Class 12 Chemistry Chapter 2 Electrochemistry

Key Concept: Reaction Equations

b) $\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-$
[Solution Description]
The Daniell cell involves the redox reaction where zinc metal is oxidized to $\text{Zn}^{2+}$ ions and copper ions are reduced to copper metal. The oxidation half-reaction occurs at the anode and can be represented as:

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

Therefore, the correct answer involves the conversion of solid zinc to zinc ions with the release of electrons.

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Key concept: Basic Concept

d) All of the above
[Solution Description] The conductivity of an electrolytic solution is influenced by several factors including the nature of the electrolyte, the size of the ions in solution, solvent viscosity, concentration of the electrolyte, and temperature. These factors affect how easily ions can move through the solution, thereby affecting its conductivity.

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Key Concepts: Basic Reaction Identification

c) $\text{NaOH}, \ \text{Cl}_2, \ \text{H}_2$

[Solution Description]

During the electrolysis of an aqueous solution of NaCl, at the cathode, water is reduced to produce hydrogen gas and hydroxide ions because it has a higher reduction potential compared to sodium ions. At the anode, chlorine gas is produced by the oxidation of chloride ions since it has a lower overpotential than oxygen evolution from water.

Therefore, the overall products are $\text{NaOH}, \text{Cl}_2, \text{H}_2$.

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.

Key Concepts: Reaction Balancing

d) $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

[Solution Description]

Consider the unbalanced reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.

The half-reactions are:

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

Reduction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

To balance the electrons, we add them such that two electrons from oxidation cancel with those used in reduction:

$$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$$

This equation is already balanced.

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Key concept: Basic Formula Application

a) $0.015 \, S \, m^2 \, mol^{-1}$
[Solution Description]
To find the molar conductivity, we use the formula for molar conductivity:

$$
Λ_m = \frac{κ}{c}
$$

where $κ$ is the conductivity and $c$ is the concentration.

Substituting the given values:

$$
Λ_m = \frac{0.150 \, S \, m^{-1}}{10 \, mol \, m^{-3}}
$$

Doing the division:

$$
Λ_m = 0.015 \, S \, m^2 \, mol^{-1}
$$

Therefore, the molar conductivity is $0.015 \, S \, m^2 \, mol^{-1}$.

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Key Concept: Basic Concept

d) Size of the container
[Solution Description] Conductivity of an ionic solution is influenced by several factors including the nature of the electrolyte, concentration, and temperature. The size of the container does not affect the conductivity.

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Key Concepts: Comparative Analysis

d) HCl (Hydrochloric acid)

[Solution Description] Strong electrolytes dissociate completely in solution, providing more ions that contribute to higher molar conductivity. Among the given options, HCl is a strong acid that dissociates fully, leading to high molar conductivity compared to other weak acids or bases.

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Key Concepts: Intermediate Calculation

a) 4.02 g
[Solution Description]
The amount of silver deposited can be calculated using Faraday’s first law of electrolysis. First, we calculate the quantity of electricity passed through:

$$ Q = I \times t $$

Here, $ I = 3 \, \text{A} $ and $ t = 20 \times 60 \, \text{s} = 1200 \, \text{s} $.

$$
Q = 3 \times 1200 = 3600 \, \text{C}
$$

Next, applying Faraday’s law, we use the formula:

Number of moles of electrons = $\frac{Q}{F}$, where $ F = 96500 \, \text{C/mol} $.

$$
\text{Moles of electrons} = \frac{3600}{96500} \approx 0.0373 \, \text{mol}
$$

According to the stoichiometry of the reaction $ Ag^+ + e^- \rightarrow Ag(s) $, 1 mole of electrons deposits 1 mole of silver.

Hence, moles of silver deposited = 0.0373 mol

Finally, the mass of silver deposited is given by:

$$
\text{Mass} = \text{moles} \times \text{atomic mass} = 0.0373 \times 107.9 = 4.0197 \, \text{g}
$$

Thus, approximately 4.02 g of silver is deposited.

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Key Concepts: Basic Reaction

a) $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
[Solution Description] During the electrolysis of copper sulfate $(CuSO_4)$ solution, copper ions $(Cu^{2+})$ are reduced at the cathode. The half-reaction occurring at the cathode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. This deposition results in the formation of pure copper metal.

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Key Concept: Half-Reaction Identification

a) $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$
[Solution Description] In a Daniell cell, zinc metal is oxidized to form zinc ions, with the release of electrons. The oxidation half-reaction occurs at the anode and involves the conversion:

$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$.

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Key Concepts: Advanced Equilibrium Concepts, Second Order Thinking

d) Assertion is false, but Reason is true.

[Solution Description]

To solve this problem, you need to understand both electrochemical equilibria and the application of the Nernst equation.

Assertion: At equilibrium, the cell potential $E(cell) = 0$, not the standard cell potential $E^\circ_{cell}$. When at equilibrium, the reaction quotient $Q$ equals the equilibrium constant $K_c$, so the assertion that the Nernst equation can directly provide $E^\circ_{cell}$ is incorrect without additional information. Therefore, the assertion is false.

Reason: This statement correctly identifies the relationship between the standard cell potential and the equilibrium constant. The formula given relates these two quantities accurately. Hence, the reason is true.

Based on this analysis, the correct response is that the assertion is false, but the reason is true.

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Key Concepts: Cell Potential Calculation

b) 0.89 V

[Solution Description]

To find the standard cell potential $(E^\circ_{\text{cell}})$, use the formula:

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

Substituting the given values:

$$E^\circ_{\text{cell}} = 1.23 \, \text{V} – 0.34 \, \text{V} = 0.89 \, \text{V}$$

Therefore, the standard cell potential is 0.89 V.

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Key Concepts: Battery Components

a) $\text{Lead} (\text{Pb})$
[Solution Description] In a lead storage battery, the anode is made up of lead $\text{Lead} (\text{Pb})$.

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Key concept: Basic Definitions

a) Ionic conductance
[Solution Description] The term used to describe the conductance of electricity through ions present in a solution is “ionic conductance.” This means that the ions in the solution facilitate the movement of electrical current.

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Key Concepts: Basic Formula Application

c) 1800 C
[Solution Description]
The formula to calculate the quantity of electricity $(Q)$ is:
$$
Q = I \times t
$$
where $I$ is the current in amperes and $t$ is the time in seconds.
Converting 10 minutes into seconds:
$$
t = 10 \times 60 = 600 \, \text{s}
$$
Substitute $I = 3 \, \text{A}$ and $t = 600 \, \text{s}$ into the equation:
$$
Q = 3 \times 600 = 1800 \, \text{C}
$$
So, the quantity of electricity passed is 1800 coulombs.

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Key Concepts: Corrosion Definition

b) An electrochemical phenomenon
[Solution Description] Corrosion is primarily an electrochemical phenomenon where metals deteriorate due to reactions with environmental components such as oxygen and moisture. This process typically involves redox reactions.

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Key Concepts: Basic Concept

b) It states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation.
[Solution Description] Kohlrausch’s Law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation.

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Key Concepts: Standard Potential Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To solve this Assertion-Reason question, we need to analyze both statements separately and determine their relationship.

Starting with the assertion: “The equilibrium constant $K_c$
for a cell reaction can be calculated from the standard cell potential using the equation $E^\circ_{\text{cell}} = \frac{0.059}{n} \log K_c$.” This statement is true and aligns perfectly with the Nernst equation’s relationship between standard cell potential and the equilibrium constant. Now, analyzing the reason: “A negative standard cell potential indicates that the reactants are favored over products at equilibrium.” This statement is also true. When the standard cell potential $E^\circ_{\text{cell}}$ is negative, it implies that the Gibbs free energy change $\Delta G^\circ$ is positive, indicating non-spontaneity and favoring reactants at equilibrium. Hence, the reason correctly explains why the given assertion holds true in terms of chemical thermodynamics.

Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation for the Assertion.

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Key Concepts: Rust Composition

b) $Fe_2O_3 \cdot xH_2O$
[Solution Description]
Rust is a compound that results from the corrosion of iron. According to the syllabus, the chemical composition of rust is represented as hydrated ferric oxide:
$$
Fe_2O_3 \cdot xH_2O
$$
This shows that rust consists of ferric oxide with variable amounts of water.

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Key concept: Complex Reaction Analysis, Electrode Potential Comparison

b) Pb
[Solution Description]
In a galvanic cell, the reducing agent is typically the species that gets oxidized at the anode (loses electrons).

Considering the given half-reactions, the reduction potential for $\mathrm{Pb^{2+}/Pb}$ is lower than that for $\mathrm{Sn^{4+}/Sn^{2+}}$, meaning $\mathrm{Pb}$ will act as the reducing agent since it will be oxidized while providing electrons to reduce $\mathrm{Sn^{4+}}$. Hence, $\mathrm{Pb}$ will get oxidized:

$$
\mathrm{Pb(s) \rightarrow Pb^{2+} + 2e^-}
$$

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Key Concepts: Electrode Potential and Gibbs Energy, Equilibrium Constant from Cell Reaction

c) $2.09 \times 10^{37}$
[Solution Description]
To find the cell potential $E^\circ_{\text{cell}}$, use:
$$
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}
= 0.34 \, \text{V} – (-0.76 \, \text{V})
= 1.10 \, \text{V}
$$
The relation between $E^\circ_{\text{cell}}$ and the equilibrium constant $K$ is given by:
$$
\Delta G^\circ = -nFE^\circ_{\text{cell}} = -RT \ln K
$$
Rearranging gives:
$$
K = e^{\frac{-nFE^\circ_{\text{cell}}}{RT}}
$$
Where $n = 2$, $F = 96500 \, \text{C/mol}$, $R = 8.314 \, \text{J/(mol cdot K)}$, and $T = 298 \, \text{K}$.

First, calculate:
$$
-nFE^\circ_{\text{cell}} = -2 \times 96500 \times 1.10 = -212300 \, \text{J/mol}
$$
Then,
$$
K = e^{\frac{212300}{8.314 \times 298}} = e^{86.03}
\approx 2.09 \times 10^{37}
$$
So, the equilibrium constant $K$ is approximately $2.09 \times 10^{37}$.

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Key concept: Basic Definitions

d) Ohm meter $(\Omega m)$
[Solution Description] The SI unit of resistivity is Ohm meter, denoted by $\Omega$ m. It represents the resistance to electric current flow per unit length and cross-sectional area of a material.

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Key Concepts: Reaction Balancing

c) $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

[Solution Description]

The overall reaction in a hydrogen-oxygen fuel cell involves the combination of hydrogen gas and oxygen gas to produce water. The given reaction is $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$, which is already balanced. Thus, the correct balanced overall reaction is $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$.

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Key Concepts: Cell Potential Calculation

a) 0.34 V
[Solution Description]
To find the cell potential $(E_{cell})$ of the galvanic cell, we need to use the formula:
$$
E_{cell} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}
$$
The standard hydrogen electrode (SHE) has an electrode potential of 0 V. Since the copper half-cell is likely the cathode due to its positive electrode potential, we have:
$$
E_{cell} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} – E^\circ_{\text{SHE}}
$$
Substituting the values:
$$
E_{cell} = 0.34 \,V – 0\,V = 0.34 \,V
$$
Therefore, the cell potential is 0.34 V.

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Key Concept: Electrode Process

d) Copper dissolves
[Solution Description] When $E_{ext} > 1.1 \, \text{V}$, the Daniell cell operates as an electrolytic cell. In this reversed process, the normal cathodic reaction $(Cu^{2+} + 2e^- \rightarrow Cu(s)) $ reverses. Therefore, instead of reduction, copper dissolves by oxidation, reverting to $Cu^{2+}$ ions.

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Key Concepts: Advanced Reaction Analysis, Non-Straightforward Path

b) Maintain $E_{ext} < 1.15V$. [Solution Description] According to the Nernst equation for the Daniell cell, the cell potential becomes dependent on ion concentrations. The equation is given by: $$ E = 1.1 - \frac{0.059}{2} \log \left(\frac{[Zn^{2+}]}{[Cu^{2+}]}\right). $$ With concentrations 0.01 M for $Zn^{2+}$ and 0.1 M for $Cu^{2+}$, substituting these into the equation yields a higher potential due to the logarithmic term. Hence, for galvanic function, $E_{ext}$ should be less than this calculated potential of 1.15 V. Thus, operating with $E_{ext} < 1.15V$ maintains the cell in its galvanic mode. Click Here to Download Notes

Key concept: Definition Recall

b) $\Delta_rG = -nFE_{\text{cell}}$
[Solution Description] The relationship between Gibbs free energy change $(\Delta_rG)$ and cell potential $(E_{\text{cell}})$ is given by the equation $\Delta_rG = -nFE_{\text{cell}}$. This indicates that the Gibbs free energy change is directly proportional to the negative of the product of the number of moles of electrons transferred, Faraday’s constant, and the cell potential.

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Key concept: Basic Concept

b) Ohm meter
[Solution Description]
Resistivity ($\rho$) is a measure of how strongly a material opposes the flow of electric current. The SI unit for resistivity is ohm meter $(\Omega \cdot m)$.

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Key Concepts:  Cell Potential Calculation

c) 3.17 V

[Solution Description] In this problem, we have to calculate the emf of the cell using the standard electrode potentials for the cathode $(E^\circ_{\text{right}})$ and anode $(E^\circ_{\text{left}})$. The oxidation occurs at the anode, which means magnesium will oxidize, and reduction occurs at the cathode, where silver ions are reduced.

The formula for calculating the emf is:

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

Substitute the given values:

$$E_{\text{cell}} = 0.80 \, \text{V} – (-2.37 \, \text{V})$$

Perform the calculation:

$$E_{\text{cell}} = 0.80 \, \text{V} + 2.37 \, \text{V} = 3.17 \, \text{V}$$

Therefore, the emf of the cell is 3.17 V.

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Key Concept: Electrode Potential

b) The potential difference between an electrode and its electrolyte under standard conditions
[Solution Description] Standard electrode potential measures the potential difference between an electrode and its electrolyte under standard conditions (1 M concentration for solutions and 1 atm pressure for gases).

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Key concept: Multi-Step Problem Solving, Thermodynamic Properties

a) 1
[Solution Description]
Using the equation to relate standard cell potential to equilibrium constant:
$$
E^\circ_{cell} = \frac{0.059}{n} \log K_c
$$
Substitute $E^\circ_{cell} = 0.00$ V and $n = 2$:
$$
0.00 = \frac{0.059}{2} \log K_c \implies \log K_c = 0
$$
Therefore, rearranging gives:
$$
K_c = 10^0 = 1
$$
So, $K_c = 1$.

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Key Concepts: Faraday’s Constant Value

d) 96500 C/mol
[Solution Description]
Faraday’s constant represents the amount of electric charge per mole of electrons. It is denoted by $F$ and is approximately:

$$
F = 96500 \ \text{C/mol}
$$

This value is crucial for calculations involving electrochemical reactions.

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Key Concepts: Sacrificial Anode

b) Magnesium
[Solution Description] Magnesium is often used as a sacrificial anode in electrochemical methods to prevent corrosion due to its higher reactivity compared to other metals like gold or silver, which allows it to oxidize instead of the protected metal.

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Key Concept: Advanced Applications, Electrochemical Efficiency

a) The Hall-Héroult process for aluminum
[Solution Description]
To achieve energy efficiency in metal extraction through electrochemical methods, one can use electrolytic reduction processes where non-spontaneous reactions are driven by electrical energy. This is common for metals that cannot be extracted from their ores via thermal means due to high melting points or reactivity with other elements. Aluminum extraction using the Hall-Héroult process exemplifies this. In this process, aluminum oxide is dissolved in molten cryolite and then reduced electrochemically at a much lower temperature than would be required if relying on direct heating.

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Key Concepts: Law Identification

d) The amounts of different substances liberated are proportional to their chemical equivalent weights.
[Solution Description]
Faraday’s Second Law of Electrolysis states that the amounts of different substances liberated by the same quantity of electricity passed through their solutions are proportional to their chemical equivalent weights. This means that if you pass the same charge through two different electrolyte solutions, the mass of substance deposited or dissolved will be directly related to its equivalent weight.

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Key Concepts: Reaction Direction

c) Towards the cathode
[Solution Description] In an electrolytic cell, cations are positively charged ions. Due to their positive charge, they are attracted to the negatively charged electrode, which is the cathode. Therefore, cations move towards the cathode during electrolysis.

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Key Concept: Electrolytic vs. Electronic Conductivity, Measurement of Conductivity

b) 0.2 S/m
[Solution Description] Conductivity is the inverse of resistivity and is given by the formula:
$$\begin{equation}
\kappa = \frac{1}{\rho}
\end{equation}$$
Given the resistivity $ \rho = 5 \, \Omega \cdot m $, substitute into the equation:
$$\begin{equation}
\kappa = \frac{1}{5} = 0.2 \, S/m
\end{equation}$$
Therefore, the conductivity of the solution is $0.2 \, \text{S/m}$.

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Key Concept: Cell Representation

b) 1.10 V
[Solution Description]
The standard cell potential $E_{cell}^\circ$ for a galvanic cell can be calculated using the formula:
$$
E_{cell}^\circ = E_{cathode}^\circ – E_{anode}^\circ
$$
Here, copper acts as the cathode and zinc as the anode.

Substituting the given values:
$$
E_{cell}^\circ = 0.34 \, \text{V} – (-0.76 \, \text{V})
$$
$$
E_{cell}^\circ = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V}
$$

Therefore, the cell potential is $1.10 \, \text{V}$.

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Key Concept: Cell Representation

c) Salt bridge
[Solution Description] In a galvanic cell representation, the salt bridge is typically placed between the anode and the cathode. It completes the circuit and allows the flow of ions to maintain charge balance.

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Key concept: Electrode Identification

a) Copper electrode
[Solution Description] In a Daniell cell, the cathode is where copper ions gain electrons and are reduced to solid copper. Therefore, the copper electrode acts as the cathode.

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Key Concepts: Quantitative Calculation, Faraday’s Law Application

b) 2.3 g
[Solution Description]
The amount of charge $(Q)$ is calculated using $Q = I \cdot t$, where $I = 5$ A and $t = 1930$ s.
$$
Q = 5 \times 1930 = 9650 \text{ C}
$$
According to Faraday’s first law, the moles of sodium deposited $(n)$ can be calculated as:
$$
n = \frac{Q}{F} = \frac{9650}{96500} = 0.1 \text{ mol}
$$
The mass of sodium is given by $m = n \times \text{atomic mass of Na}$:
$$
m = 0.1 \times 23 = 2.3 \text{ grams}
$$

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Key Concepts: Comparative Analysis, Theoretical vs Practical

d) Assertion is false, but Reason is true.

[Solution Description]

The assertion states that the limiting molar conductivity $\Lambda_m^0$ for weak electrolytes can be found using extrapolation of $\Lambda_m$ against $c^{1/2}$. This technique is typically used for strong electrolytes due to their complete dissociation, but not directly applied to weak electrolytes. Instead, for weak electrolytes, $\Lambda_m^0$ is derived from Kohlrausch’s law by considering individual ionic contributions.

The reason correctly points out that weak electrolytes do not fully dissociate except at infinite dilution, creating practical difficulties in measuring $\Lambda_m$ accurately. While this statement about weak electrolyte behavior is true, it does not support or explain the incorrect method mentioned in the assertion for determining $\Lambda_m^0$ in practice.

Therefore, while both statements contain elements of truth, they are unrelated in terms of explanation.

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Key Concepts: Gibbs Energy Calculation

a) $-150.64 \, \text{kJ mol}^{-1}$

[Solution Description] The formula to calculate standard Gibbs energy change is $\Delta G^\circ = -nFE^\circ_{\text{cell}}$. Substituting the given values:

$$\Delta G^\circ = -(2)(96487 \, \text{C} \, \text{mol}^{-1})(0.78 \, \text{V})$$

$$= -150643.08 \, \text{J mol}^{-1}$$

Converting Joules to kilojoules:

$$= -150.64 \, \text{kJ mol}^{-1}$$

Thus, the standard Gibbs energy change is $-150.64 \, \text{kJ mol}^{-1}$.

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Key Concepts: Cathode Reaction

b) $\text{O}_2 (g) + 4 \text{H}^+ (aq) + 4 e^- \rightarrow 2 \text{H}_2 \text{O} (l)$

[Solution Description] The cathodic reaction during the rusting of iron involves the reduction of oxygen in acidic conditions: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$. This reaction occurs simultaneously with the anodic oxidation of iron.

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Key Concepts: Graph Interpretation, Complex Calculation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] For weak electrolytes, the molar conductivity $(\Lambda_m)$ increases as the solution is diluted because the degree of dissociation increases, leading to more ions being available per mole of electrolyte. This enhances the ionic mobility and thus increases $\Lambda_m$. Therefore, both the assertion and reason are true. Moreover, the explanation in the reason directly accounts for the increase observed in the assertion due to increased ionization at lower concentrations.

When a weak electrolyte is diluted, its dissociation into ions increases according to Ostwald’s dilution law. At high dilutions $(low c)$, almost complete dissociation occurs, reaching a state closer to $\Lambda_m^0$, hence increasing molar conductivity.

Therefore, the correct answer is that both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

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