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Electrochemistry Complete NCERT Resource Guide
Welcome to your comprehensive study hub, Lucky! Physical chemistry can look complex with its intense mathematical formulas, but we are here to break it down step by step. This page provides the highly accurate, comprehensive Updated NCERT Solutions for Class 12 Chemistry Chapter 3. We will explore Electrochemistry Class 12 Solutions with logical breakdowns, exact cell reactions, clear derivations, and high-yield Important Questions to ensure you score full marks in your 2026 Board Exams and competitive assessments like NEET or JEE.
Chapter NameElectrochemistry
SubjectChemistry
ClassClass 12
BoardCBSE & State Boards (NCERT)
Important TopicsGalvanic Cells, Nernst Equation, Kohlrausch's Law, Faraday's Laws of Electrolysis, Lead Storage Battery, Fuel Cells
Difficulty LevelModerate to High (Requires numerical practice)
Exam WeightageApprox. 7 to 8 Marks
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Learning Objectives
After completing this chapter, students will be able to:
- Understand the construction, working, and representation of Electrochemical and Electrolytic cells.
- Apply the Nernst Equation to calculate cell potentials ($E_{\text{cell}}$) under non-standard conditions.
- Relate standard cell potential ($E^\circ_{\text{cell}}$) to Gibbs Free Energy ($\Delta G^\circ$) and the Equilibrium Constant ($K_c$).
- Differentiate between Conductivity ($\kappa$), Molar Conductivity ($\Lambda_m$), and their trends upon dilution.
- Use Kohlrausch's Law to calculate limiting molar conductivities for weak electrolytes.
- Solve quantitative problems related to electrolysis using Faraday's First and Second Laws.
- Understand the operational reactions behind commercial batteries, fuel cells, and the mechanism of corrosion.
Core Formulas, Definitions & Rules
Before moving to the comprehensive text solutions, review these high-yield operational equations required for problem-solving:
- Standard Cell Potential ($E^\circ_{\text{cell}}$): Calculated from standard reduction potentials as: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$
- The Nernst Equation (at $298\text{ K}$): Used to determine the potential of a cell under non-standard concentrations: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]}$$
- Free Energy and Cell Potential Connection: Relates the electrical work done to thermodynamic spontaneity: $$\Delta G = -nFE_{\text{cell}} \quad \text{and} \quad \Delta G^\circ = -nFE^\circ_{\text{cell}}$$ Where $F = 96487 \approx 96500\text{ C mol}^{-1}$ is Faraday's Constant.
- Equilibrium Constant Relationship: Derived from the Nernst equation when the cell reaches electrical equilibrium ($E_{\text{cell}} = 0$): $$\log K_c = \frac{n E^\circ_{\text{cell}}}{0.0591}$$
- Molar Conductivity Equation ($\Lambda_m$): Relates the conductivity ($\kappa$) to concentration molarity ($C$): $$\Lambda_m = \frac{\kappa \times 1000}{C}$$ Units are expressed in $\text{S cm}^2\text{ mol}^{-1}$ when $\kappa$ is in $\text{S cm}^{-1}$.
- Kohlrausch's Law Formulation: For an electrolyte $A_x B_y$, the limiting molar conductivity is given by: $$\Lambda_m^\circ = x\lambda_+^\circ + y\lambda_-^\circ$$
- Faraday's First Law of Electrolysis: Mass ($m$) of substance deposited or liberated is directly proportional to the charge ($Q$) passed: $$m = Z \cdot Q = Z \cdot I \cdot t = \frac{\text{Molar Mass}}{n \times 96500} \times I \times t$$
Full NCERT Solutions (In-Depth, Step-by-Step)
Below are the comprehensive solutions for Class 12 Chemistry Chapter 3 Text Exercises, presented in an easy-to-follow, exam-aligned format.
Question 3.1: How would you determine the standard electrode potential of the system $\text{Mg}^{2+}|\text{Mg}$?
Step 1: Set up the electrochemical cell. To determine the standard electrode potential of a magnesium electrode ($\text{Mg}^{2+}|\text{Mg}$), it is connected as an anode to a Standard Hydrogen Electrode (SHE), which acts as the cathode.
Step 2: Define standard conditions. The magnesium half-cell consists of a pure magnesium wire dipped in a $1\text{ M }\text{MgSO}_4$ solution. The SHE consists of platinum foil dipped in a $1\text{ M }\text{H}^+$ ion solution with pure hydrogen gas bubbled through it at $1\text{ bar}$ pressure and a constant temperature of $298\text{ K}$.
Step 3: Represent the cell diagram. The constructed galvanic cell is represented as: $$\text{Mg}(s) | \text{Mg}^{2+}(1\text{ M}) \parallel \text{H}^+(1\text{ M}) | \text{H}_2(g, 1\text{ bar}) | \text{Pt}(s)$$
Step 4: Measure and compute the potential. The electromotive force (EMF) of this cell is measured using a voltmeter. Since the standard potential of the SHE ($E^\circ_{\text{H}^+/\text{H}_2}$) is defined as $0.00\text{ V}$, the formula gives: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\text{H}^+/\text{H}_2} - E^\circ_{\text{Mg}^{2+}/\text{Mg}}$$ $$E^\circ_{\text{cell}} = 0.00\text{ V} - E^\circ_{\text{Mg}^{2+}/\text{Mg}} \implies E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -E^\circ_{\text{cell}}$$ The measured voltmeter value is $2.36\text{ V}$, confirming that the standard reduction potential of Magnesium is $-2.36\text{ V}$.
Step 2: Define standard conditions. The magnesium half-cell consists of a pure magnesium wire dipped in a $1\text{ M }\text{MgSO}_4$ solution. The SHE consists of platinum foil dipped in a $1\text{ M }\text{H}^+$ ion solution with pure hydrogen gas bubbled through it at $1\text{ bar}$ pressure and a constant temperature of $298\text{ K}$.
Step 3: Represent the cell diagram. The constructed galvanic cell is represented as: $$\text{Mg}(s) | \text{Mg}^{2+}(1\text{ M}) \parallel \text{H}^+(1\text{ M}) | \text{H}_2(g, 1\text{ bar}) | \text{Pt}(s)$$
Step 4: Measure and compute the potential. The electromotive force (EMF) of this cell is measured using a voltmeter. Since the standard potential of the SHE ($E^\circ_{\text{H}^+/\text{H}_2}$) is defined as $0.00\text{ V}$, the formula gives: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\text{H}^+/\text{H}_2} - E^\circ_{\text{Mg}^{2+}/\text{Mg}}$$ $$E^\circ_{\text{cell}} = 0.00\text{ V} - E^\circ_{\text{Mg}^{2+}/\text{Mg}} \implies E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -E^\circ_{\text{cell}}$$ The measured voltmeter value is $2.36\text{ V}$, confirming that the standard reduction potential of Magnesium is $-2.36\text{ V}$.
Question 3.2: Can you store copper sulfate solution in a zinc pot?
Step 1: Compare the standard standard reduction potentials. Look up the standard reduction values from the electrochemical series:
$$E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$$
$$E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$$
Step 2: Determine relative reactivity. A lower standard reduction potential indicates that zinc has a stronger tendency to lose electrons and undergo oxidation compared to copper. Zinc is therefore more reactive than copper.
Step 3: Analyze the single displacement reaction. If $\text{CuSO}_4$ is stored in a zinc container, a spontaneous redox reaction occurs where zinc dissolves into solution and precipitates copper metal: $$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$$ Let's double-check the cell potential to confirm spontaneity: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.76\text{ V}) = +1.10\text{ V}$$ Since $E^\circ_{\text{cell}}$ is positive ($\Delta G^\circ < 0$), the reaction is highly spontaneous. The zinc pot would develop holes and corrode, so **copper sulfate solution cannot be stored in a zinc pot**.
Step 2: Determine relative reactivity. A lower standard reduction potential indicates that zinc has a stronger tendency to lose electrons and undergo oxidation compared to copper. Zinc is therefore more reactive than copper.
Step 3: Analyze the single displacement reaction. If $\text{CuSO}_4$ is stored in a zinc container, a spontaneous redox reaction occurs where zinc dissolves into solution and precipitates copper metal: $$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$$ Let's double-check the cell potential to confirm spontaneity: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.76\text{ V}) = +1.10\text{ V}$$ Since $E^\circ_{\text{cell}}$ is positive ($\Delta G^\circ < 0$), the reaction is highly spontaneous. The zinc pot would develop holes and corrode, so **copper sulfate solution cannot be stored in a zinc pot**.
Question 3.3: Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.
Step 1: State the target chemical oxidation half-reaction. We want to oxidize ferrous ions ($\text{Fe}^{2+}$) into ferric ions ($\text{Fe}^{3+}$):
$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \quad (E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77\text{ V})$$
Step 2: Identify the criteria for the oxidizing agent. Any substance capable of oxidizing $\text{Fe}^{2+}$ must be a stronger oxidizing agent than $\text{Fe}^{3+}$. This means it must have a standard reduction potential ($E^\circ$) greater than $+0.77\text{ V}$.
Step 3: Choose three substances from the standard potentials table. From the electrochemical series, the following common species have values above $+0.77\text{ V}$:
Step 2: Identify the criteria for the oxidizing agent. Any substance capable of oxidizing $\text{Fe}^{2+}$ must be a stronger oxidizing agent than $\text{Fe}^{3+}$. This means it must have a standard reduction potential ($E^\circ$) greater than $+0.77\text{ V}$.
Step 3: Choose three substances from the standard potentials table. From the electrochemical series, the following common species have values above $+0.77\text{ V}$:
- Chlorine gas ($\text{Cl}_2$): $E^\circ = +1.36\text{ V}$
- Permanganate ions ($\text{MnO}_4^-$ in acidic medium): $E^\circ = +1.51\text{ V}$
- Fluorine gas ($\text{F}_2$): $E^\circ = +2.87\text{ V}$
Question 3.4: Calculate the potential of a hydrogen electrode in contact with a solution whose pH is 10.
Step 1: Determine the hydronium ion concentration from the pH. By definition:
$$\text{pH} = -\log[\text{H}^+] \implies 10 = -\log[\text{H}^+] \implies [\text{H}^+] = 10^{-10}\text{ M}$$
Step 2: Write out the standard hydrogen reduction half-reaction. $$2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \quad (\text{where } n = 2, \text{ and } E^\circ_{\text{H}^+/\text{H}_2} = 0.00\text{ V})$$ Alternatively, using a single electron system: $$\text{H}^+(aq) + e^- \rightarrow \frac{1}{2}\text{H}_2(g) \quad (\text{where } n = 1)$$
Step 3: Apply the Nernst equation assuming standard hydrogen gas pressure ($1\text{ bar}$). $$E_{\text{H}^+/\text{H}_2} = E^\circ_{\text{H}^+/\text{H}_2} - \frac{0.0591}{1} \log \frac{1}{[\text{H}^+]}$$ $$E_{\text{H}^+/\text{H}_2} = 0.00 - 0.0591 \log \left(\frac{1}{10^{-10}}\right)$$ $$E_{\text{H}^+/\text{H}_2} = -0.0591 \log(10^{10}) = -0.0591 \times 10 = -0.591\text{ V}$$ The calculated potential of the hydrogen electrode at $\text{pH} = 10$ is **$-0.591\text{ V}$**.
Step 2: Write out the standard hydrogen reduction half-reaction. $$2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \quad (\text{where } n = 2, \text{ and } E^\circ_{\text{H}^+/\text{H}_2} = 0.00\text{ V})$$ Alternatively, using a single electron system: $$\text{H}^+(aq) + e^- \rightarrow \frac{1}{2}\text{H}_2(g) \quad (\text{where } n = 1)$$
Step 3: Apply the Nernst equation assuming standard hydrogen gas pressure ($1\text{ bar}$). $$E_{\text{H}^+/\text{H}_2} = E^\circ_{\text{H}^+/\text{H}_2} - \frac{0.0591}{1} \log \frac{1}{[\text{H}^+]}$$ $$E_{\text{H}^+/\text{H}_2} = 0.00 - 0.0591 \log \left(\frac{1}{10^{-10}}\right)$$ $$E_{\text{H}^+/\text{H}_2} = -0.0591 \log(10^{10}) = -0.0591 \times 10 = -0.591\text{ V}$$ The calculated potential of the hydrogen electrode at $\text{pH} = 10$ is **$-0.591\text{ V}$**.
Question 3.5: Calculate the EMF of the cell in which the following reaction takes place:
$\text{Ni}(s) + 2\text{Ag}^+(0.002\text{ M}) \rightarrow \text{Ni}^{2+}(0.160\text{ M}) + 2\text{Ag}(s)$ given that $E^\circ_{\text{cell}} = 1.05\text{ V}$.
Step 1: Extract the reaction parameters. From the balanced equation, nickel undergoes a 2-electron oxidation ($\text{Ni} \rightarrow \text{Ni}^{2+} + 2e^-$) and two silver ions undergo reduction. The total number of transferred electrons is $n = 2$.
Given concentrations: $[\text{Ni}^{2+}] = 0.160\text{ M}$ and $[\text{Ag}^+] = 0.002\text{ M} = 2 \times 10^{-3}\text{ M}$.
Step 2: Formulate the Nernst equation for this cell setup. $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$$
Step 3: Substitute the numerical values and solve. $$E_{\text{cell}} = 1.05 - \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}$$ $$E_{\text{cell}} = 1.05 - 0.02955 \log \frac{0.160}{4 \times 10^{-6}}$$ $$E_{\text{cell}} = 1.05 - 0.02955 \log(40000) = 1.05 - 0.02955 \log(4 \times 10^4)$$ Using log rules ($\log 4 \approx 0.6021$): $$\log(40000) = 4 + 0.6021 = 4.6021$$ $$E_{\text{cell}} = 1.05 - (0.02955 \times 4.6021) = 1.05 - 0.136\text{ V} = 0.914\text{ V}$$ The calculated real-time EMF of the cell is **$0.914\text{ V}$**.
Given concentrations: $[\text{Ni}^{2+}] = 0.160\text{ M}$ and $[\text{Ag}^+] = 0.002\text{ M} = 2 \times 10^{-3}\text{ M}$.
Step 2: Formulate the Nernst equation for this cell setup. $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$$
Step 3: Substitute the numerical values and solve. $$E_{\text{cell}} = 1.05 - \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}$$ $$E_{\text{cell}} = 1.05 - 0.02955 \log \frac{0.160}{4 \times 10^{-6}}$$ $$E_{\text{cell}} = 1.05 - 0.02955 \log(40000) = 1.05 - 0.02955 \log(4 \times 10^4)$$ Using log rules ($\log 4 \approx 0.6021$): $$\log(40000) = 4 + 0.6021 = 4.6021$$ $$E_{\text{cell}} = 1.05 - (0.02955 \times 4.6021) = 1.05 - 0.136\text{ V} = 0.914\text{ V}$$ The calculated real-time EMF of the cell is **$0.914\text{ V}$**.
Question 3.6: A galvanic cell is designed with the following half-cells: $\text{Fe}^{2+}|\text{Fe}$ ($E^\circ = -0.44\text{ V}$) and $\text{Ag}^+|\text{Ag}$ ($E^\circ = +0.80\text{ V}$). Calculate its standard free energy change ($\Delta G^\circ$) and equilibrium constant at $298\text{ K}$.
Step 1: Find the standard cell potential ($E^\circ_{\text{cell}}$). Identify the cathode (higher reduction potential) and anode (lower reduction potential):
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.44\text{ V}) = +1.24\text{ V}$$
Step 2: Balance the redox equation to find the electron transfer value ($n$). $$\text{Anode oxidation: } \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$$ $$\text{Cathode reduction: } 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$$ The net reaction shows that $n = 2$ moles of electrons are transferred.
Step 3: Calculate the standard Gibbs Free Energy change ($\Delta G^\circ$). $$\Delta G^\circ = -nFE^\circ_{\text{cell}} = -2 \times 96500\text{ C mol}^{-1} \times 1.24\text{ V}$$ $$\Delta G^\circ = -239320\text{ J mol}^{-1} = -239.32\text{ kJ mol}^{-1}$$
Step 4: Calculate the equilibrium constant ($K_c$). Using the equilibrium expression: $$\log K_c = \frac{n E^\circ_{\text{cell}}}{0.0591} = \frac{2 \times 1.24}{0.0591} = \frac{2.48}{0.0591} \approx 41.9627$$ $$K_c = \text{antilog}(41.9627) \approx 9.18 \times 10^{41}$$ The standard free energy change is **$-239.32\text{ kJ mol}^{-1}$** and the equilibrium constant is **$9.18 \times 10^{41}$**.
Step 2: Balance the redox equation to find the electron transfer value ($n$). $$\text{Anode oxidation: } \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$$ $$\text{Cathode reduction: } 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$$ The net reaction shows that $n = 2$ moles of electrons are transferred.
Step 3: Calculate the standard Gibbs Free Energy change ($\Delta G^\circ$). $$\Delta G^\circ = -nFE^\circ_{\text{cell}} = -2 \times 96500\text{ C mol}^{-1} \times 1.24\text{ V}$$ $$\Delta G^\circ = -239320\text{ J mol}^{-1} = -239.32\text{ kJ mol}^{-1}$$
Step 4: Calculate the equilibrium constant ($K_c$). Using the equilibrium expression: $$\log K_c = \frac{n E^\circ_{\text{cell}}}{0.0591} = \frac{2 \times 1.24}{0.0591} = \frac{2.48}{0.0591} \approx 41.9627$$ $$K_c = \text{antilog}(41.9627) \approx 9.18 \times 10^{41}$$ The standard free energy change is **$-239.32\text{ kJ mol}^{-1}$** and the equilibrium constant is **$9.18 \times 10^{41}$**.
Question 3.7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Step 1: Define Conductivity ($\kappa$). Conductivity (or specific conductance) is defined as the conductance of a solution kept between two parallel electrodes of unit cross-sectional area separated by a unit length. Its SI unit is $\text{S m}^{-1}$ (often expressed as $\text{S cm}^{-1}$).
Step 2: Define Molar Conductivity ($\Lambda_m$). Molar conductivity is the conducting power of all the ions produced by dissolving exactly one mole of an electrolyte in a given volume ($V$) of solution: $$\Lambda_m = \frac{\kappa}{C}$$
Step 3: Explain the variation of Conductivity with concentration. For both strong and weak electrolytes, conductivity **decreases** as the concentration decreases (upon dilution). This occurs because dilution reduces the total number of current-carrying ions present per unit volume of the solution.
Step 4: Explain the variation of Molar Conductivity with concentration. Molar conductivity **increases** as concentration decreases (upon dilution).
Step 2: Define Molar Conductivity ($\Lambda_m$). Molar conductivity is the conducting power of all the ions produced by dissolving exactly one mole of an electrolyte in a given volume ($V$) of solution: $$\Lambda_m = \frac{\kappa}{C}$$
Step 3: Explain the variation of Conductivity with concentration. For both strong and weak electrolytes, conductivity **decreases** as the concentration decreases (upon dilution). This occurs because dilution reduces the total number of current-carrying ions present per unit volume of the solution.
Step 4: Explain the variation of Molar Conductivity with concentration. Molar conductivity **increases** as concentration decreases (upon dilution).
- For strong electrolytes, dilution reduces inter-ionic attractions, allowing ions to move more freely. This follows the Debye-Hückel-Onsager equation: $\Lambda_m = \Lambda_m^\circ - A\sqrt{C}$.
- For weak electrolytes, dilution shifts the dissociation equilibrium forward, significantly increasing the degree of ionization ($\alpha$) and the total number of ions, as described by Ostwald's Dilution Law.
Question 3.8: The conductivity of $0.001028\text{ M}$ acetic acid is $4.95 \times 10^{-5}\text{ S cm}^{-1}$. Calculate its molar conductivity. If $\Lambda_m^\circ$ for acetic acid is $390.5\text{ S cm}^2\text{ mol}^{-1}$, calculate its dissociation constant.
Step 1: Calculate the molar conductivity ($\Lambda_m$). Using the standard concentration relationship:
$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{4.95 \times 10^{-5}\text{ S cm}^{-1} \times 1000}{0.001028\text{ mol L}^{-1}}$$
$$\Lambda_m = \frac{0.0495}{0.001028} = 48.15\text{ S cm}^2\text{ mol}^{-1}$$
Step 2: Determine the degree of dissociation ($\alpha$). $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{48.15}{390.5} \approx 0.1233$$
Step 3: Calculate the acid dissociation constant ($K_a$). Using Ostwald's formulation for weak weak monoprotic systems: $$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233}$$ $$K_a = \frac{0.001028 \times 0.015203}{0.8767} = \frac{1.5629 \times 10^{-5}}{0.8767} \approx 1.78 \times 10^{-5}\text{ mol L}^{-1}$$ The molar conductivity is **$48.15\text{ S cm}^2\text{ mol}^{-1}$** and the dissociation constant is **$1.78 \times 10^{-5}\text{ mol L}^{-1}$**.
Step 2: Determine the degree of dissociation ($\alpha$). $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{48.15}{390.5} \approx 0.1233$$
Step 3: Calculate the acid dissociation constant ($K_a$). Using Ostwald's formulation for weak weak monoprotic systems: $$K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233}$$ $$K_a = \frac{0.001028 \times 0.015203}{0.8767} = \frac{1.5629 \times 10^{-5}}{0.8767} \approx 1.78 \times 10^{-5}\text{ mol L}^{-1}$$ The molar conductivity is **$48.15\text{ S cm}^2\text{ mol}^{-1}$** and the dissociation constant is **$1.78 \times 10^{-5}\text{ mol L}^{-1}$**.
Question 3.9: State Kohlrausch's law of independent migration of ions. Write its important applications.
Step 1: State the fundamental law. Kohlrausch's Law states that at infinite dilution, when an electrolyte dissociates completely, each ion migrates independently and contributes a definite amount to the total molar conductivity of the electrolyte, regardless of the identity of the other ion with which it is associated.
Step 2: Outline Key Application 1. *Calculation of limiting molar conductivity ($\Lambda_m^\circ$) for weak electrolytes:* Weak electrolytes do not dissociate completely at real concentrations, making it impossible to find their limiting molar conductivity by extrapolating experimental data. Kohlrausch's law circumvents this by summing the limiting conductivities of their component ions using data from strong electrolytes: $$\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl})$$
Step 3: Outline Key Application 2. *Determining the degree of dissociation ($\alpha$):* It can be found at any concentration by comparing the measured molar conductivity to the limiting value: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
Step 4: Outline Key Application 3. *Calculating the solubility product ($K_{\text{sp}}$) of sparingly soluble salts:* Since these solutions are extremely dilute even when saturated, we can approximate $\Lambda_m \approx \Lambda_m^\circ$. This allows us to calculate solubility ($S$) using: $S = \frac{\kappa \times 1000}{\Lambda_m^\circ}$.
Step 2: Outline Key Application 1. *Calculation of limiting molar conductivity ($\Lambda_m^\circ$) for weak electrolytes:* Weak electrolytes do not dissociate completely at real concentrations, making it impossible to find their limiting molar conductivity by extrapolating experimental data. Kohlrausch's law circumvents this by summing the limiting conductivities of their component ions using data from strong electrolytes: $$\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl})$$
Step 3: Outline Key Application 2. *Determining the degree of dissociation ($\alpha$):* It can be found at any concentration by comparing the measured molar conductivity to the limiting value: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
Step 4: Outline Key Application 3. *Calculating the solubility product ($K_{\text{sp}}$) of sparingly soluble salts:* Since these solutions are extremely dilute even when saturated, we can approximate $\Lambda_m \approx \Lambda_m^\circ$. This allows us to calculate solubility ($S$) using: $S = \frac{\kappa \times 1000}{\Lambda_m^\circ}$.
Question 3.10: Three electrolytic cells A, B, C containing solutions of $\text{ZnSO}_4$, $\text{AgNO}_3$ and $\text{CuSO}_4$ 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?
Step 1: Calculate the total electrical charge passed using Cell B data. The reduction reaction for silver is:
$$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$$
This shows that $1\text{ mole}$ of electrons ($96500\text{ C}$) deposits $108\text{ g}$ of silver metal.
Using Faraday's First Law: $$m = \frac{\text{Molar Mass}}{n \times 96500} \times I \times t \implies 1.45 = \frac{108}{1 \times 96500} \times 1.5 \times t$$ $$t = \frac{1.45 \times 96500}{108 \times 1.5} = \frac{139925}{162} \approx 863.73\text{ seconds}$$
Step 2: Calculate the mass of copper deposited in Cell C. The reduction reaction for copper is: $$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad (\text{Molar mass } = 63.5\text{ g mol}^{-1}, n = 2)$$ Using the calculated time ($t = 863.73\text{ s}$): $$m_{\text{Cu}} = \frac{63.5}{2 \times 96500} \times 1.5 \times 863.73 = \frac{63.5 \times 1295.6}{193000} \approx 0.426\text{ g}$$
Step 3: Calculate the mass of zinc deposited in Cell A. The reduction reaction for zinc is: $$\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad (\text{Molar mass } = 65.4\text{ g mol}^{-1}, n = 2)$$ $$m_{\text{Zn}} = \frac{65.4}{2 \times 96500} \times 1.5 \times 863.73 = \frac{65.4 \times 1295.6}{193000} \approx 0.439\text{ g}$$ The current flowed for **$863.73\text{ seconds}$**, depositing **$0.426\text{ g}$ of copper** and **$0.439\text{ g}$ of zinc**.
Using Faraday's First Law: $$m = \frac{\text{Molar Mass}}{n \times 96500} \times I \times t \implies 1.45 = \frac{108}{1 \times 96500} \times 1.5 \times t$$ $$t = \frac{1.45 \times 96500}{108 \times 1.5} = \frac{139925}{162} \approx 863.73\text{ seconds}$$
Step 2: Calculate the mass of copper deposited in Cell C. The reduction reaction for copper is: $$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad (\text{Molar mass } = 63.5\text{ g mol}^{-1}, n = 2)$$ Using the calculated time ($t = 863.73\text{ s}$): $$m_{\text{Cu}} = \frac{63.5}{2 \times 96500} \times 1.5 \times 863.73 = \frac{63.5 \times 1295.6}{193000} \approx 0.426\text{ g}$$
Step 3: Calculate the mass of zinc deposited in Cell A. The reduction reaction for zinc is: $$\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad (\text{Molar mass } = 65.4\text{ g mol}^{-1}, n = 2)$$ $$m_{\text{Zn}} = \frac{65.4}{2 \times 96500} \times 1.5 \times 863.73 = \frac{65.4 \times 1295.6}{193000} \approx 0.439\text{ g}$$ The current flowed for **$863.73\text{ seconds}$**, depositing **$0.426\text{ g}$ of copper** and **$0.439\text{ g}$ of zinc**.
Question 3.11: Predict the products of electrolysis in each of the following: (i) An aqueous solution of $\text{AgNO}_3$ with silver electrodes. (ii) An aqueous solution of $\text{AgNO}_3$ with platinum electrodes.
Step 1: Analyze Case (i) - Aqueous $\text{AgNO}_3$ with active silver electrodes.
In this setup, silver is an active electrode that can participate in the reaction.
Step 2: Analyze Case (ii) - Aqueous $\text{AgNO}_3$ with inert platinum electrodes. In this setup, the platinum electrodes are inert and will not oxidize or reduce.
- At Cathode: Both $\text{Ag}^+$ ions and $\text{H}_2\text{O}$ are available for reduction. Because silver has a higher reduction potential ($E^\circ = +0.80\text{ V}$) than water ($E^\circ = -0.83\text{ V}$), silver ions are preferentially reduced, depositing silver metal: $$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$$
- At Anode: The silver metal of the anode has a lower reduction potential than water or nitrate ions, meaning it oxidizes more easily. The anode dissolves, releasing silver ions into solution: $$\text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-$$
Step 2: Analyze Case (ii) - Aqueous $\text{AgNO}_3$ with inert platinum electrodes. In this setup, the platinum electrodes are inert and will not oxidize or reduce.
- At Cathode: The reduction potentials remain unchanged. Silver ions are reduced preferentially over water, depositing silver metal: $$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$$
- At Anode: Water molecules and nitrate ions ($\text{Sir, }\text{NO}_3^-$) compete for oxidation. Nitrate ions are already in their highest oxidation state (+5) and are very difficult to oxidize. Water is oxidized instead, releasing oxygen gas: $$2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$$
Question 3.12: Write the operational cell reactions occurring during the discharge phase of a standard commercial Lead Storage Battery. Is it a primary or secondary cell?
Step 1: Define the cell category. A lead storage battery can be recharged by passing an external electrical current through it in the reverse direction, meaning it can undergo multiple charge-discharge cycles. It is therefore a **secondary cell**.
Step 2: State the chemical components. The anode is made of spongy lead ($\text{Pb}$), the cathode consists of a grid of lead packed with lead dioxide ($\text{PbO}_2$), and the electrolyte is a $38\%$ solution of sulfuric acid ($\text{H}_2\text{SO}_4$).
Step 3: Write the discharge reactions. During use (discharge), the following half-cell reactions take place: $$\text{At Anode (Oxidation): } \text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-$$ $$\text{At Cathode (Reduction): } \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$$
Step 4: Write the net cell reaction. Combining the two half-reactions gives: $$\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$$ Notice that sulfuric acid is consumed and water is produced during discharge, which lowers the density of the electrolyte solution.
Step 2: State the chemical components. The anode is made of spongy lead ($\text{Pb}$), the cathode consists of a grid of lead packed with lead dioxide ($\text{PbO}_2$), and the electrolyte is a $38\%$ solution of sulfuric acid ($\text{H}_2\text{SO}_4$).
Step 3: Write the discharge reactions. During use (discharge), the following half-cell reactions take place: $$\text{At Anode (Oxidation): } \text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-$$ $$\text{At Cathode (Reduction): } \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$$
Step 4: Write the net cell reaction. Combining the two half-reactions gives: $$\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$$ Notice that sulfuric acid is consumed and water is produced during discharge, which lowers the density of the electrolyte solution.
Question 3.13: Explain the composition and working mechanism of a Hydrogen-Oxygen Fuel Cell. Highlight its key advantages over conventional thermal plants.
Step 1: Describe the basic construction. Galvanic cells designed to convert the chemical energy from the combustion of fuels (like hydrogen, methane, or methanol) directly into electrical energy are called fuel cells. In a standard $\text{H}_2-\text{O}_2$ fuel cell, hydrogen and oxygen gases are bubbled through porous carbon electrodes into a concentrated aqueous potassium hydroxide ($\text{KOH}$) electrolyte solution. Finely divided platinum or palladium catalysts are embedded in the electrodes to speed up the reactions.
Step 2: Write out the electrode reactions. $$\text{At Anode (Oxidation): } 2\text{H}_2(g) + 4\text{OH}^-(aq) \rightarrow 4\text{H}_2\text{O}(l) + 4e^-$$ $$\text{At Cathode (Reduction): } \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)$$ Net overall reaction: $$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$$
Step 3: List the advantages.
Step 2: Write out the electrode reactions. $$\text{At Anode (Oxidation): } 2\text{H}_2(g) + 4\text{OH}^-(aq) \rightarrow 4\text{H}_2\text{O}(l) + 4e^-$$ $$\text{At Cathode (Reduction): } \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)$$ Net overall reaction: $$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$$
Step 3: List the advantages.
- High Efficiency: Fuel cells convert chemical energy directly to electrical energy, achieving an operational efficiency of around $70\%$. This is significantly higher than conventional thermal power plants ($40\%$), which waste energy turning heat into mechanical work.
- Eco-Friendly Operation: The only product of the reaction is water vapor, making the cell completely pollution-free. It was famously used to provide electricity and drinking water for astronauts in the Apollo space program.
- Continuous Supply: Unlike traditional batteries, fuel cells do not go dead over time. They continue to produce power indefinitely as long as fuel and oxidant are continuously supplied to the electrodes.
Extra Practice Problems (Exam-Style Matrix)
Prepare for high-level board assessments with these curated exam-style practice questions.
Section A: Multiple Choice Questions (MCQs)
Q1. If the standard reduction potential of an electrode is negative, it implies that:
A) It acts as a stronger reducing agent than hydrogen gas.
B) It is less reactive than hydrogen.
C) It undergoes reduction more easily than hydronium ions.
D) It cannot form a galvanic half-cell configuration.
Correct Answer Choice: A) It acts as a stronger reducing agent than hydrogen gas.
Explanation: A negative standard reduction potential means the species loses electrons more readily than hydrogen gas, making it a stronger reducing agent. (Difficulty Level: Easy)
Explanation: A negative standard reduction potential means the species loses electrons more readily than hydrogen gas, making it a stronger reducing agent. (Difficulty Level: Easy)
Q2. Which parameter increases when a solution of a weak electrolyte is diluted?
A) Specific Conductance ($\kappa$)
B) Molar Molarity ($M$)
C) Molar Conductivity ($\Lambda_m$)
D) Total Ionic Concentration Density
Correct Answer Choice: C) Molar Conductivity ($\Lambda_m$)
Explanation: Dilution increases the degree of dissociation ($\alpha$) for weak electrolytes, causing a significant increase in molar conductivity ($\Lambda_m$) that outweighs the increase in volume. (Difficulty Level: Easy)
Explanation: Dilution increases the degree of dissociation ($\alpha$) for weak electrolytes, causing a significant increase in molar conductivity ($\Lambda_m$) that outweighs the increase in volume. (Difficulty Level: Easy)
Section B: Assertion-Reason Questions
Directions: Choose the option that best fits the statements:
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Q3. Assertion (A): Copper metal cannot displace hydrogen gas from a dilute mineral acid solution.
Reason (R): Copper has a positive standard reduction potential ($E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$).
Correct Answer Choice: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Explanation: Because copper has a positive reduction potential, it is less reactive than hydrogen and cannot be oxidized by $\text{H}^+$ ions to release hydrogen gas. (Difficulty Level: Medium)
Explanation: Because copper has a positive reduction potential, it is less reactive than hydrogen and cannot be oxidized by $\text{H}^+$ ions to release hydrogen gas. (Difficulty Level: Medium)
Section C: Short Answer Questions
Q4. Write out the chemical reactions involved in the electrochemical mechanism of rusting/corrosion of iron.
Step 1: Identify the anodic site. Rusting behaves like an electrochemical cell on the surface of the metal. At a specific spot, iron is oxidized to ferrous ions:
$$\text{Anode reaction: } 2\text{Fe}(s) \rightarrow 2\text{Fe}^{2+} + 4e^- \quad (E^\circ = -0.44\text{ V})$$
Step 2: Identify the cathodic site. Electrons released at the anode travel through the metal to another spot on the surface. There, they reduce atmospheric oxygen in the presence of $\text{H}^+$ ions (derived from dissolved $\text{CO}_2$ or $\text{SO}_2$ in moisture): $$\text{Cathode reaction: } \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l) \quad (E^\circ = +1.23\text{ V})$$
Step 3: State the overall and subsequent rust formation steps. The combined cell reaction is: $$2\text{Fe}(s) + \text{O}_2(g) + 4\text{H}^+(aq) \rightarrow 2\text{Fe}^{2+}(aq) + 2\text{H}_2\text{O}(l)$$ The ferrous ions ($\text{Fe}^{2+}$) migrate and are further oxidized by atmospheric oxygen to form hydrated ferric oxide, which we know as rust: $$4\text{Fe}^{2+}(aq) + \text{O}_2(g) + 4\text{H}_2\text{O}(l) \rightarrow 2\text{Fe}_2\text{O}_3(s) + 8\text{H}^+(aq)$$ $$\text{Fe}_2\text{O}_3(s) + x\text{H}_2\text{O}(l) \rightarrow \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \text{ (Hydrated Rust Matrix)}$$ (Difficulty Level: Medium)
Step 2: Identify the cathodic site. Electrons released at the anode travel through the metal to another spot on the surface. There, they reduce atmospheric oxygen in the presence of $\text{H}^+$ ions (derived from dissolved $\text{CO}_2$ or $\text{SO}_2$ in moisture): $$\text{Cathode reaction: } \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l) \quad (E^\circ = +1.23\text{ V})$$
Step 3: State the overall and subsequent rust formation steps. The combined cell reaction is: $$2\text{Fe}(s) + \text{O}_2(g) + 4\text{H}^+(aq) \rightarrow 2\text{Fe}^{2+}(aq) + 2\text{H}_2\text{O}(l)$$ The ferrous ions ($\text{Fe}^{2+}$) migrate and are further oxidized by atmospheric oxygen to form hydrated ferric oxide, which we know as rust: $$4\text{Fe}^{2+}(aq) + \text{O}_2(g) + 4\text{H}_2\text{O}(l) \rightarrow 2\text{Fe}_2\text{O}_3(s) + 8\text{H}^+(aq)$$ $$\text{Fe}_2\text{O}_3(s) + x\text{H}_2\text{O}(l) \rightarrow \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \text{ (Hydrated Rust Matrix)}$$ (Difficulty Level: Medium)
Q5. Calculate the mass of aluminum deposited when a current of 5.0 A is passed through a molten $\text{Al}_2\text{O}_3$ bath for 20 minutes. (Atomic weight of $\text{Al} = 27\text{ u}$).
Step 1: Extract the parameters and convert time to seconds.
$$I = 5.0\text{ A}, \quad t = 20 \times 60 = 1200\text{ seconds}$$
Step 2: Determine the number of transferred electrons ($n$). Aluminum ion reduction follows: $$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \implies n = 3$$
Step 3: Apply Faraday's First Law. $$m = \frac{\text{Molar Mass}}{n \times 96500} \times I \times t$$ $$m = \frac{27}{3 \times 96500} \times 5.0 \times 1200 = \frac{9}{96500} \times 6000 = \frac{54000}{96500} \approx 0.559\text{ g}$$ The mass of aluminum deposited is **$0.559\text{ g}$**. (Difficulty Level: Medium)
Step 2: Determine the number of transferred electrons ($n$). Aluminum ion reduction follows: $$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \implies n = 3$$
Step 3: Apply Faraday's First Law. $$m = \frac{\text{Molar Mass}}{n \times 96500} \times I \times t$$ $$m = \frac{27}{3 \times 96500} \times 5.0 \times 1200 = \frac{9}{96500} \times 6000 = \frac{54000}{96500} \approx 0.559\text{ g}$$ The mass of aluminum deposited is **$0.559\text{ g}$**. (Difficulty Level: Medium)
Section D: Long Answer Problems
Q6. Given the following standard reduction half-cell potentials:
$\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \quad (E^\circ = -2.37\text{ V})$
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = +0.34\text{ V})$
1. Design a functional electrochemical cell using these components and write the balanced net reaction.
2. Calculate the cell potential at $298\text{ K}$ if $[\text{Mg}^{2+}] = 1.0 \times 10^{-3}\text{ M}$ and $[\text{Cu}^{2+}] = 1.0 \times 10^{-1}\text{ M}$.
Step 1: Identify the electrodes and calculate $E^\circ_{\text{cell}}$. The copper electrode has a higher reduction potential, so it acts as the cathode. The magnesium electrode acts as the anode.
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34 - (-2.37) = +2.71\text{ V}$$
Step 2: Write out the balanced net chemical reaction. $$\text{Mg}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu}(s) \quad (\text{where } n = 2)$$
Step 3: Apply the Nernst Equation to find the non-standard potential. $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}$$ $$E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log \left(\frac{1.0 \times 10^{-3}}{1.0 \times 10^{-1}}\right)$$ $$E_{\text{cell}} = 2.71 - 0.02955 \log(10^{-2})$$ $$\log(10^{-2}) = -2$$ $$E_{\text{cell}} = 2.71 - (0.02955 \times (-2)) = 2.71 + 0.0591 = 2.769\text{ V}$$ The standard cell potential is **$+2.71\text{ V}$** and the non-standard potential is **$2.769\text{ V}$**. (Difficulty Level: Hard)
Step 2: Write out the balanced net chemical reaction. $$\text{Mg}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu}(s) \quad (\text{where } n = 2)$$
Step 3: Apply the Nernst Equation to find the non-standard potential. $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}$$ $$E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log \left(\frac{1.0 \times 10^{-3}}{1.0 \times 10^{-1}}\right)$$ $$E_{\text{cell}} = 2.71 - 0.02955 \log(10^{-2})$$ $$\log(10^{-2}) = -2$$ $$E_{\text{cell}} = 2.71 - (0.02955 \times (-2)) = 2.71 + 0.0591 = 2.769\text{ V}$$ The standard cell potential is **$+2.71\text{ V}$** and the non-standard potential is **$2.769\text{ V}$**. (Difficulty Level: Hard)
Section E: Case-Based Multi-Layer Scenario
Q7. Read the paragraph below and answer the following questions: Measurement of electrical conductivity of electrolytic solutions faces two major technical obstacles. First, passing a direct current (DC) causes electrolysis, which alters the composition and concentration of the solution near the electrodes. Second, a standard column of liquid cannot be easily connected to a typical Wheatstone bridge instrument. To overcome these hurdles, chemists use an alternating current (AC) power source combined with a specially designed containment vessel called a conductivity cell. Once the cell constant ($G^* = \frac{l}{A}$) is determined using a standard calibrating reference solution like $\text{KCl}$, it can be used to measure the properties of any target electrolyte.
1. Why is an alternating current (AC) source used instead of direct current (DC) when measuring solution resistance?
2. If the resistance of a conductivity cell filled with $0.05\text{ M }\text{KCl}$ is $100\ \Omega$ and the cell constant is $0.5\text{ cm}^{-1}$, calculate the conductivity ($\kappa$) and molar conductivity ($\Lambda_m$) of the solution.
Step 1: Answer part 1. An alternating current (AC) source constantly reverses its direction, which prevents net electrolysis and keeps the concentration of the solution near the electrodes uniform.
Step 2: Calculate conductivity ($\kappa$) for part 2. Using the relationship between resistance and cell constant: $$\kappa = \frac{\text{Cell Constant } (G^*)}{R} = \frac{0.5\text{ cm}^{-1}}{100\ \Omega} = 0.005\text{ S cm}^{-1} = 5.0 \times 10^{-3}\text{ S cm}^{-1}$$
Step 3: Calculate the molar conductivity ($\Lambda_m$). $$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{5.0 \times 10^{-3} \times 1000}{0.05} = \frac{5.0}{0.05} = 100\text{ S cm}^2\text{ mol}^{-1}$$ The conductivity is **$5.0 \times 10^{-3}\text{ S cm}^{-1}$** and the molar conductivity is **$100\text{ S cm}^2\text{ mol}^{-1}$**. (Difficulty Level: Hard)
Step 2: Calculate conductivity ($\kappa$) for part 2. Using the relationship between resistance and cell constant: $$\kappa = \frac{\text{Cell Constant } (G^*)}{R} = \frac{0.5\text{ cm}^{-1}}{100\ \Omega} = 0.005\text{ S cm}^{-1} = 5.0 \times 10^{-3}\text{ S cm}^{-1}$$
Step 3: Calculate the molar conductivity ($\Lambda_m$). $$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{5.0 \times 10^{-3} \times 1000}{0.05} = \frac{5.0}{0.05} = 100\text{ S cm}^2\text{ mol}^{-1}$$ The conductivity is **$5.0 \times 10^{-3}\text{ S cm}^{-1}$** and the molar conductivity is **$100\text{ S cm}^2\text{ mol}^{-1}$**. (Difficulty Level: Hard)
Common Mistakes Students Make
- Incorrect Identification of the Anode vs. Cathode: Do not just assume the left electrode is always the anode. Compare their standard reduction potentials ($E^\circ$). The electrode with the lower, more negative reduction potential is always the anode in a galvanic cell.
- Squaring Concentrations in the Nernst Equation: Remember to raise the concentrations in the log term to the power of their stoichiometric coefficients from the balanced net cell equation (for example, $[\text{Ag}^+]^2$).
- Unit Mismatch with Molar Conductivity: When using the formula $\Lambda_m = \frac{\kappa \times 1000}{C}$, ensure your conductivity ($\kappa$) is in $\text{S cm}^{-1}$ and concentration ($C$) is in $\text{mol L}^{-1}$. If $\kappa$ is given in SI units ($\text{S m}^{-1}$), use the alternative formula $\Lambda_m = \frac{\kappa}{1000 \times C}$.
Exam Preparation Tips
- Master Sign Conventions: Remember that in a Galvanic cell, the anode is negative and the cathode is positive. In an Electrolytic cell, the anode is positive and the cathode is negative. In both types of cells, oxidation always occurs at the anode and reduction always occurs at the cathode.
- Practice Log and Antilog Math: The Nernst equation and equilibrium constant calculations require working with logs. Practice estimating these values to save time during exams.
- Review Battery Reactions: Make sure you can write the balanced equations for commercial cells, especially the lead storage battery and the $\text{H}_2-\text{O}_2$ fuel cell, as these are frequently asked.
Frequently Asked Questions (FAQs)
1. Is Electrochemistry a high-weightage chapter for the CBSE Class 12 board exam?
Yes, it is a high-weightage chapter in Class 12 Physical Chemistry, typically contributing around 7 to 8 marks. It includes both theoretical definitions and numerical problems based on the Nernst Equation and Faraday's Laws.
2. What is the difference between an electrochemical cell and an electrolytic cell?
An electrochemical cell (Galvanic/Voltaic cell) converts chemical energy from a spontaneous redox reaction into electrical energy. An electrolytic cell uses external electrical energy to drive a non-spontaneous chemical redox reaction.
3. What does Kohlrausch's Law of Independent Migration of Ions state?
Kohlrausch's Law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of its constituent anions and cations at infinite dilution.
4. Why does conductivity decrease while molar conductivity increases with dilution?
Conductivity is the conductance of a unit volume of solution. On dilution, the number of current-carrying ions per unit volume decreases, lowering conductivity. However, molar conductivity increases because the total volume containing one mole of electrolyte increases significantly, outweighing the decrease in ion density.
5. Where can I find official sample papers for CBSE Class 12?
You can download them along with their official marking schemes directly from the CBSE Academic portal.
Conclusion: Mastering Electrochemistry comes down to understanding electron flow and balance. Focus on the relationship between standard potentials and thermodynamics rather than just memorizing equations. Keep practicing these calculations regularly. You've got this, Lucky!
More Class 12 Chemistry Chapters
Ch 1: The Solid State
Ch 2: Solutions
Ch 3: Electrochemistry
Ch 4: Chemical Kinetics
Ch 5: Surface Chemistry
Ch 6: Isolation of Elements
Ch 7: The p-Block Elements
Ch 8: The d- & f-Block Elements
Ch 9: Coordination Compounds
Ch 10: Haloalkanes
Ch 11: Alcohols & Phenols
Ch 12: Aldehydes & Ketones
Ch 13: Amines
Ch 14: Biomolecules
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