Ag}^+(0.002 \, \text{M}) \rightarrow \text{Ni}^{2+}(0.160 \, \text{M}) + 2\text{Ag(s)}$

Ni(s)+2Ag+(0.002M)→Ni2+(0.160M)+2Ag(s) Given

$E^\circ_{\text{cell}} = 1.05 \, \text{V}$

Solution:
Using the Nernst equation:

$E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$ $E_{\text{cell}} = 1.05 – \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}$

​

$E_{\text{cell}} = 1.05 – 0.02955 \log \left( \frac{0.160}{0.000004} \right)$

 

$E_{\text{cell}} = 1.05 – 0.02955 \log 40000$

 

$E_{\text{cell}} = 1.05 – 0.02955 \times 4.602$

 

$E_{\text{cell}} = 1.05 – 0.136$

 

$E_{\text{cell}} = 0.914 \, \text{V}$

Therefore, the emf of the cell is

$0.914 \, \text{V}$

---

Question 6: The cell in which the following reactions occur has $E^\circ_{\text{cell}} = 0.236 \, \text{V}$

 

at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution:
Given

$E^\circ_{\text{cell}} = 0.236 \, \text{V}$

and

$T = 298 \, \text{K}$

T=298K:

1. Standard Gibbs Energy
   $\Delta G^\circ$ 

   ΔG∘:

   $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ 

   ΔG∘=−nFEcell∘​ Assuming

   $n = 2$ 

   $n=2$ (since it is not specified, but generally

   $n$is 2):

   $\Delta G^\circ = -2 \times 96487 \times 0.236 \, \text{V}$ 

    

   $\Delta G^\circ = -45548.472 \, \text{J/mol}$ 

    

   $\Delta G^\circ = -45.55 \, \text{kJ/mol}$ 

    

2. Equilibrium Constant
   $K$ 

   The relation between $\Delta G^\circ$ 

   ΔG∘ and

   $K$ 

   $K$ is:

   $\Delta G^\circ = -RT \ln K$ 

   ΔG∘=−RTlnK

   $\ln K = \frac{-\Delta G^\circ}{RT}$ 

   lnK=RT−ΔG∘​

   $\ln K = \frac{45548.472}{8.314 \times 298}$ 

   lnK=8.314×29845548.472​

   $\ln K = \frac{45548.472}{2476.372}$ 

   ​

   $\ln K = 18.39$ 

    

   $K = e^{18.39} \approx 9.8 \times 10^7$ 

    

Thus,

$\Delta G^\circ = -45.55 \, \text{kJ/mol}$

and

$K = 9.8 \times 10^7$

 

---

Question 7: Why does the conductivity of a solution decrease with dilution?

Solution:
The conductivity of a solution depends on the number of ions present in a unit volume of the solution. When the solution is diluted, the concentration of ions per unit volume decreases. Although the degree of dissociation might increase, the overall number of ions in a given volume decreases, leading to a decrease in conductivity.

---

Question 8: Suggest a way to determine the $\Lambda^\circ_m$

 

 

​ value of water.

Solution:
Using Kohlrausch’s law of independent migration of ions, the

$\Lambda^\circ_m$

∘​ value of water can be determined as follows:

$\Lambda^\circ_m(\text{H}_2\text{O}) = \Lambda^\circ_m(\text{HCl}) + \Lambda^\circ_m(\text{NaOH}) – \Lambda^\circ_m(\text{NaCl})$

Λm∘​(H2​O)

By knowing the

$\Lambda^\circ_m$

∘​ values of HCl, NaOH, and NaCl, the

$\Lambda^\circ_m$

value of water can be calculated.

---

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 $\lambda^\circ$

 

 

λ∘(H⁺) = 349.6 S cm² mol⁻¹ and $\lambda^\circ$

 

 

λ∘(HCOO⁻) = 54.6 S cm² mol⁻¹.

Solution:
Given:

• 
  $C = 0.025 \, \text{mol L}^{-1}$ 
• 
  $\Lambda_m = 46.1 \, \text{S cm}^2 \text{mol}^{-1}$ 
• 
  $\Lambda^\circ_m = 349.6 + 54.6 = 404.2 \, \text{S cm}^2 \text{mol}^{-1}$ 

Degree of Dissociation

$\alpha$

:

$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{46.1}{404.2} \approx 0.114$

α=Λm

Dissociation Constant

$K_a$

:

$K_a = C \alpha^2 = 0.025 \times (0.114)^2 \approx 0.00032 \, \text{mol L}^{-1}$

Thus, the degree of dissociation is 0.114, and the dissociation constant is

$3.2 \times 10^{-4} \, \text{mol L}^{-1}$

 

---

Question 10: If a current of 0.5 ampere flows through a metallic wire for 2 hours, how many electrons would flow through the wire?

Solution:
Given:

• Current,
  $I = 0.5 \, \text{A}$ 

  $I=0.5\text{A}$

• Time,
  $t = 2 \, \text{hours} = 2 \times 3600 \, \text{seconds} = 7200 \, \text{seconds}$ 

  $t=2\text{hours}=2\times 3600\text{seconds}=7200\text{seconds}$

The charge

$Q$

is given by:

$Q = I \times t = 0.5 \times 7200 = 3600 \, \text{C}$

 

We know:

• 1 Faraday
  $= 96487 \, \text{C}$ 

  $=96487\text{C}$

• 
  $6.022 \times 10^{23}$ 

  6.022×1023 electrons correspond to

  $96487 \, \text{C}$ 

  $96487\text{C}$

Thus:

$\text{Number of electrons} = \frac{3600 \times 6.022 \times 10^{23}}{96487} \approx 2.25 \times 10^{22}$

 

Therefore,

$2.25 \times 10^{22}$

electrons will flow through the wire.

Electrochemistry Class 12 NCERT solutions Chapter 2 Questions 11 to 20

Question 12: What is the quantity of electricity in coulombs needed to reduce 1 mol of $\text{Cr}_2\text{O}_7^{2-}$

 

Cr2​O72−​?

Solution:
The reaction for the reduction of

$\text{Cr}_2\text{O}_7^{2-}$

Cr2​O72−​ is as follows:

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Cr2​O72−​+14H++6e−→2Cr3++7H2​O

For 1 mole of

$\text{Cr}_2\text{O}_7^{2-}$

Cr2​O72−​, 6 moles of electrons are required. The quantity of electricity required is:

$Q = n \times F$

Q=n×F Where:

• 
  $n = 6$ 

  $n=6$ (number of electrons)

• 
  $F = 96487 \, \text{C/mol}$ 

  $F=96487\text{C/mol}$ (Faraday’s constant)

Thus:

$Q = 6 \times 96487 = 578922 \, \text{C}$

Q=6×96487=578922C

So, 578922 coulombs of electricity are required to reduce 1 mole of

$\text{Cr}_2\text{O}_7^{2-}$

Cr2​O72−​.

---

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

Solution:
A lead storage battery consists of lead (

$\text{Pb}$

Pb) as the anode, lead dioxide (

$\text{PbO}_2$

PbO2​) as the cathode, and sulfuric acid (

$\text{H}_2\text{SO}_4$

H2​SO4​) as the electrolyte.

During recharging:

• The lead sulfate (
  $\text{PbSO}_4$ 

  PbSO4​) formed at both electrodes during discharge is converted back to lead ( $\text{Pb}$ 

  $\text{Pb}$) at the anode and lead dioxide ( $\text{PbO}_2$ 

  PbO2​) at the cathode.

• The overall reaction is:
  $\text{PbSO}_4 + 2\text{H}_2\text{O} + \text{PbSO}_4 \rightarrow \text{Pb} + \text{PbO}_2 + 4\text{H}^+ + 2\text{SO}_4^{2-}$ 

  PbSO4​+2H2​O+PbSO4​→Pb+PbO2​+4H++2SO42−​

In summary:

• At the anode:
  $\text{PbSO}_4$ 

  PbSO4​ is reduced to $\text{Pb}$ 

  $\text{Pb}$.

• At the cathode:
  $\text{PbSO}_4$ 

  PbSO4​ is oxidized to $\text{PbO}_2$ 

  PbO2​.

• Sulfuric acid (
  $\text{H}_2\text{SO}_4$ 

  H2​SO4​) is regenerated.

---

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

Solution:
Two materials that can be used as fuels in fuel cells, other than hydrogen, are:

1. Methanol (
   $\text{CH}_3\text{OH}$ 

   CH3​OH): Used in direct methanol fuel cells (DMFCs) where it reacts with oxygen to produce electricity, water, and carbon dioxide.

2. Methane (
   $\text{CH}_4$ 

   CH4​): Used in solid oxide fuel cells (SOFCs) where it undergoes internal reforming to produce hydrogen, which then reacts to produce electricity.

Both methanol and methane are efficient alternatives to hydrogen in certain types of fuel cells.

---

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

Solution:
Rusting of iron can be viewed as the setup of an electrochemical cell, where:

1. Anode Reaction (Oxidation):
   • Iron acts as the anode.
   • The iron metal is oxidized to ferrous ions (
     $\text{Fe}^{2+}$ 

     Fe2+) and releases electrons. $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$ 

     Fe→Fe2++2e−

2. Cathode Reaction (Reduction):
   • The electrons released at the anode move through the iron and reduce oxygen in the presence of water at a different site on the iron surface (the cathode).
   • The reduction of oxygen occurs as follows:
     $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$ 

     O2​+4H++4e−→2H2​O

   • In neutral or slightly alkaline environments, the reduction can occur as:
     $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$ 

     O2​+2H2​O+4e−→4OH−

3. Overall Reaction:
   • The ferrous ions further react with oxygen to form ferric ions (
     $\text{Fe}^{3+}$ 

     Fe3+), which combine with water to form hydrated ferric oxide ( $\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ 

     Fe2​O3​⋅xH2​O), known as rust.

Thus, the process of rusting involves both oxidation and reduction reactions, similar to 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 order in which the given metals displace each other from the solution of their salts is based on their reactivity, which can be deduced from the electrochemical series. The order of reactivity is:

 

$\text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Cu}$

Mg>Al>Zn>Fe>Cu

Thus, magnesium (Mg) can displace all the other metals, and copper (Cu) can be displaced by all the other metals.

---

Question 17: Given the standard electrode potentials, arrange the following metals in their increasing order of reducing power: K⁺/K = −2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = −0.74 V.

Solution:
The reducing power of a metal is inversely proportional to its standard electrode potential. The lower the reduction potential, the stronger the reducing agent. Based on the given potentials:

1. K⁺/K = −2.93 V: Strongest reducing agent
2. Mg²⁺/Mg = −2.37 V
3. Cr³⁺/Cr = −0.74 V
4. Hg²⁺/Hg = 0.79 V
5. Ag⁺/Ag = 0.80 V: Weakest reducing agent

Thus, the order of increasing reducing power is:

$\text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K}$

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 electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

Solution:
Galvanic Cell Representation:

$\text{Zn(s)} | \text{Zn}^{2+}(aq) || \text{Ag}^{+}(aq) | \text{Ag(s)}$

Zn(s)∣Zn2+(aq)∣∣Ag+(aq)∣Ag(s)

(i) Which electrode is negatively charged?

• The zinc electrode (Zn) is the anode and is negatively charged because it is where oxidation occurs, releasing electrons.

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

• The electrons flow from the zinc electrode (anode) to the silver electrode (cathode) through the external circuit.
• The ions in the solution carry the current within the cell:
  $\text{Zn}^{2+}$ 

  Zn2+ ions move toward the cathode, and $\text{Ag}^+$ 

  Ag+ ions move toward the anode.

(iii) Individual reaction at each electrode:

• Anode (Oxidation):
  $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^-$ 

  Zn(s)→Zn2+(aq)+2e−

• Cathode (Reduction):
  $2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag(s)}$ 

  2Ag+(aq)+2e−→2Ag(s)

---

Question 19: Calculate the standard cell potentials of the galvanic cell in which the following reactions take place: (i) $2\text{Cr(s)} + 3\text{Cd}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cd(s)}$

 

, (ii) $\text{Fe}^{2+}(aq) + \text{Ag}^+(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ag(s)}$

 

. Calculate the $\Delta_r G^\circ$

 

and equilibrium constant of the reactions.

Solution:

(i) For

$2\text{Cr(s)} + 3\text{Cd}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Cd(s)}$

 

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}$

​

• 
  $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V}$ 
• 
  $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \, \text{V}$ 

  
  

 

$$E^\circ_{\text{cell}} = (-0.40) – (-0.74) = 0.34 \, \text{V}$$

 

$$\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -6 \times 96487 \times 0.34 = -196860 \, \text{J/mol} = -196.86 \, \text{kJ/mol}$$

 

Equilibrium constant

$K_c$

Kc​:

 

$$\log K_c = \frac{nE^\circ_{\text{cell}}}{0.059} = \frac{6 \times 0.34}{0.059} = 34.576$$

logKc​=0.059nEcell∘​​=0.0596×0.34​=34.576

$$K_c = 3.76 \times 10^{34}$$

Kc​=3.76×1034

(ii) For

$\text{Fe}^{2+}(aq) + \text{Ag}^+(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Ag(s)}$

 

• 
  $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V}$ 

  
  

• 
  $E_{{\text{Ag}}^{+}\mathrm{/}\text{Ag}}^{\circ }=+0.80\text{V}$ 

  $$E^\circ_{\text{cell}} = 0.80 – 0.77 = 0.03 \, \text{V}$$ 

  
  $$\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -1 \times 96487 \times 0.03 = -2895 \, \text{J/mol} = -2.895 \, \text{kJ/mol}$$ 

  
  

Equilibrium constant

$K_c$

Kc​:

 

$$\log K_c = \frac{1 \times 0.03}{0.059} = 0.508$$

 

$$K_c = 3.22$$

 

---

Question 20: Write the Nernst equation and emf of the following cells at 298 K: (i) $\text{Mg(s)} | \text{Mg}^{2+}(0.001 \, \text{M}) \parallel \text{Cu}^{2+}(0.0001 \, \text{M}) | \text{Cu(s)}$

 

Mg(s)∣Mg2+(0.001M)∥Cu2+(0.0001M)∣Cu(s), (ii) $\text{Fe(s)} | \text{Fe}^{2+}(0.001 \, \text{M}) \parallel \text{H}^+(1 \, \text{M}) | \text{H}_2(\text{g})(1 \, \text{bar}) | \text{Pt(s)}$

 

Fe(s)∣Fe2+(0.001M)∥H+(1M)∣H2​(g)(1bar)∣Pt(s), (iii) $\text{Sn(s)} | \text{Sn}^{2+}(0.050 \, \text{M}) \parallel \text{H}^+(0.020 \, \text{M}) | \text{H}_2(\text{g})(1 \, \text{bar}) | \text{Pt(s)}$

 

Sn(s)∣Sn2+(0.050M)∥H+(0.020M)∣H2​(g)(1bar)∣Pt(s), (iv) $\text{Pt(s)} | \text{Br}_2(\text{l}) | \text{Br}^- (0.010 \, \text{M}) \parallel \text{H}^+(0.030 \, \text{M}) | \text{H}_2(\text{g})(1 \, \text{bar}) | \text{Pt(s)}$

 

Pt(s)∣Br2​(l)∣Br−(0.010M)∥H+(0.030M)∣H2​(g)(1bar)∣Pt(s).

Solution:

(i)

 

$$E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} – E^\circ_{\text{Mg}^{2+}/\text{Mg}} = 0.34 – (-2.37) = 2.71 \, \text{V}$$

 

$$E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.059}{2} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} = 2.71 – \frac{0.059}{2} \log \frac{0.001}{0.0001} = 2.71 – 0.02955 \times 1 = 2.6805 \, \text{V}$$

 

(ii)

 

$$E^\circ_{\text{cell}} = 0.00 – (-0.44) = 0.44 \, \text{V}$$

 

$$E_{\text{cell}} = 0.44 – \frac{0.059}{2} \log \frac{0.001}{(1)^2} = 0.44 + 0.02955 \times 3 = 0.5285 \, \text{V}$$

 

(iii)

 

$$E^\circ_{\text{cell}} = 0.00 – (-0.14) = 0.14 \, \text{V}$$

 

$$E_{\text{cell}} = 0.14 – \frac{0.059}{2} \log \frac{0.050}{(0.020)^2} = 0.14 – 0.02955 \times 0.5 = 0.08 \, \text{V}$$

 

(iv)

 

$$E^\circ_{\text{cell}} = 0.00 – 1.08 = -1.08 \, \text{V}$$

 

$$E_{\text{cell}} = -1.08 – \frac{0.059}{2} \log \frac{1}{(0.01)^2(0.030)^2} = -1.08 – 0.0591 \times 2 = -1.288 \, \text{V}$$

 

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.

Solution

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.
$\Lambda =\frac{k}{C}$
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 $S{m}^2/mol$
The common unit of molar conductivity is $Sc{m}^2/mol$
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.

 

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.

Conductivity of 0.00241 M acetic acid s 7.896 x 10 ${}^{}$ S cm, calculate its molar conductivity. If $\lambda$^{0}${}_m$ for acetic acid is 390.5 S cm${}^{}$ mol ${}^{}$, what is its dissociation constant?

Solution : Given: $K=7.896\times {10}^{-5}Sc{m}^{-1}$ $C=M=0.00241mol{L}^{-1}$

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:
(i) $20g$ of $Ca$ from molten $CaC{l}_2$
(ii) $40g$ of $Al$ from molten $A{l}_2{O}_3$

[Given : Molar mass of calcium & Aluminium are 40 g $mo{l}^{-1}$ & 27 g $mo{l}^{-1}$ respectively]

Solution

$\left(1\right)$ $C{a}^{2+}+2e^{-}\to Ca$$1$ mole ($40$ g) of $Ca$ will require $2$ F of electricity.
$20$ g ($0.5$ mole) of $Ca$ will require $1$ F of electricity.

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 $A,B,C$ containing solutions of $ZnS{O}_4,AgN{O}_3$ and $CuS{O}_3$, respectively are connected in series. A steady current of $1.5$ amperes was passed through them until $1.45g$ of silver deposited at the cathode of cell $B$. How long did the current flow? What mass of copper and zinc were deposited?

Solution

Cell $B$: $A{g}^{+}+e^{-}\rightleftharpoons Ag$ at cathode.

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.