GSEB Class 12 Chemistry Solutions Chapter 3 Electrochemistry

GSEB Class 12 Chemistry Electrochemistry InText Questions and Answers

Question 1. How would you determine the standard electrode potential of the system \( \text{Mg}^{2+}|\text{Mg} \)?
Answer: The standard electrode potential, \(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}\), can be ascertained by constructing an electrochemical cell that includes a standard hydrogen electrode (SHE). The cell configuration would be represented as: \( \text{Mg(s) | Mg}^{2+}\text{(1 M) || H}^{+}\text{(g) | H}_2\text{ (1 atm) Pt.} \)
In simple words: To find the standard potential of magnesium, pair it with a standard hydrogen electrode and measure the voltage of the resulting cell.
🎯 Exam Tip: Understanding cell notation and its components is crucial. Remember to use standard concentrations and pressures when determining standard electrode potentials.

Question 2. Can you store copper sulphate solutions in a zinc pot?
Answer: Storing copper sulfate solution in a zinc pot is not advisable. The standard electrode potential for copper, \(\mathrm{E}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^{\circ}\), is \(+0.34 \text{ V}\), while for zinc, \(\mathrm{E}_{\mathrm{Zn}^{2+}/\mathrm{Zn}}^{\circ}\), it is \(-0.76 \text{ V}\). Since zinc has a lower (more negative) electrode potential compared to copper, it is more reactive and will undergo oxidation. Consequently, the zinc pot would react with the copper sulfate solution, displacing copper ions and leading to the corrosion of the pot.
In simple words: No, because zinc is more reactive than copper and will corrode, preventing storage of copper sulfate solution.
🎯 Exam Tip: Compare the standard electrode potentials to predict the feasibility of a redox reaction. A more reactive metal (lower electrode potential) will displace ions of a less reactive metal from its salt solution.

Question 3. Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions?
Answer: Only chemical species possessing a standard reduction potential greater than \(+0.77\text{ V}\) are capable of oxidizing ferrous ions (\(\text{Fe}^{2+}\)). Therefore, appropriate oxidizing agents for ferrous ions would include fluorine (\(\text{F}_2\)), chlorine (\(\text{Cl}_2\)), and bromine (\(\text{Br}_2\)).
In simple words: Look for substances with a higher reduction potential than ferrous ions to oxidize them; examples are fluorine, chlorine, and bromine.
🎯 Exam Tip: For oxidation to occur, the oxidizing agent must have a standard reduction potential higher than the species being oxidized. This indicates a stronger tendency to gain electrons.

Question 4. Calculate the potential of the hydrogen electrode in contact with a solution whose pH is 10.
Answer: The reduction half-reaction for a hydrogen electrode is: \( \text{H}^{+} + \text{1e}^{-} \rightarrow \frac{1}{2}\text{H}_2 \)
The Nernst equation for this electrode potential is:
\( E = \text{E}^{\circ} - \frac{0.0591}{n} \log \frac{1}{\left[\mathrm{H}^{+}\right]} \)
Given that the standard electrode potential for hydrogen, \(\text{E}^{\circ}\), is \(0\text{ V}\) and \(n=1\). Also, \(- \log \left[\mathrm{H}^{+}\right] = \text{pH}\).
So, the equation simplifies to:
\( E = \text{E}^{\circ} - \frac{0.0591}{1} \text{ pH} \)
Substituting the given values:
\( E = 0 - 0.0591 \times 10 \)
\( E = -0.591 \text{ V} \)
In simple words: Using the Nernst equation and the given pH of 10, the hydrogen electrode potential is calculated to be -0.591 V.
🎯 Exam Tip: Remember that pH is related to hydrogen ion concentration, which directly impacts electrode potential calculations via the Nernst equation. For a standard hydrogen electrode, \(\text{E}^{\circ}\) is always 0.

Question 5. Calculate the emf of the cell in which the following reaction takes place
\( \text{Ni(s)} + \text{2Ag}^{+}\text{ (0.002 M)} \rightarrow \text{Ni}^{2+}\text{ (0.160 M)} + \text{2Ag(s)} \)
Given that \(\text{E}_{\text{cell}}^{\circ} = 1.05 \text{ V}\)
Answer: The Nernst equation for the cell potential is given by:
\( E = \text{E}_{\text{cell}}^{\circ} - \frac{0.0591}{n} \log \frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \)
In this reaction, \(n=2\) (as two electrons are transferred).
Substituting the given values:
\( E = 1.05 - \frac{0.0591}{2} \log \frac{(0.16)}{(0.002)^2} \)
\( E = 1.05 - \frac{0.0591}{2} \log \frac{0.16}{0.000004} \)
\( E = 1.05 - \frac{0.0591}{2} \log (40000) \)
\( E = 1.05 - \frac{0.0591}{2} \times 4.602 \)
\( E = 1.05 - 0.02955 \times 4.602 \)
\( E = 1.05 - 0.136 \)
\( E = 0.914 \text{ V} \) (rounding to three decimal places from the provided \(0.915 \text{ V}\))
In simple words: Using the Nernst equation, we calculate the cell's electromotive force (EMF) by subtracting a concentration-dependent term from its standard EMF.
🎯 Exam Tip: Pay close attention to the stoichiometry of the reaction to correctly determine 'n' and the exponents in the concentration term of the Nernst equation.

Question 6. The cell in which the following reaction occurs:
\( \text{2Fe}^{3+}\text{(aq)} + \text{2I}^{-}\text{(aq)} \rightarrow \text{2Fe}^{2+}\text{(aq)} + \text{I}_{2}\text{(s)} \)
has \(\text{E}_{\text{cell}}^{\circ} = 0.236 \text{ V at 298 K}\). Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction?
Answer: To calculate the standard Gibbs energy (\(\Delta\text{G}^{\circ}\)):
\( \Delta\text{G}^{\circ} = -\text{nFE}_{\text{cell}}^{\circ} \)
Given \(n=2\) (two electrons transferred in the reaction), \(F = 96500 \text{ C mol}^{-1}\), and \(\text{E}_{\text{cell}}^{\circ} = 0.236 \text{ V}\).
\( \Delta\text{G}^{\circ} = -2 \times 96500\text{ C} \times 0.236\text{ V} \)
\( \Delta\text{G}^{\circ} = -45548 \text{ J mol}^{-1} \)
To calculate the equilibrium constant (\(\text{K}_{\text{c}}\)):
The relationship between standard cell potential and equilibrium constant is:
\( \text{E}_{\text{cell}}^{\circ} = \frac{0.059}{n} \log \text{K}_{\text{c}} \)
Rearranging to find \(\log \text{K}_{\text{c}}\):
\( \log \text{K}_{\text{c}} = \frac{n\text{E}_{\text{cell}}^{\circ}}{0.0591} \)
\( \log \text{K}_{\text{c}} = \frac{2 \times 0.236}{0.0591} \)
\( \log \text{K}_{\text{c}} = \frac{0.472}{0.0591} \)
\( \log \text{K}_{\text{c}} \approx 7.986 \approx 8 \)
Therefore, \(\text{K}_{\text{c}} = \text{antilog (8)} \)
\( \text{K}_{\text{c}} = 10^8 \)
In simple words: The standard Gibbs energy is calculated from the cell potential and electron transfer, while the equilibrium constant is determined using the relationship between cell potential and log K.
🎯 Exam Tip: Remember the sign convention for Gibbs energy: a negative \(\Delta\text{G}^{\circ}\) indicates a spontaneous reaction. Also, be precise with constants and the number of electrons transferred ('n').

Question 7. Why does the conductivity of a solution decrease with dilution?
Answer: The conductivity of a solution diminishes upon dilution because the concentration of ions per unit volume (\(\text{cm}^3\)) of the solution decreases. As the total number of charge-carrying ions in a given volume reduces, the solution's ability to conduct electricity is lowered.
In simple words: Conductivity drops with dilution because there are fewer ions per unit volume to carry electric charge.
🎯 Exam Tip: Differentiate between conductivity (specific conductance) and molar conductivity. While conductivity decreases with dilution, molar conductivity generally increases due to increased ionic mobility.

Question 8. Suggest a way to determine the \(\Lambda_{\mathrm{m}}^{\circ}\) value of water?
Answer: The limiting molar conductivity of water (\(\Lambda_{\text{m(H}_2\text{O})}^{\circ}\)) can be calculated using Kohlrausch's law of independent migration of ions. This law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. Since water is a weak electrolyte, its \(\Lambda_{\text{m}}^{\circ}\) cannot be directly determined by extrapolation from a conductivity plot. Instead, it can be derived from the \(\Lambda_{\text{m}}^{\circ}\) values of strong electrolytes that share its constituent ions. For example:
\( \Lambda_{\text{m(H}_2\text{O})}^{\circ} = \Lambda_{\text{m(NaOH})}^{\circ} + \Lambda_{\text{m(HCl})}^{\circ} - \Lambda_{\text{m(NaCl})}^{\circ} \)
In simple words: We can find water's molar conductivity at infinite dilution by adding the molar conductivities of strong acids and bases and subtracting that of their salt (like NaOH + HCl - NaCl).
🎯 Exam Tip: Kohlrausch's law is particularly useful for determining the limiting molar conductivity of weak electrolytes, which cannot be found by direct extrapolation from conductivity versus concentration graphs.

Question 9. The molar conductivity of \(0.025 \text{ mol L}^{-1}\) methanoic acid is \(46.1 \text{ S cm}^{2}\text{mol}^{-1}\). Calculate its degree of dissociation and dissociation constant. Given \( \lambda_{\left(\mathrm{H}^{+}\right)}^{\circ} = 349.6 \text{ S cm}^{2}\text{ mol}^{-1} \) and \( \lambda_{\left(\mathrm{HCOO}^{-}\right)}^{\circ} = 54.6 \text{ S cm}^{2}\text{mol}^{-1} \).
Answer: First, calculate the limiting molar conductivity of methanoic acid (\(\text{HCOOH}\)) using Kohlrausch's Law:
\( \Lambda_{\text{m(HCOOH})}^{\circ} = \lambda_{\left(\mathrm{H}^{+}\right)}^{\circ} + \lambda_{\left(\mathrm{HCOO}^{-}\right)}^{\circ} \)
\( \Lambda_{\text{m(HCOOH})}^{\circ} = 349.6 + 54.6 = 404.2 \text{ S cm}^{2}\text{mol}^{-1} \)
Next, calculate the degree of dissociation (\(\alpha\)):
\( \alpha = \frac{\Lambda_{\text{m}}^{\text{c}}}{\Lambda_{\text{m}}^{\circ}} \)
Given molar conductivity at concentration 'c' (\(\Lambda_{\text{m}}^{\text{c}}\)) is \(46.1 \text{ S cm}^{2}\text{mol}^{-1}\).
\( \alpha = \frac{46.1}{404.2} = 0.114 \)
Finally, calculate the dissociation constant (\(\text{K}_{\text{a}}\)):
For a weak electrolyte, \(\text{K}_{\text{a}} = \text{C}\alpha^2 \)
Given concentration (\(\text{C}\)) is \(0.025 \text{ mol L}^{-1}\).
\( \text{K}_{\text{a}} = 0.025 \times (0.114)^2 \)
\( \text{K}_{\text{a}} = 0.025 \times 0.012996 \)
\( \text{K}_{\text{a}} = 0.0003249 \)
\( \text{K}_{\text{a}} = 3.249 \times 10^{-4} \)
In simple words: We calculate the total molar conductivity at infinite dilution, then use it to find the degree of dissociation and finally the dissociation constant for methanoic acid.
🎯 Exam Tip: Remember to apply Kohlrausch's law correctly to find the limiting molar conductivity for weak electrolytes before calculating the degree of dissociation and dissociation constant.

Question 10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Answer: First, calculate the total charge (Q) passed:
\( \text{Q = It} \)
Where I = 0.5 A, t = 2 hours = \(2 \times 60 \times 60 \text{ s} = 7200 \text{ s}\).
\( \text{Q} = 0.5 \times 7200 = 3600 \text{ C} \)
Next, relate the total charge to the number of electrons (n):
\( \text{Q = ne} \)
Where e = charge of one electron = \(1.6 \times 10^{-19} \text{ C}\).
So, \( 3600 \text{ C} = n \times 1.6 \times 10^{-19} \text{ C} \)
Rearranging to find n:
\( n = \frac{3600}{1.6 \times 10^{-19}} \)
\( n = 2250 \times 10^{19} \)
\( n = 2.25 \times 10^{22} \text{ electrons} \)
In simple words: Calculate the total charge passed using current and time, then divide this total charge by the charge of a single electron to find the total number of electrons.
🎯 Exam Tip: Ensure all units are consistent (seconds for time, Coulombs for charge) and remember Faraday's constant relates moles of electrons to charge if a mole calculation is needed.

Question 11. Suggest a list of metals that are extracted electrolytically?
Answer: Highly reactive metals, which have a strong tendency to lose electrons and form positive ions, are typically extracted electrolytically. Examples of such metals include sodium (\(\text{Na}\)), potassium (\(\text{K}\)), magnesium (\(\text{Mg}\)), aluminum (\(\text{Al}\)), and calcium (\(\text{Ca}\)). These metals are obtained by the electrolysis of their molten salts.
In simple words: Very reactive metals like sodium, potassium, magnesium, aluminum, and calcium are extracted using electrolysis.
🎯 Exam Tip: Electrolytic extraction is preferred for highly reactive metals because their reduction potentials are too low for chemical reduction methods to be practical or efficient.

Question 12. Consider the reaction: \( \text{Cr}_2\text{O}_7^{2-} + \text{14H}^{+} + \text{6e}^{-} \rightarrow \text{2Cr}^{3+} + \text{7H}_2\text{O} \). What is the quantity of electricity in coulombs needed to reduce 1 mol of \( \text{Cr}_2\text{O}_7^{2-} \)?
Answer: The given reaction indicates that the reduction of 1 mole of dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) necessitates the involvement of 6 moles of electrons. According to Faraday's laws of electrolysis, 1 mole of electrons carries a charge of approximately \(96500 \text{ C}\) (Faraday's constant).
Therefore, the total quantity of electricity required is:
\( \text{Quantity of electricity} = 6 \text{ mol e}^{-} \times 96500 \text{ C/mol e}^{-} \)
\( \text{Quantity of electricity} = 579000 \text{ C} \)
(The calculation \(6 \times 6.02 \times 10^{23} \times 1.6 \times 10^{-19}\text{ C} = 577920 \text{ C}\) uses Avogadro's number and elementary charge, which is another way to derive Faraday's constant.)
In simple words: To reduce one mole of dichromate, 6 moles of electrons are needed, which translates to 6 times Faraday's constant in Coulombs.
🎯 Exam Tip: The number of electrons (n) in the balanced redox reaction directly determines the amount of electricity (in Faradays or Coulombs) required for the reduction or oxidation of a specific amount of substance.

Question 13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. During this process, lead is deposited at the cathode, lead oxide is formed at the anode and \(\text{H}_2\text{SO}_4\) is generated in the cell?
Answer: During the recharging of a lead storage battery, the electrochemical reactions that occurred during discharge are reversed. An external electric current is applied, forcing the products of the discharge reaction back into their original forms. The overall reaction for recharging is:
\( \text{2PbSO}_4\text{(s)} + \text{2H}_2\text{O(l)} \rightarrow \text{Pb(s)} + \text{PbO}_2\text{(s)} + \text{2H}_2\text{SO}_4\text{(aq)} \)
At the cathode, lead sulfate is reduced to lead:
\( \text{PbSO}_4\text{(s)} + \text{2e}^{-} \rightarrow \text{Pb(s)} + \text{SO}_4^{2-}\text{(aq)} \)
At the anode, lead sulfate is oxidized to lead dioxide:
\( \text{PbSO}_4\text{(s)} + \text{2H}_2\text{O(l)} \rightarrow \text{PbO}_2\text{(s)} + \text{SO}_4^{2-}\text{(aq)} + \text{4H}^{+}\text{(aq)} + \text{2e}^{-} \)
During this process, solid lead (\(\text{Pb(s)}\)) is redeposited on the cathode, lead dioxide (\(\text{PbO}_2\text{(s)}\)) reforms on the anode, and sulfuric acid (\(\text{H}_2\text{SO}_4\)) is regenerated, increasing the electrolyte's concentration.
In simple words: Recharging a lead battery reverses the discharge reactions: lead sulfate turns back into lead at the cathode, lead dioxide at the anode, and sulfuric acid is regenerated in the electrolyte.
🎯 Exam Tip: Be sure to distinguish between the discharge and charge reactions of a lead storage battery. The charge reaction is the reverse of discharge, involving external electrical energy input.

Question 14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells?
Answer: Besides hydrogen, several other materials can serve as fuels in fuel cells, offering different advantages for various applications. Two common examples include:
1. Methanol (\(\text{CH}_3\text{OH}\)): This is a liquid fuel that is relatively easy to store and transport, making it suitable for portable power applications.
2. Propane (\(\text{C}_3\text{H}_8\)): A readily available hydrocarbon gas that can be used in fuel cells, particularly where hydrogen infrastructure is not developed.
In simple words: Methanol and propane are alternative fuels for fuel cells, offering practical benefits over hydrogen in certain situations.
🎯 Exam Tip: Fuel cells can utilize various fuels, but the choice often depends on factors like availability, ease of storage, and environmental impact. Methanol and propane are good examples of liquid and gaseous alternatives to pure hydrogen.

Question 15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell?
Answer: Rusting of iron is an electrochemical process where the impure surface of iron behaves like a tiny galvanic cell, often termed a "corrosion cell," in the presence of water and oxygen. This process involves the slow degradation of the metal through redox reactions.
In a corrosion cell:
- **Anodic Regions:** Pure iron spots act as the anode, where iron undergoes oxidation, losing electrons and forming ferrous ions.
\( \text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{2e}^{-} \) (\(\text{E}_{\text{Fe}^{2+}/\text{Fe}}^{\circ} = -0.44 \text{ V}\))
- **Cathodic Regions:** Impure iron spots or regions with higher oxygen concentration act as the cathode, where oxygen molecules in the solution are reduced. This reduction typically occurs in the presence of \(\text{H}^{+}\) ions (often formed from the dissolution of \(\text{CO}_2\) in water to create carbonic acid, \(\text{H}_2\text{CO}_3\)).
\( \text{O}_2\text{(g)} + \text{4H}^{+}\text{(aq)} + \text{4e}^{-} \rightarrow \text{2H}_2\text{O(l)} \) (\(\text{E}_{\text{H}^{+}|\text{O}_2|\text{H}_2\text{O}}^{\circ} = 1.23 \text{ V}\))
The electrons released at the anode flow through the iron metal (acting as an external circuit) to the cathodic regions. Ions flow through the water film (electrolyte) to complete the circuit. The overall reaction for the formation of ferrous ions is:
\( \text{2Fe(s)} + \text{O}_2\text{(g)} + \text{4H}^{+}\text{(aq)} \rightarrow \text{2Fe}^{2+}\text{(aq)} + \text{2H}_2\text{O(l)} \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र लोहे के जंग लगने की प्रक्रिया को एक विद्युत रासायनिक सेल के रूप में दर्शाता है। इसमें दिखाया गया है कि लोहे की सतह पर कैसे एनोडिक और कैथोडिक क्षेत्र बनते हैं, जहां लोहा ऑक्सीकृत होकर आयरन आयन बनाता है और ऑक्सीजन पानी में अपचयित होता है। यह एक जल की बूंद की उपस्थिति में होता है जो एक इलेक्ट्रोलाइट के रूप में कार्य करती है और इलेक्ट्रॉन लोहे के माध्यम से प्रवाहित होते हैं, जिससे जंग लगने की प्रक्रिया शुरू होती है।

The ferrous ions (\(\text{Fe}^{2+}\)) subsequently undergo further oxidation by atmospheric oxygen to form hydrated ferric oxide, which is known as rust (\(\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}\)).
\( \text{4Fe}^{2+}\text{(aq)} + \text{O}_2\text{(g)} + \text{4H}_2\text{O(l)} \rightarrow \text{2Fe}_2\text{O}_3\text{(s)} + \text{8H}^{+}\text{(aq)} \)
\( \text{Fe}_2\text{O}_3\text{(s)} + x\text{H}_2\text{O(l)} \rightarrow \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O(s)} \) (Rust)
**Factors enhancing corrosion:**
- **Impurities:** Pure iron resists rusting more than impure iron, which provides distinct anodic and cathodic sites.
- **Air and moisture:** Essential for the electrochemical process.
- **Electrolytes:** Salts (e.g., in seawater) accelerate corrosion by increasing the conductivity of the aqueous film.
- **Strains:** Dents, scratches, or stress points on the metal surface can create anodic regions.
**Prevention of corrosion:**
(i) **Barrier protection:** Coating the metal surface with paints, grease, or less reactive metals (like tin or nickel) creates a physical barrier against oxygen and moisture.
(ii) **Sacrificial protection (Galvanization):** Coating iron with a more active metal (e.g., zinc) ensures that the active metal corrodes preferentially, protecting the iron. Zinc acts as the anode and is consumed, while iron becomes the cathode.
\( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{2e}^{-} \) (Anodic region)
At the iron surface, oxygen reduction still occurs:
\( \frac{1}{2}\text{O}_2\text{(g)} + \text{2H}^{+}\text{(aq)} + \text{2e}^{-} \rightarrow \text{H}_2\text{O(l)} \) (Cathodic region)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गैल्वेनाइजेशन की प्रक्रिया को दर्शाता है, जहाँ लोहे की सतह को जिंक की एक पतली परत से ढका जाता है। जिंक, अधिक सक्रिय धातु होने के कारण, एनोड के रूप में कार्य करता है और ऑक्सीकृत होकर जिंक आयन बनाता है। लोहा कैथोड के रूप में कार्य करता है जहाँ ऑक्सीजन और हाइड्रोजन आयन मिलकर पानी बनाते हैं। यह व्यवस्था लोहे को जंग लगने से बचाती है क्योंकि जिंक पहले corrode होता है।

Zinc (\(\text{E}_{\text{Zn}^{2+}/\text{Zn}}^{\circ} = -0.76 \text{ V}\)) is more electropositive than iron (\(\text{E}_{\text{Fe}^{2+}/\text{Fe}}^{\circ} = -0.44 \text{ V}\)). If the zinc layer is scratched, atmospheric air and moisture dissolve zinc, not iron, as zinc continues to act as the anode.
(iii) **Cathodic protection:** Connecting iron to an even more active metal (like magnesium or zinc) makes the iron the cathode, preventing it from losing electrons and dissolving. The more active metal acts as the sacrificial anode.
\( \text{Mg(s)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{2}\bar{\text{e}} \) (Anode)
(iv) **Anti-rust solutions:** Solutions containing alkaline phosphate or chromate, when applied to iron objects, form insoluble, heat-resistant coatings that prevent rusting. Their alkalinity also removes \(\text{H}^{+}\) ions, further retarding the corrosion process. These solutions are used in car radiators.
In simple words: Rusting is an electrochemical process where iron acts as an anode, losing electrons, and oxygen acts as a cathode, gaining electrons, forming rust; it can be prevented by barriers, sacrificial metals, or cathodic protection.
🎯 Exam Tip: Be able to identify the anodic and cathodic reactions in the rusting process and explain various methods of corrosion prevention, particularly sacrificial protection and cathodic protection.

Question 1. 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.
Answer: The ability of a metal to displace another metal from its salt solution is determined by its position in the electrochemical series, specifically by its standard reduction potential. A metal with a lower (more negative) standard reduction potential is more reactive and can displace metals with higher (less negative or positive) standard reduction potentials.
Based on their standard reduction potentials, the metals can be arranged in increasing order of their ability to displace each other (i.e., decreasing reactivity or increasing standard reduction potential):
Mg > Al > Zn > Fe > Cu
Magnesium is the most reactive, followed by aluminum, zinc, iron, and then copper, which is the least reactive among them.
In simple words: Metals are arranged by their ability to displace others from salt solutions, from most reactive (magnesium) to least reactive (copper).
🎯 Exam Tip: Remember the order of reactivity of metals; more reactive metals (with more negative electrode potentials) displace less reactive metals from their salt solutions.

Question 2. Standard electrode potentials, \( \text{K}^{+}/\text{K} = -2.93 \text{ V} \), \( \text{Ag}^{+}/\text{Ag} = 0.80\text{V} \), \( \text{Hg}^{2+}/\text{Hg} = 0.79 \text{ V} \), \( \text{Mg}^{2+}/\text{Mg} = -2.37 \text{ V} \), \( \text{Cr}^{3+}/\text{Cr} = -0.74\text{V} \). Arrange these metals in their increasing order of reducing power?
Answer: Reducing power is inversely related to standard reduction potential. A lower (more negative) standard reduction potential indicates a stronger tendency to lose electrons and thus a greater reducing power. Conversely, a higher (more positive) standard reduction potential indicates weaker reducing power.
Given standard reduction potentials:
\( \text{K}^{+}/\text{K} = -2.93 \text{ V} \)
\( \text{Ag}^{+}/\text{Ag} = 0.80 \text{ V} \)
\( \text{Hg}^{2+}/\text{Hg} = 0.79 \text{ V} \)
\( \text{Mg}^{2+}/\text{Mg} = -2.37 \text{ V} \)
\( \text{Cr}^{3+}/\text{Cr} = -0.74 \text{ V} \)
Arranging these from least reducing power (most positive \(\text{E}^{\circ}\)) to greatest reducing power (most negative \(\text{E}^{\circ}\)):
\( \text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K} \)
Therefore, potassium has the strongest reducing power, and silver has the weakest.
In simple words: The metals are ordered by increasing reducing power, which means ordering them from highest (most positive) to lowest (most negative) standard electrode potential.
🎯 Exam Tip: A metal with a more negative standard reduction potential acts as a stronger reducing agent, meaning it readily gets oxidized (loses electrons).

Question 3. Depict the galvanic cell in which the reaction \( \text{Zn(s)} + \text{2Ag}^{+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{2e}^{-} \) takes place. Further show.
1. which of the electrode is negatively charged?
2. the carriers of the current in the cell?
3. individual reaction at each electrode?
Answer: The galvanic cell for the given reaction can be represented using standard cell notation:
\( \text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Ag}^{+}\text{(aq)}|\text{Ag(s)} \)
1. **Which electrode is negatively charged?**
In a galvanic cell, oxidation occurs at the anode, and reduction occurs at the cathode. Zinc is oxidized, acting as the anode, and silver ions are reduced, with silver acting as the cathode. Electrons flow from the anode to the cathode. Therefore, the **zinc electrode (anode) will be negatively charged** as it is the source of electrons.
2. **The carriers of the current in the cell?**
- **External Circuit:** Electrons are the carriers of current, flowing from the zinc anode to the silver cathode through the external wire.
- **Internal Circuit (Electrolyte):** Ions are the carriers of current within the salt bridge and the electrolyte solutions, maintaining charge neutrality.
3. **Individual reaction at each electrode?**
- **At anode (Zinc electrode - negative):** Oxidation of zinc metal occurs.
\( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{2e}^{-} \)
- **At cathode (Silver electrode - positive):** Reduction of silver ions occurs.
\( \text{Ag}^{+}\text{(aq)} + \text{1e}^{-} \rightarrow \text{Ag(s)} \)
In simple words: The cell involves zinc losing electrons at its negative electrode (anode) and silver ions gaining electrons at the positive silver electrode (cathode), with electrons flowing externally and ions internally to conduct current.
🎯 Exam Tip: Remember that in a galvanic cell, the anode is the negative terminal and the cathode is the positive terminal. Oxidation always occurs at the anode, and reduction at the cathode.

Question 4. Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) \( \text{2Cr(s)} + \text{3Cd}^{2+}\text{(aq)} \rightarrow \text{2Cr}^{3+}\text{(aq)} + \text{3Cd(s)} \)
(ii) \( \text{Fe}^{3+}\text{(aq)} + \text{Ag(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Ag}^{+}\text{(aq)} \)
Calculate the \(\Delta\text{G}^{\circ}\) and equilibrium constant of the reactions.
Answer: To calculate the standard cell potential (\(\text{E}_{\text{cell}}^{\circ}\)), standard Gibbs energy (\(\Delta\text{G}^{\circ}\)), and equilibrium constant (\(\text{K}_{\text{c}}\)), we need the standard electrode potentials:
Relevant standard electrode potentials (approximate values, typically provided in a table):
\( \text{E}_{\text{Cr}^{3+}/\text{Cr}}^{\circ} = -0.74 \text{ V} \)
\( \text{E}_{\text{Cd}^{2+}/\text{Cd}}^{\circ} = -0.40 \text{ V} \)
\( \text{E}_{\text{Ag}^{+}/\text{Ag}}^{\circ} = +0.80 \text{ V} \)
\( \text{E}_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} = +0.77 \text{ V} \)

i. For the reaction: \( \text{2Cr(s)} + \text{3Cd}^{2+}\text{(aq)} \rightarrow \text{2Cr}^{3+}\text{(aq)} + \text{3Cd(s)} \)
- **Standard Cell Potential (\(\text{E}_{\text{cell}}^{\circ}\)):**
Chromium is oxidized (\(\text{Cr} \rightarrow \text{Cr}^{3+}\)), so it acts as the anode. Cadmium ions are reduced (\(\text{Cd}^{2+} \rightarrow \text{Cd}\)), so it acts as the cathode.
\( \text{E}_{\text{cell}}^{\circ} = \text{E}_{\text{cathode}}^{\circ} - \text{E}_{\text{anode}}^{\circ} = \text{E}_{\text{Cd}^{2+}/\text{Cd}}^{\circ} - \text{E}_{\text{Cr}^{3+}/\text{Cr}}^{\circ} \)
\( \text{E}_{\text{cell}}^{\circ} = -0.40\text{V} - (-0.74\text{V}) = +0.34 \text{ V} \)
- **Standard Gibbs Energy (\(\Delta\text{G}^{\circ}\)):**
The number of electrons transferred, \(n=6\) (e.g., \(2\text{Cr} \rightarrow 2\text{Cr}^{3+} + 6\text{e}^{-}\) and \(3\text{Cd}^{2+} + 6\text{e}^{-} \rightarrow 3\text{Cd}\)).
\( \Delta\text{G}^{\circ} = -\text{nFE}_{\text{cell}}^{\circ} \)
\( \Delta\text{G}^{\circ} = -6 \times 96500 \text{ C mol}^{-1} \times 0.34 \text{ V} \)
\( \Delta\text{G}^{\circ} = -196860 \text{ J mol}^{-1} \)
- **Equilibrium Constant (\(\text{K}_{\text{c}}\)):**
\( \text{E}_{\text{cell}}^{\circ} = \frac{0.0591}{n} \log \text{K}_{\text{c}} \)
\( 0.34 = \frac{0.0591}{6} \log \text{K}_{\text{c}} \)
\( \log \text{K}_{\text{c}} = \frac{0.34 \times 6}{0.0591} = \frac{2.04}{0.0591} \approx 34.52 \)
\( \text{K}_{\text{c}} = 10^{34.52} \approx 3.3 \times 10^{34} \)

ii. For the reaction: \( \text{Fe}^{3+}\text{(aq)} + \text{Ag(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{Ag}^{+}\text{(aq)} \)
- **Standard Cell Potential (\(\text{E}_{\text{cell}}^{\circ}\)):**
Silver is oxidized (\(\text{Ag} \rightarrow \text{Ag}^{+}\)), so it acts as the anode. Ferric ions are reduced (\(\text{Fe}^{3+} \rightarrow \text{Fe}^{2+}\)), so it acts as the cathode.
\( \text{E}_{\text{cell}}^{\circ} = \text{E}_{\text{cathode}}^{\circ} - \text{E}_{\text{anode}}^{\circ} = \text{E}_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - \text{E}_{\text{Ag}^{+}/\text{Ag}}^{\circ} \)
\( \text{E}_{\text{cell}}^{\circ} = 0.77\text{V} - 0.80\text{V} = -0.03 \text{ V} \)
- **Standard Gibbs Energy (\(\Delta\text{G}^{\circ}\)):**
The number of electrons transferred, \(n=1\) (e.g., \( \text{Fe}^{3+} + 1\text{e}^{-} \rightarrow \text{Fe}^{2+} \) and \( \text{Ag} \rightarrow \text{Ag}^{+} + 1\text{e}^{-} \)).
\( \Delta\text{G}^{\circ} = -\text{nFE}_{\text{cell}}^{\circ} \)
\( \Delta\text{G}^{\circ} = -1 \times 96500 \text{ C mol}^{-1} \times (-0.03 \text{ V}) \)
\( \Delta\text{G}^{\circ} = +2895 \text{ J mol}^{-1} \)
- **Equilibrium Constant (\(\text{K}_{\text{c}}\)):**
Since \(\text{E}_{\text{cell}}^{\circ}\) is negative, \(\Delta\text{G}^{\circ}\) is positive, indicating a non-spontaneous reaction under standard conditions. Thus, the equilibrium constant will be less than 1.
\( \text{E}_{\text{cell}}^{\circ} = \frac{0.0591}{n} \log \text{K}_{\text{c}} \)
\( -0.03 = \frac{0.0591}{1} \log \text{K}_{\text{c}} \)
\( \log \text{K}_{\text{c}} = \frac{-0.03}{0.0591} \approx -0.507 \)
\( \text{K}_{\text{c}} = 10^{-0.507} \approx 0.31 \)
In simple words: For each reaction, we calculate the standard cell potential, then use it to find the standard Gibbs energy and the equilibrium constant, indicating spontaneity and equilibrium position.
🎯 Exam Tip: Remember that a positive \(\text{E}_{\text{cell}}^{\circ}\) and negative \(\Delta\text{G}^{\circ}\) correspond to a spontaneous reaction and \(\text{K}_{\text{c}} > 1\). Conversely, a negative \(\text{E}_{\text{cell}}^{\circ}\) and positive \(\Delta\text{G}^{\circ}\) mean a non-spontaneous reaction and \(\text{K}_{\text{c}} < 1\).

Question 17. Predict the products of electrolysis in each of the following:
(1) An aqueous solution of AgNO3 with silver electrodes.
(2) An aqueous solution of AgNO3 with platinum electrodes.
(3) A dilute solution of H2SO4 with platinum electrodes.
(4) An aqueous solution of CuCl2 with platinum electrodes.
Answer:
(1) For an aqueous silver nitrate (\( \mathrm{AgNO}_{3} \)) solution with silver electrodes, the dissolution of silver metal occurs at the anode, while silver ions (\( \mathrm{Ag}^{+} \)) are deposited as silver metal at the cathode, as silver ions have a lower reduction potential than hydrogen ions.
At anode: \( \mathrm{Ag}_{(\mathrm{s})} \rightarrow \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{e}^{-} \)
At cathode: \( \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}_{(\mathrm{s})} \)

(2) In an aqueous silver nitrate (\( \mathrm{AgNO}_{3} \)) solution with platinum electrodes, at the cathode, silver ions (\( \mathrm{Ag}^{+} \)) will be deposited as silver metal due to their lower reduction potential compared to hydrogen ions. At the anode, since platinum is inert, hydroxide ions (\( \mathrm{OH}^{-} \)) from water will be oxidized to oxygen gas, as they have a lower discharge potential than nitrate ions (\( \mathrm{NO}_{3}^{-} \)).
At cathode: \( \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}_{(\mathrm{s})} \)
At anode: \( 4\mathrm{OH}^{-}_{(\mathrm{aq})} \rightarrow 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} + \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{e}^{-} \)

(3) When a dilute sulfuric acid (\( \mathrm{H}_{2}\mathrm{SO}_{4} \)) solution is electrolyzed using platinum electrodes, at the cathode, hydrogen ions (\( \mathrm{H}^{+} \)) are reduced to hydrogen gas. At the anode, hydroxide ions (\( \mathrm{OH}^{-} \)) from water are oxidized to oxygen gas. Thus, hydrogen gas is liberated at the cathode and oxygen gas at the anode.
At cathode: \( 2\mathrm{H}^{+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2(\mathrm{g})} \)
At anode: \( 4\mathrm{OH}^{-}_{(\mathrm{aq})} \rightarrow 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} + \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{e}^{-} \)

(4) For an aqueous copper(II) chloride (\( \mathrm{CuCl}_{2} \)) solution with platinum electrodes, at the cathode, copper ions (\( \mathrm{Cu}^{2+} \)) are preferentially reduced to copper metal over hydrogen ions. At the anode, chloride ions (\( \mathrm{Cl}^{-} \)) are oxidized to chlorine gas in preference to hydroxide ions (\( \mathrm{OH}^{-} \)).
At cathode: \( \mathrm{Cu}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})} \)
At anode: \( 2\mathrm{Cl}^{-}_{(\mathrm{aq})} \rightarrow \mathrm{Cl}_{2(\mathrm{g})} + 2\mathrm{e}^{-} \)
In simple words: The products of electrolysis depend on the electrolyte and electrode type. At the cathode, the species with higher reduction potential gets reduced, while at the anode, the species with lower oxidation potential gets oxidized.

🎯 Exam Tip: Understanding the standard electrode potentials and the nature of electrodes (inert vs. active) is crucial for correctly predicting electrolysis products.

GSEB Class 12 Chemistry Electrochemistry Additional Important Questions And Answers

Question 1. Define molar conductivity of a solution.
Answer:
Molar conductivity of an electrolyte solution at a specific concentration is defined as the conductance of a solution volume containing one mole of the electrolyte, positioned between two electrodes separated by unit length and having a unit cross-sectional area. If the electrodes have a unit cross-sectional area and are separated by a unit length, the volume 'V' of the solution containing one mole of the electrolyte is equivalent to the reciprocal of concentration. Therefore, molar conductivity (\( \Lambda_{\mathrm{m}} \)) is given by \( \Lambda_{\mathrm{m}} = \kappa \mathrm{V} \), where \( \kappa \) is the conductivity.
In simple words: Molar conductivity measures how well a solution conducts electricity per mole of electrolyte, assuming it's between two electrodes a unit distance apart with unit area.

🎯 Exam Tip: Remember the formula \( \Lambda_{\mathrm{m}} = \kappa \mathrm{V} \) and the inverse relationship between volume and concentration when defining molar conductivity.

Question 2. A galvanic cell is constructed out of \( \mathrm{Zn}^{2+} | \mathrm{Zn} \) and \( \mathrm{Ag}^{+} | \mathrm{Ag} \) half cells. Given: \( \mathrm{E}^{\circ}_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76 V \) and \( \mathrm{E}^{\circ}_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.80 V \). Answer the following.
(1) What is the cell reaction?
(2) What is the e.m.f of the cell?
(3) What are the signs of the two electrodes in the galvanic cell?
(4) Which ion is the powerful oxidising agent?
(5) Which metal is the most powerful reducing agent?
Answer:
(1) The overall cell reaction involves zinc being oxidized at the anode and silver ions being reduced at the cathode:
Oxidation half-cell (anode): \( \mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \)
Reduction half-cell (cathode): \( 2\mathrm{Ag}^{+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}_{(\mathrm{s})} \)
Overall cell reaction: \( \mathrm{Zn}_{(\mathrm{s})} + 2\mathrm{Ag}^{+}_{(\mathrm{aq})} \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2\mathrm{Ag}_{(\mathrm{s})} \)

(2) The electromotive force (e.m.f.) of the cell, or standard cell potential, is calculated as the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode:
\( \mathrm{E}^{\circ}_{\mathrm{cell}} = \mathrm{E}^{\circ}_{\mathrm{red(cathode)}} - \mathrm{E}^{\circ}_{\mathrm{red(anode)}} \)
\( \mathrm{E}^{\circ}_{\mathrm{cell}} = \mathrm{E}^{\circ}_{\mathrm{Ag}^{+}/\mathrm{Ag}} - \mathrm{E}^{\circ}_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = +0.80 V - (-0.76 V) = +1.56 V \)

(3) In this galvanic cell, the zinc electrode (anode) serves as the negative terminal, and the silver electrode (cathode) acts as the positive terminal.

(4) The silver ion (\( \mathrm{Ag}^{+} \)) is the most potent oxidizing agent among the species present, as it readily accepts electrons.

(5) Zinc metal (\( \mathrm{Zn} \)) is the most powerful reducing agent in this cell, as it readily loses electrons to undergo oxidation.
In simple words: Zinc loses electrons to silver ions, producing a voltage of 1.56 V. Zinc is the negative electrode and the best reducing agent, while silver ions are the positive electrode and the best oxidizing agent.

🎯 Exam Tip: To determine the cell reaction and its e.m.f., always identify the species with higher standard reduction potential as the cathode and the one with lower potential as the anode. The more negative the standard electrode potential, the stronger the reducing agent, and vice-versa for oxidizing agents.

Question 3. Ramu found that the cell which he had used in torches are not rechargeable while that used in car are chargeable?
(1) What are the basic differences between them?
(2) Explain the working of batteries used in torches?
(3) A dry cell becomes dead after a long time even if it has not been used. Give reason?
(4) Which battery is used in digital watches? What are the advantages of them?
Answer:
(1) The primary difference is that the cell used in torches is a primary cell (dry cell), which means its electrode reactions are irreversible and it cannot be recharged. In contrast, the battery in a car is a secondary cell (lead storage cell), where the chemical reactions can be reversed by applying an external electric current, making it rechargeable.

(2) A dry cell, commonly used in torches, consists of a zinc container acting as the anode, and a central graphite rod as the cathode. The space between these electrodes is filled with a moist paste of zinc chloride (\( \mathrm{ZnCl}_{2} \)) and ammonium chloride (\( \mathrm{NH}_{4}\mathrm{Cl} \)), along with powdered manganese dioxide (\( \mathrm{MnO}_{2} \)) and carbon. The cell typically produces an e.m.f. of about 1.45 V.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक शुष्क सेल की आंतरिक संरचना को दर्शाता है, जिसमें एक केंद्रीय ग्रेफाइट कैथोड एक जस्ते के कप के भीतर स्थित होता है। जस्ते का कप एनोड के रूप में कार्य करता है और अमोनियम क्लोराइड, मैंगनीज डाइऑक्साइड और कार्बन के नम पेस्ट से भरा होता है, जो इलेक्ट्रोलाइट के रूप में कार्य करता है। यह व्यवस्था सेल के विद्युत प्रवाह उत्पादन को दर्शाती है।
The cell reactions are:
At anode: \( \mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \)
At cathode: \( 2\mathrm{MnO}_{2(\mathrm{s})} + 2\mathrm{NH}^{+}_{4(\mathrm{aq})} + 2\mathrm{e}^{-} \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3(\mathrm{s})} + 2\mathrm{NH}_{3(\mathrm{g})} + \mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \)

(3) A dry cell becomes inactive over a long period, even without use, because the acidic ammonium chloride (\( \mathrm{NH}_{4}\mathrm{Cl} \)) paste corrodes the zinc container, eventually leading to its breakdown and loss of cell integrity.

(4) Digital watches commonly utilize mercury cells. A key advantage of these cells is their constant e.m.f. (approximately 1.35 V) throughout their operational life. This stability occurs because the overall cell reaction does not involve any species whose concentration changes significantly in the solution.
In simple words: Torch batteries are single-use dry cells where zinc corrodes. Car batteries are rechargeable lead-acid cells. Digital watches use mercury cells for stable voltage.

🎯 Exam Tip: Differentiate between primary and secondary cells based on rechargeability. For dry cells, remember the components and their roles, especially the acidic nature of \( \mathrm{NH}_{4}\mathrm{Cl} \) and its corrosive effect. For mercury cells, emphasize the stable voltage output and the reason behind it.

Question 4. A car battery is made of lead dipping into sulphuric acid (Lead Accumulator). It can be recharged?
(1) It is said to be a secondary cell. Do you agree? Substantiate your answer.
(2) When the battery is used, a reaction takes place. Identify the change to this reaction when the battery is recharged and represent the reactions at the anode of the cell.
(3) Is there any change in the concentration of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) during discharging and recharging the cell? Explain.
Answer:
(1) Yes, a car battery is indeed a secondary cell. This classification is due to its ability to be recharged and reused multiple times. Unlike primary cells, the chemical reactions that occur during discharge can be reversed by supplying external electrical energy, restoring the battery to its original state.

(2) During the discharge process (when the battery is in use), oxidation occurs at the anode, where lead is converted to lead sulfate:
Discharging (Anode, Oxidation): \( \mathrm{Pb}_{(\mathrm{s})} \rightarrow \mathrm{Pb}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \)
During recharging, the reverse reaction takes place at the same electrode, where lead sulfate is reduced back to lead:
Recharging (Anode, Reduction): \( \mathrm{Pb}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \rightarrow \mathrm{Pb}_{(\mathrm{s})} \)

(3) Yes, there is a significant change in the concentration of sulfuric acid (\( \mathrm{H}_{2}\mathrm{SO}_{4} \)) during both discharging and recharging. During discharge, sulfuric acid is consumed as lead and lead dioxide react to form lead sulfate and water, leading to a decrease in the acid's concentration:
Discharging: \( \mathrm{Pb}_{(\mathrm{s})} + \mathrm{PbO}_{2(\mathrm{s})} + 2\mathrm{H}_{2}\mathrm{SO}_{4(\mathrm{aq})} \rightarrow 2\mathrm{PbSO}_{4(\mathrm{s})} + 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \)
Conversely, during recharging, the lead sulfate and water are converted back to lead, lead dioxide, and sulfuric acid, which results in an increase in the \( \mathrm{H}_{2}\mathrm{SO}_{4} \) concentration:
Recharging: \( 2\mathrm{PbSO}_{4(\mathrm{s})} + 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{Pb}_{(\mathrm{s})} + \mathrm{PbO}_{2(\mathrm{s})} + 2\mathrm{H}_{2}\mathrm{SO}_{4(\mathrm{aq})} \)
In simple words: Car batteries are secondary cells because they can be recharged. When used, lead turns into lead sulfate and sulfuric acid is consumed. When recharged, lead sulfate turns back into lead and lead dioxide, increasing sulfuric acid concentration.

🎯 Exam Tip: Focus on the reversibility of reactions in secondary cells. Remember that \( \mathrm{H}_{2}\mathrm{SO}_{4} \) concentration decreases during discharge and increases during recharge, which is a key indicator of the battery's state.

Question 5. In Apollo space programme the fuel cell are used not only for electric power but also to prepare H2O for drinking?
(1) What do you understand by a fuel cell?
(2) Explain the working of an \( \mathrm{H}_{2} - \mathrm{O}_{2} \) fuel cell?
(3) What are the advantages of fuel cells over ordinary cells?
Answer:
(1) A fuel cell is an electrochemical device that efficiently converts the chemical energy from a fuel (like hydrogen, methane, or carbon monoxide) and an oxidant (like oxygen) into electrical energy. The most common type is the hydrogen-oxygen fuel cell.

(2) The hydrogen-oxygen fuel cell operates with porous carbon electrodes, which are often impregnated with catalysts such as platinum, silver, or cobalt oxide to facilitate reactions. A concentrated solution of sodium hydroxide (\( \mathrm{NaOH} \)) or potassium hydroxide (\( \mathrm{KOH} \)) acts as the electrolyte between these electrodes. The reactions are:
At anode (oxidation): \( 2\mathrm{H}_{2(\mathrm{g})} + 4\mathrm{OH}^{-}_{(\mathrm{aq})} \rightarrow 4\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} + 4\mathrm{e}^{-} \)
At cathode (reduction): \( \mathrm{O}_{2(\mathrm{g})} + 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} + 4\mathrm{e}^{-} \rightarrow 4\mathrm{OH}^{-}_{(\mathrm{aq})} \)
Overall cell reaction: \( 2\mathrm{H}_{2(\mathrm{g})} + \mathrm{O}_{2(\mathrm{g})} \rightarrow 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \)
This cell typically produces an e.m.f. of about 0.9 V.

(3) Fuel cells offer several advantages over traditional galvanic cells: they are highly efficient at generating energy, often exceeding the efficiency of combustion engines; and they are environmentally friendly as they do not produce polluting byproducts, with water being the primary product in the case of hydrogen-oxygen fuel cells.
In simple words: Fuel cells turn fuel directly into electricity cleanly and efficiently, producing water as a byproduct. Hydrogen-oxygen fuel cells use porous electrodes with catalysts and an alkaline electrolyte to generate power.

🎯 Exam Tip: When describing fuel cells, emphasize their high efficiency and eco-friendliness. For the \( \mathrm{H}_{2} - \mathrm{O}_{2} \) fuel cell, remember the specific reactions at the anode and cathode, and the overall formation of water.

Question 6. Ramu found that an iron sample in contact with sand is corroded?
(1) What is corrosion?
(2) Give the mechanism of rusting of iron on the basis of electrochemical theory?
(3) It is found that an alkaline medium inhibits the rusting of iron. Explain?
Answer:
(1) Corrosion is the gradual degradation of a metallic material due to unwanted electrochemical reactions with its surrounding environment, typically initiating at the metal's surface. In the context of iron, this process is specifically termed rusting.

(2) Rusting of iron is an electrochemical process. Impurities on the iron surface act as tiny electrochemical cells in the presence of moisture and other gases. Pure iron functions as the anode, while impure iron serves as the cathode.
At the anodic regions, iron undergoes oxidation:
\( \mathrm{Fe}_{(\mathrm{s})} \rightarrow \mathrm{Fe}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \), with a standard electrode potential \( \mathrm{E}^{\circ}_{\mathrm{Fe}^{2+}/\mathrm{Fe}} = -0.44 V \). (1)
At the cathodic regions, oxygen molecules in the dissolved solution are reduced:
\( \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{H}^{+}_{(\mathrm{aq})} + 4\mathrm{e}^{-} \rightarrow 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \), with a standard electrode potential \( \mathrm{E}^{\circ}_{\mathrm{H}^{+}|\mathrm{O}_{2}|\mathrm{H}_{2}\mathrm{O}} = 1.23 V \). (2)
(Hydrogen ions are believed to be derived from carbonic acid, which forms when \( \mathrm{CO}_{2} \) dissolves in water.)
Combining these half-cell reactions (multiplying equation (1) by 2):
\( 2\mathrm{Fe}_{(\mathrm{s})} + \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{H}^{+}_{(\mathrm{aq})} \rightarrow 2\mathrm{Fe}^{2+}_{(\mathrm{aq})} + 2\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \). (3)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र लोहे में जंग लगने की प्रक्रिया का एक सरलीकृत आरेख है। इसमें दिखाया गया है कि पानी की बूंदों के संपर्क में आने पर लोहे की सतह पर एनोडिक (Fe ऑक्सीकरण) और कैथोडिक (ऑक्सीजन और \( \mathrm{H}^{+} \) का अपचयन) क्षेत्र कैसे बनते हैं। इलेक्ट्रॉन लोहे के माध्यम से स्थानांतरित होते हैं, जिससे जंग लगने की विद्युत रासायनिक प्रक्रिया पूरी होती है।

The ferrous ions (\( \mathrm{Fe}^{2+} \)) generated are further oxidized by atmospheric oxygen to form hydrated ferric oxide, commonly known as rust (\( \mathrm{Fe}_{2}\mathrm{O}_{3}\mathrm{.xH}_{2}\mathrm{O} \)):
\( 4\mathrm{Fe}^{2+}_{(\mathrm{aq})} + \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3(\mathrm{s})} + 8\mathrm{H}^{+}_{(\mathrm{aq})} \). (4)
\( \mathrm{Fe}_{2}\mathrm{O}_{3(\mathrm{s})} + \mathrm{xH}_{2}\mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}\mathrm{.xH}_{2}\mathrm{O}_{(\mathrm{s})} \) (Rust). (5)
These two half-reactions occur at distinct locations, confirming that corrosion proceeds via an electrochemical mechanism. The iron surface is conceptualized as a network of minute voltaic cells, where iron is oxidized at anodic sites and oxygen is reduced at cathodic sites.
Electrons migrate through the iron, which acts as the external circuit, while ions flowing through the water solution or the protective film on the iron complete the electrical circuit. Highly pure iron resists rusting effectively. However, the presence of acids, salts, electrolytes, air, and moisture significantly accelerates the rusting process. For example, a completely dry metal surface is not susceptible to oxygen attack (due to an incomplete electrical circuit), and corrosion occurs more rapidly in seawater than in freshwater because seawater provides necessary ions for conduction.

(3) An alkaline medium retards the rusting of iron. This is because the hydroxyl ions (\( \mathrm{OH}^{-} \)) provided by the alkaline environment neutralize the hydrogen ions (\( \mathrm{H}^{+} \)) involved in the cathodic reaction (\( \mathrm{O}_{2} + 4\mathrm{H}^{+} + 4\mathrm{e}^{-} \rightarrow 2\mathrm{H}_{2}\mathrm{O} \)). This removal of \( \mathrm{H}^{+} \) ions slows down the cathodic process, thereby inhibiting the overall oxidation of iron and consequently preventing rusting.
Prevention of corrosion can be achieved through various methods:
(i) Barrier protection: This involves applying a physical barrier (e.g., paints, grease, or coatings of metals like zinc, copper, tin, or nickel) to prevent direct contact between the iron and the atmosphere. Adherent films of iron phosphate or chemicals like bisphenol also serve this purpose.
(ii) Sacrificial protection: This method protects iron by coating it with a more reactive metal (e.g., zinc, chromium, or nickel). The more active metal preferentially loses electrons, undergoing oxidation and protecting the iron. The process of coating iron with a thin layer of zinc is known as galvanization.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख गैल्वनीकरण द्वारा लोहे के जंग लगने से बचाव को दर्शाता है। इसमें दिखाया गया है कि कैसे एक जस्ता एनोड (Zn) लोहे के कैथोड (Iron Cathode) को सुरक्षा प्रदान करता है। जस्ता ऑक्सीकृत होकर इलेक्ट्रॉन खोता है, जो ऑक्सीजन और हाइड्रोजन आयनों के अपचयन के लिए उपलब्ध होते हैं, इस प्रकार लोहे को जंग लगने से बचाता है।
Zinc (\( \mathrm{E}^{\circ}_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76 V \)) is more electropositive than iron (\( \mathrm{E}^{\circ}_{\mathrm{Fe}^{2+}/\mathrm{Fe}} = -0.44 V \)), meaning zinc has a greater tendency to lose electrons than iron. Thus, zinc sacrifices itself to protect iron from rusting. If the protective zinc layer is scratched, atmospheric air and moisture will dissolve zinc, not iron. Here, zinc acts as the anodic region, and iron acts as the cathodic region.
Anodic reaction (zinc): \( \mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-} \)
Cathodic reaction (iron surface): \( \mathrm{H}_{2}\mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{OH}^{-}_{(\mathrm{aq})} \)
\( 2\mathrm{H}^{+}_{(\mathrm{Fe})} + 2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2(\mathrm{g})} \)
(iii) Cathodic protection: In this method, iron is intentionally made the cathode of an electrochemical cell, preventing it from losing electrons and dissolving. This is achieved by connecting it to a more active metal, such as magnesium or zinc, which then acts as the sacrificial anode, losing electrons while the iron remains protected as the cathode.
\( (\mathrm{Mg}_{(\mathrm{s})} \rightarrow \mathrm{Mg}^{2+}_{(\mathrm{aq})} + 2\mathrm{e}^{-}) \)
(iv) Anti-rust solutions: Applying solutions of alkaline phosphate or alkaline chromate to iron objects forms an insoluble, heat-resistant coating of iron phosphate or chromate, which prevents rusting. The alkalinity of these solutions further contributes to rust prevention by removing \( \mathrm{H}^{+} \) ions, thereby decreasing the rate of rusting. Such solutions are commonly used in car radiators.
In simple words: Corrosion is metal damage from electrochemical reactions; rusting is iron corrosion. It happens when iron acts as an anode and oxygen as a cathode, facilitated by moisture and acids. Alkaline conditions inhibit rusting by neutralizing acid that drives the cathodic reaction. Prevention methods include protective coatings and sacrificial or cathodic protection.

🎯 Exam Tip: For corrosion, remember it's an electrochemical process. Be ready to explain the anodic and cathodic reactions, and how factors like pH (alkaline vs. acidic) and presence of other metals (sacrificial protection) influence the rate of rusting. Clearly differentiate between types of corrosion prevention.

Question 7. Michael Faraday after studying the decomposition of electrolytes by electric current, put forward the laws of electrolysis?
(1) State and explain Faraday's laws of electrolysis.
(2) Explain the term electrochemical equivalent.
(3) Calculate the charge required to deposit 40.5 g of Al when the electrode reaction is \( \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al} \)
Answer:
(1) **Faraday's First Law of Electrolysis:**
This law states that the amount of any substance liberated or deposited at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
Mathematically, \( \mathrm{w} \propto \mathrm{Q} \), or \( \mathrm{w} = \mathrm{zQ} \).
Since \( \mathrm{Q} = \mathrm{It} \) (where I is current in amperes and t is time in seconds), the equation becomes \( \mathrm{w} = \mathrm{zIt} \). Here, 'w' represents the mass of the substance liberated, and 'z' is the electrochemical equivalent.

**Faraday's Second Law of Electrolysis:**
This law states that when the same quantity of electricity is passed through different electrolytic solutions connected in series, the amounts of different substances liberated at the electrodes are directly proportional to their chemical equivalent weights.
\[ \begin{array}{|l|l|} \hline \textbf{Equivalent weight of metal} & \textbf{Atomic mass of metal} \\ \cline{2-2} \textbf{Number of electrons required to reduce the cation} \\ \hline \textbf{Mass of A deposited} & \textbf{Eq. mass of A} \\ \cline{2-2} \textbf{Mass of B deposited} & \textbf{Eq. mass of B} \\ \hline \end{array} \]
A key constant derived from Faraday's laws is the Faraday's constant (F). This constant represents the quantity of electricity required to liberate one gram equivalent of any substance. It has been experimentally determined to be approximately 96,500 coulombs (1 Faraday = 96,500 C). For example, 1.008 g (1 gram equivalent) of hydrogen or 108 g (1 gram equivalent) of silver requires 96,500 coulombs.
Combining both laws, the amount of electricity or charge needed for oxidation or reduction is dependent on the stoichiometry of the electrode reaction. For instance, in the reduction of a silver ion: \( \mathrm{Ag}^{+}_{(\mathrm{aq})} + 1\mathrm{e}^{-} \rightarrow \mathrm{Ag}_{(\mathrm{s})} \), one mole of electrons is required per mole of silver ions. Knowing that the charge on a single electron is \( 1.6021 \times 10^{-19} \mathrm{C} \), the charge on one mole of electrons is approximately \( 1.6021 \times 10^{-19} \mathrm{C} \times 6.022 \times 10^{23} \mathrm{mol}^{-1} \approx 96487 \mathrm{C}\ \mathrm{mol}^{-1} \), which is rounded to 96,500 C, the Faraday constant (F). Thus, for reactions like \( \mathrm{Mg}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Mg}_{(\mathrm{s})} \), 2 moles of electrons (2F) are needed, and for \( \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}_{(\mathrm{s})} \), 3 moles of electrons (3F) are required.

(2) **Electrochemical Equivalent (z):**
The electrochemical equivalent of a substance is defined as the mass of that substance (in grams) liberated or deposited at an electrode when one ampere of current is passed for one second (i.e., one coulomb of charge). Its unit is grams per coulomb (g/C).

(3) **Calculation for depositing 40.5 g of Al:**
Given the electrode reaction: \( \mathrm{Al}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Al} \)
From the stoichiometry, 1 mole of aluminum (27 g) requires 3 moles of electrons, which corresponds to \( 3 \times 96500 \mathrm{C} \) of charge.
Charge required to deposit 1 mole (27 g) of Al = \( 3 \times 96500 \mathrm{C} \).
To deposit 40.5 g of Al, the charge required can be calculated using a proportion:
Charge \( = \frac{\text{Mass of Al desired}}{\text{Molar mass of Al}} \times \text{Charge per mole of Al} \)
Charge \( = \frac{40.5 \mathrm{g}}{27 \mathrm{g/mol}} \times (3 \times 96500 \mathrm{C/mol}) \)
Charge \( = 1.5 \mathrm{mol} \times (289500 \mathrm{C/mol}) \)
Charge \( = 434250 \mathrm{C} \)
Therefore, the charge required to deposit 40.5 g of Al is 434,250 C.
In simple words: Faraday's laws describe how much substance is deposited during electrolysis based on electricity passed. The electrochemical equivalent is the mass deposited per coulomb. To deposit 40.5g of aluminum, 434,250 coulombs are needed.

🎯 Exam Tip: Clearly state and differentiate between Faraday's first and second laws. Remember the definition of electrochemical equivalent and Faraday's constant (96,500 C). For calculations, always balance the half-reaction to determine the number of electrons (n) involved per mole.

Question 8. What are the factors that affect the conductivity of an anionic (electrolytic) solution?
Answer:
The conductivity of an electrolytic solution is influenced by several key factors:
- Temperature: Higher temperatures generally increase ionic mobility, thus increasing conductivity.
- Concentration of electrolyte: Conductivity initially increases with concentration due to more ions, but can decrease at very high concentrations due to increased ionic interactions.
- Nature of the electrolyte added: Strong electrolytes dissociate completely, leading to higher conductivity, while weak electrolytes dissociate partially, resulting in lower conductivity.
- Nature of solvent and its viscosity: Solvents with higher dielectric constants and lower viscosity allow for better ion dissociation and mobility, leading to higher conductivity.
- Size of the ions produced and their solvation: Smaller, less solvated ions can move more freely, contributing to higher conductivity.
In simple words: The ability of a solution to conduct electricity depends on temperature, how much and what kind of electrolyte is dissolved, the type of solvent, its stickiness, and the size and hydration of the ions.

🎯 Exam Tip: List at least 3-4 factors and briefly explain their effect on conductivity. Remember that conductivity is directly related to the number and mobility of ions.

Question 9. The graph showing the variation of molar conductance with concentration for acetic acid and hydrochloride acid is given below. Analyse the graph and answer the following questions?
(1) Why does molar conductance increase with dilution?
(2) Why the graphs obtained in each case has such a large difference?
(3) Give the equation for the variation of molar conductivity with a concentration of strong electrolytes.
(4) What is meant by molar conductance at infinite dilution? Suggest methods to find out \( \Lambda^{\circ}_{\mathrm{m}} \) for HCl and \( \mathrm{CH}_{3}\mathrm{COOH} \).
Answer:
(1) Molar conductance of a solution increases with dilution because as the solution becomes more dilute, the inter-ionic attractive forces between ions decrease. This reduction in attraction allows ions to move more freely and quickly, increasing their mobility and, consequently, the molar conductivity.

(2) There is a significant difference between the graphs for strong electrolytes (like HCl) and weak electrolytes (like acetic acid). Strong electrolytes are almost completely ionized even at moderate concentrations, so their molar conductivity increases gradually with dilution as inter-ionic interactions decrease. In contrast, weak electrolytes are only partially ionized in concentrated solutions. Upon dilution, their degree of ionization increases substantially, releasing more ions into the solution and leading to a more pronounced increase in molar conductance.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तनुकरण के साथ मोलर चालकता (\( \Lambda_{\mathrm{m}} \)) के परिवर्तन को दर्शाता है, जिसमें \( \sqrt{\mathrm{C}} \) (सांद्रता का वर्गमूल) के विरुद्ध \( \Lambda_{\mathrm{m}} \) का प्लॉट किया गया है। यह मजबूत इलेक्ट्रोलाइट्स (जैसे HCl और KCl) के लिए एक रैखिक संबंध दिखाता है, जो सांद्रता घटने पर मोलर चालकता में वृद्धि दर्शाता है।

(3) For strong electrolytes, the variation of molar conductivity (\( \Lambda_{\mathrm{m}} \)) with concentration (\( \mathrm{C} \)) is described by the Debye-Hückel-Onsager equation:
\( \Lambda_{\mathrm{m}} = \Lambda^{\circ}_{\mathrm{m}} - \mathrm{b}\sqrt{\mathrm{C}} \)
Here, \( \Lambda^{\circ}_{\mathrm{m}} \) is the molar conductivity at infinite dilution (when \( \mathrm{C} \rightarrow 0 \)), and 'b' is a constant that depends on the nature of the electrolyte, solvent, and temperature.

(4) Molar conductance at infinite dilution (\( \Lambda^{\circ}_{\mathrm{m}} \)), also known as limiting molar conductivity, is the molar conductivity of an electrolyte when its concentration approaches zero. At infinite dilution, inter-ionic interactions are negligible, and each ion contributes independently to the total molar conductivity.
To determine \( \Lambda^{\circ}_{\mathrm{m}} \):
- For HCl (a strong electrolyte): A graph of molar conductivity (\( \Lambda_{\mathrm{m}} \)) versus the square root of concentration (\( \sqrt{\mathrm{C}} \)) is plotted. For strong electrolytes, this plot is linear at low concentrations. Extrapolating this linear curve to \( \sqrt{\mathrm{C}} = 0 \) yields the value of \( \Lambda^{\circ}_{\mathrm{m}} \).
- For \( \mathrm{CH}_{3}\mathrm{COOH} \) (acetic acid, a weak electrolyte): Direct extrapolation of the \( \Lambda_{\mathrm{m}} \) vs \( \sqrt{\mathrm{C}} \) graph is not possible because it's non-linear. Instead, \( \Lambda^{\circ}_{\mathrm{m}} \) for weak electrolytes is determined using Kohlrausch's Law of Independent Migration of Ions. This law states that \( \Lambda^{\circ}_{\mathrm{m}} \) for a weak electrolyte can be calculated from the limiting molar conductivities of strong electrolytes. For example:
\( \Lambda^{\circ}_{\mathrm{m}}(\mathrm{CH}_{3}\mathrm{COOH}) = \Lambda^{\circ}_{\mathrm{m}}(\mathrm{CH}_{3}\mathrm{COONa}) + \Lambda^{\circ}_{\mathrm{m}}(\mathrm{HCl}) - \Lambda^{\circ}_{\mathrm{m}}(\mathrm{NaCl}) \)
In simple words: Molar conductivity increases with dilution because ions move more freely. Strong electrolytes show a linear increase, while weak electrolytes show a sharper increase due to more ionization. Infinite dilution molar conductivity is found by extrapolating for strong electrolytes or using Kohlrausch's law for weak ones.

🎯 Exam Tip: Remember that the behavior of strong and weak electrolytes differs significantly with dilution. For strong electrolytes, use the Debye-Hückel-Onsager equation and graphical extrapolation. For weak electrolytes, recall Kohlrausch's Law as the primary method to determine \( \Lambda^{\circ}_{\mathrm{m}} \).