Chapter 5: Electrochemistry
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. Define Electrochemistry. What are the two main types of electrochemical cells?
Electrochemistry: It is the branch of physical chemistry that deals with the study of the relationship between electrical energy and chemical energy, and their inter-conversion.
Two types of cells:
- Electrolytic cell: Converts electrical energy into chemical energy (drives a non-spontaneous reaction).
- Galvanic / Voltaic cell: Converts chemical energy into electrical energy (drives a spontaneous reaction).
2. Differentiate between metallic conduction and electrolytic conduction.
| Metallic Conduction | Electrolytic Conduction |
|---|---|
| Flow of electricity is due to the movement of free electrons. | Flow of electricity is due to the movement of ions. |
| No chemical decomposition occurs. | Chemical decomposition of the electrolyte occurs. |
| Conductivity decreases with an increase in temperature. | Conductivity increases with an increase in temperature. |
| No transfer of matter takes place. | Transfer of matter (ions) takes place. |
3. Define Resistance ($R$) and Conductance ($G$). State their units.
Resistance ($R$): It is the opposition offered to the flow of electric current through a conductor. Unit: Ohm ($\Omega$).
Conductance ($G$): It is the reciprocal of electrical resistance ($G = 1/R$). It represents the ease with which current flows. Unit: Ohm$^{-1}$ ($\Omega^{-1}$), mho, or Siemens (S).
4. Define Conductivity (Specific Conductance, $\kappa$). Mention its SI unit.
Conductivity ($\kappa$): It is the reciprocal of resistivity ($\rho$). It is defined as the conductance of a conductor of unit length and unit area of cross-section. For solutions, it is the conductance of 1 cm$^3$ (or 1 m$^3$) of the solution.
$$\kappa = \frac{1}{\rho} = G \cdot \frac{l}{A}$$
SI Unit: Siemens per meter (S m$^{-1}$) or commonly used S cm$^{-1}$ ($\Omega^{-1}$ cm$^{-1}$).
5. Define Cell Constant. Write its formula and unit.
Cell Constant ($b$): It is the ratio of the distance between the electrodes ($l$) to the area of cross-section ($A$) of the electrodes in a conductivity cell.
$$\text{Cell Constant} (b) = \frac{l}{A}$$
Also, $\text{Cell Constant} = \text{Conductivity } (\kappa) \times \text{Resistance } (R)$
Unit: m$^{-1}$ or cm$^{-1}$.
6. Define Molar Conductivity ($\Lambda_m$). State its formula and SI unit.
Molar Conductivity ($\Lambda_m$): It is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a given volume ($V$) of solution.
$$\Lambda_m = \frac{\kappa}{C}$$
If $\kappa$ is in S cm$^{-1}$ and $C$ is molarity in mol L$^{-1}$:
$$\Lambda_m = \frac{1000 \times \kappa}{C}$$
SI Unit: S m$^2$ mol$^{-1}$. (Commonly used: S cm$^2$ mol$^{-1}$).
7. How does the conductivity ($\kappa$) of an electrolytic solution vary with dilution? Why?
Conductivity ($\kappa$) decreases with an increase in dilution (decrease in concentration) for both strong and weak electrolytes.
Reason: Conductivity is the conductance of 1 cm$^3$ of the solution. On dilution, the volume of the solution increases, so the number of current-carrying ions per unit volume (per cm$^3$) decreases. Hence, conductivity decreases.
8. How does the molar conductivity ($\Lambda_m$) of a strong electrolyte vary with dilution?
Molar conductivity of a strong electrolyte increases slowly with dilution and reaches a limiting value at infinite dilution ($\Lambda_m^\circ$).
Reason: Strong electrolytes are completely dissociated. On dilution, the distance between the ions increases, which decreases the interionic attractive forces. This increases the mobility of the ions, leading to a gradual increase in molar conductivity.
It follows the Kohlrausch equation: $\Lambda_m = \Lambda_m^\circ - A\sqrt{C}$
9. How does the molar conductivity ($\Lambda_m$) of a weak electrolyte vary with dilution?
Molar conductivity of a weak electrolyte increases steeply at higher dilutions (low concentrations) but does not reach a definite limiting value at infinite dilution.
Reason: Weak electrolytes are partially dissociated at higher concentrations. As dilution increases, their degree of dissociation ($\alpha$) increases drastically (Ostwald's dilution law). This results in a massive increase in the total number of ions, causing a steep rise in $\Lambda_m$.
10. State Kohlrausch's law of independent migration of ions.
Kohlrausch's Law: At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards the molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated.
Mathematically, the molar conductivity of an electrolyte at infinite dilution ($\Lambda_m^\circ$) is the sum of the limiting molar conductivities of its constituent cations and anions.
$$\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ$$
Where $\nu_+$ and $\nu_-$ are the number of cations and anions per formula unit, and $\lambda_+^\circ$ and $\lambda_-^\circ$ are their limiting molar conductivities.
11. State any two applications of Kohlrausch's law.
- Calculation of $\Lambda_m^\circ$ for weak electrolytes: Since it cannot be found by extrapolating the graph, Kohlrausch's law allows calculating it using values from strong electrolytes. (e.g., $\Lambda_m^\circ(CH_3COOH) = \Lambda_m^\circ(CH_3COONa) + \Lambda_m^\circ(HCl) - \Lambda_m^\circ(NaCl)$).
- Calculation of degree of dissociation ($\alpha$): For a weak electrolyte at concentration $C$, $\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}$.
12. Define Electrolysis.
Electrolysis: The process of decomposition of an electrolyte by the passage of direct electric current through its aqueous solution or molten (fused) state.
During electrolysis, cations move to the cathode (reduction) and anions move to the anode (oxidation).
13. Describe the electrolysis of molten NaCl. Write the electrode reactions.
Molten $NaCl$ contains only $Na^+$ and $Cl^-$ ions.
- At Cathode (Reduction): $Na^+$ ions are reduced to sodium metal.
$Na^+_{(l)} + e^- \rightarrow Na_{(s)}$ - At Anode (Oxidation): $Cl^-$ ions are oxidized to chlorine gas.
$2Cl^-_{(l)} \rightarrow Cl_{2(g)} + 2e^-$
Overall reaction: $2Na^+_{(l)} + 2Cl^-_{(l)} \rightarrow 2Na_{(s)} + Cl_{2(g)}$
14. Describe the electrolysis of aqueous NaCl solution. Write the electrode reactions.
Aqueous $NaCl$ contains $Na^+, Cl^-, H_2O, H^+,$ and $OH^-$ ions.
- At Cathode (Reduction): Water is reduced preferentially over $Na^+$ because it has a higher reduction potential.
$2H_2O_{(l)} + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$ - At Anode (Oxidation): $Cl^-$ is oxidized to $Cl_2$ gas instead of water due to overvoltage of oxygen.
$2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$
Overall: $2NaCl_{(aq)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq)} + H_{2(g)} + Cl_{2(g)}$
15. What is overpotential (overvoltage)?
Overpotential: The extra voltage that must be applied beyond the theoretically calculated reversible standard electrode potential to cause an electrochemical reaction to occur at a practical rate. It is often required for reactions involving gases (like $O_2$ evolution).
16. State Faraday's First Law of Electrolysis. Give its mathematical form.
Faraday's First Law: The mass ($W$) of a substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte.
$$W \propto Q$$
Since $Q = I \times t$ (Current $\times$ Time),
$$W = Z \cdot I \cdot t$$
Where $Z$ is the electrochemical equivalent of the substance.
17. Define Electrochemical Equivalent ($Z$).
Electrochemical Equivalent ($Z$): It is the mass of a substance deposited or liberated at an electrode when one ampere of current is passed for one second (i.e., one Coulomb of electricity).
Formula: $Z = \frac{\text{Molar Mass}}{\text{Number of electrons } (n) \times 96500}$. Unit: $\text{g C}^{-1}$.
18. State Faraday's Second Law of Electrolysis.
Faraday's Second Law: When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the masses of the substances produced at the electrodes are directly proportional to their equivalent weights.
$$\frac{W_1}{W_2} = \frac{E_1}{E_2}$$
Where $W$ is mass deposited and $E$ is equivalent weight.
19. Differentiate between Electrolytic cells and Galvanic cells.
| Electrolytic Cell | Galvanic (Voltaic) Cell |
|---|---|
| Converts electrical energy to chemical energy. | Converts chemical energy to electrical energy. |
| Non-spontaneous reaction is driven. | Spontaneous reaction occurs. |
| Anode is Positive, Cathode is Negative. | Anode is Negative, Cathode is Positive. |
| Both electrodes usually in the same container. | Electrodes are in different half-cells connected by a salt bridge. |
20. What is a Salt Bridge? How is it prepared?
Salt Bridge: It is a U-shaped glass tube containing a saturated solution of an inert electrolyte (like $KCl, KNO_3$, or $NH_4NO_3$) in a semi-solid gel like agar-agar, which connects the two half-cells of a galvanic cell.
Preparation: A hot saturated solution of $KCl$ or $KNO_3$ is mixed with 5% agar-agar powder and boiled. It is poured into a U-tube and allowed to cool and set into a gel. The ends are plugged with glass wool.
21. What are the functions of a salt bridge?
- It connects the two half-cells internally and completes the electrical circuit.
- It maintains electrical neutrality in both half-cells by allowing the migration of ions ($K^+$ and $Cl^-$) to counteract the charge accumulation.
- It prevents the mechanical mixing of the two electrolytes.
- It prevents liquid junction potential.
22. How is a Galvanic cell represented (IUPAC convention)?
- Anode (oxidation half-cell) is written on the left.
- Cathode (reduction half-cell) is written on the right.
- A single vertical line ($|$) represents a phase boundary (e.g., solid electrode and aqueous solution).
- A double vertical line ($||$) represents a salt bridge.
- Concentrations or pressures are written in brackets.
Example (Daniell Cell): $Zn_{(s)} | Zn^{2+}_{(aq, 1M)} || Cu^{2+}_{(aq, 1M)} | Cu_{(s)}$
23. Define Standard Electrode Potential ($E^\circ$).
Standard Electrode Potential: It is the potential difference developed between an electrode and its electrolyte when the concentration of the electrolyte is exactly 1 M (or gas at 1 atm pressure) at a standard temperature of 298 K. According to IUPAC, standard reduction potentials are used as standard electrode potentials.
24. Describe the construction of Standard Hydrogen Electrode (SHE).
Construction: It consists of a platinum wire sealed in a glass tube, ending in a platinum foil coated with finely divided platinum black. This foil acts as the electrode surface. It is immersed in a 1 M $H^+$ solution (like 1M $HCl$). Pure $H_2$ gas at 1 atm pressure is bubbled over the platinum foil constantly at 298 K.
Cell Representation: $Pt, H_{2(g, 1 atm)} | H^+_{(aq, 1M)}$
By convention, its standard electrode potential ($E^\circ$) is arbitrarily assigned a value of exactly 0.00 V.
25. What are the difficulties in setting up and using the Standard Hydrogen Electrode (SHE)?
- It is difficult to maintain the pure $H_2$ gas pressure at exactly 1 atm throughout the experiment.
- It is difficult to maintain the $H^+$ ion concentration at exactly 1 M because water can evaporate.
- The platinum electrode is easily "poisoned" by impurities present in the hydrogen gas.
26. Write the Nernst equation for a single electrode at 298 K.
For a general reduction reaction: $M^{n+}_{(aq)} + ne^- \rightarrow M_{(s)}$
The Nernst equation at 298 K is:
$$E = E^\circ - \frac{0.0592}{n} \log_{10} \frac{1}{[M^{n+}]}$$
Since the concentration of solid metal $M$ is taken as 1. ($E$ is electrode potential, $E^\circ$ is standard electrode potential, $n$ is number of electrons).
27. Write the Nernst equation for a complete galvanic cell.
For a general cell reaction: $aA + bB \rightleftharpoons cC + dD$
The Nernst equation at 298 K is:
$$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} \frac{[C]^c [D]^d}{[A]^a [B]^b}$$
Where $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ and $n$ is the number of moles of electrons transferred.
28. How is standard cell potential ($E^\circ_{cell}$) related to standard Gibbs free energy ($\Delta G^\circ$)?
The electrical work done by a cell is equal to the decrease in Gibbs free energy.
$$\Delta G^\circ = -nFE^\circ_{cell}$$
Where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant (96500 C/mol), and $E^\circ_{cell}$ is the standard cell potential.
For a spontaneous reaction, $E^\circ_{cell}$ must be positive, making $\Delta G^\circ$ negative.
29. Derive the relationship between $E^\circ_{cell}$ and the equilibrium constant ($K_c$).
At equilibrium, the cell potential $E_{cell} = 0$, and the reaction quotient $Q = K_c$.
From the Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{2.303 RT}{nF} \log_{10} Q$
Substituting $E_{cell} = 0$ and $Q = K_c$:
$$0 = E^\circ_{cell} - \frac{2.303 RT}{nF} \log_{10} K_c$$
$$E^\circ_{cell} = \frac{2.303 RT}{nF} \log_{10} K_c$$
At 298 K, $E^\circ_{cell} = \frac{0.0592}{n} \log_{10} K_c$.
30. What is an Electrochemical Series (EMF series)?
Electrochemical Series: The arrangement of various elements (electrodes) in the increasing or decreasing order of their standard reduction potentials ($E^\circ$) is called the electrochemical series.
Elements at the top (negative $E^\circ$, e.g., Li, K) are strong reducing agents. Elements at the bottom (positive $E^\circ$, e.g., F$_2$, Au) are strong oxidizing agents.
31. State three applications of the electrochemical series.
- Comparing oxidizing/reducing power: Lower $E^\circ$ means better reducing agent; higher $E^\circ$ means better oxidizing agent.
- Predicting spontaneity of redox reactions: If $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} > 0$, the reaction is spontaneous.
- Predicting metal displacement: A metal with a lower reduction potential can displace a metal with a higher reduction potential from its salt solution (e.g., Zn displaces Cu$^{2+}$).
32. Classify commercial cells (batteries). Give an example of each.
- Primary Cells: These cells cannot be recharged and become dead after the chemicals are consumed. The cell reaction is irreversible. Example: Dry cell (Leclanche cell), Button cell.
- Secondary Cells: These cells can be recharged by passing a direct current through them in the opposite direction. The cell reaction is reversible. Example: Lead accumulator (Lead storage battery), Nickel-Cadmium cell.
33. Describe the construction of a Dry Cell (Leclanche cell).
Construction: It consists of a Zinc cylinder which acts as the anode (-). A graphite (carbon) rod placed in the center acts as the cathode (+). The graphite rod is surrounded by a powder mixture of Manganese dioxide ($MnO_2$) and carbon black. The remaining space inside the zinc cylinder is filled with a moist paste of Ammonium chloride ($NH_4Cl$) and Zinc chloride ($ZnCl_2$), which acts as the electrolyte. The cell is sealed at the top.
34. Write the anode and cathode reactions in a Dry Cell.
- At Anode (Oxidation): Zinc is oxidized.
$Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$ - At Cathode (Reduction): $MnO_2$ is reduced, and ammonia is liberated.
$2MnO_{2(s)} + 2NH_4^+_{(aq)} + 2e^- \rightarrow Mn_2O_{3(s)} + 2NH_{3(g)} + H_2O_{(l)}$
The $NH_3$ gas is not released but combines with $Zn^{2+}$ to form a complex ion $[Zn(NH_3)_4]^{2+}$, preventing the buildup of gas pressure.
35. Describe the construction of a Lead Accumulator (Lead Storage Battery).
It is a secondary battery. It consists of multiple cells connected in series (usually 6 cells giving 12V).
- Anode: A grid of lead packed with spongy lead ($Pb$).
- Cathode: A grid of lead packed with lead dioxide ($PbO_2$).
- Electrolyte: An aqueous solution of 38% by mass sulfuric acid ($H_2SO_4$).
36. Write the discharging reactions of a Lead Accumulator.
During discharging, it acts as a Galvanic cell.
- At Anode: $Pb_{(s)} + SO_4^{2-}_{(aq)} \rightarrow PbSO_{4(s)} + 2e^-$
- At Cathode: $PbO_{2(s)} + 4H^+_{(aq)} + SO_4^{2-}_{(aq)} + 2e^- \rightarrow PbSO_{4(s)} + 2H_2O_{(l)}$
Overall: $Pb_{(s)} + PbO_{2(s)} + 2H_2SO_{4(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)}$
Note: $H_2SO_4$ is consumed, so its density decreases.
37. Write the recharging reactions of a Lead Accumulator.
During recharging, external current is applied in the opposite direction. It acts as an electrolytic cell. The reactions are reversed.
Overall recharging reaction:
$$2PbSO_{4(s)} + 2H_2O_{(l)} \xrightarrow{\text{charging}} Pb_{(s)} + PbO_{2(s)} + 2H_2SO_{4(aq)}$$
Lead sulfate is converted back to lead and lead dioxide, and $H_2SO_4$ is regenerated, increasing its density.
38. What is a Fuel Cell? Give an example.
Fuel Cell: A galvanic cell in which the chemical energy of continuously supplied fuels (like hydrogen, methane, methanol) and oxidants (like oxygen) is directly converted into electrical energy.
Example: Hydrogen-Oxygen ($H_2 - O_2$) fuel cell used in the Apollo space program.
39. Write the electrode reactions for the Hydrogen-Oxygen fuel cell.
The electrolyte is a concentrated aqueous $KOH$ solution. Electrodes are porous carbon containing catalysts (Pt/Pd).
- Anode (Hydrogen gas bubbled):
$2H_{2(g)} + 4OH^-_{(aq)} \rightarrow 4H_2O_{(l)} + 4e^-$ - Cathode (Oxygen gas bubbled):
$O_{2(g)} + 2H_2O_{(l)} + 4e^- \rightarrow 4OH^-_{(aq)}$
Overall Reaction: $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$
40. What are the advantages of fuel cells?
- High Efficiency: They convert about 70-75% of chemical energy into electrical energy, which is much higher than thermal power plants (~40%).
- Pollution Free: The only by-product is pure water vapor, so they do not cause air or noise pollution.
- Continuous Energy: They provide power continuously as long as the fuel and oxidant are supplied. They do not need to be "recharged" like batteries.
41. Define Corrosion. Give an example.
Corrosion: The slow, spontaneous degradation or destruction of metals due to chemical or electrochemical reactions with their environment (moisture and gases like $O_2, CO_2$).
Example: Rusting of iron (formation of reddish-brown hydrated ferric oxide, $Fe_2O_3 \cdot xH_2O$).
42. Explain the electrochemical theory of rusting of iron. (Write anode and cathode reactions).
Rusting is an electrochemical phenomenon where a miniature galvanic cell is formed on the iron surface.
- Anode region: Pure iron acts as the anode and is oxidized.
$Fe_{(s)} \rightarrow Fe^{2+}_{(aq)} + 2e^-$ - Cathode region: Electrons travel through the metal to another spot where oxygen is reduced in the presence of $H^+$ (from $H_2CO_3$ or moisture).
$O_{2(g)} + 4H^+_{(aq)} + 4e^- \rightarrow 2H_2O_{(l)}$
The $Fe^{2+}$ ions are further oxidized by atmospheric oxygen to $Fe^{3+}$ which form rust:
$2Fe^{2+} + \frac{1}{2}O_2 + 2H_2O \rightarrow Fe_2O_3 + 4H^+$
43. How is corrosion prevented by cathodic protection (sacrificial anode)?
In cathodic protection, the iron object (like underground pipes or ship hulls) is connected by a wire to a more reactive metal like Magnesium ($Mg$) or Zinc ($Zn$).
Because $Mg$ or $Zn$ has a lower reduction potential (more easily oxidized) than iron, it acts as the anode and gets corroded (sacrificed) while forcing the iron to act as the cathode. As long as the active metal is present, the iron is protected from rusting.
44. What is galvanization?
Galvanization: It is the process of coating the surface of iron or steel with a thin layer of Zinc to prevent it from rusting. Zinc is more reactive than iron, so even if the coating is scratched, zinc acts as a sacrificial anode and protects the iron.
45. The resistance of a conductivity cell containing 0.001 M KCl solution is 1500 $\Omega$. What is the cell constant if conductivity of 0.001 M KCl at 298 K is $0.146 \times 10^{-3} \text{ S cm}^{-1}$?
Given: $R = 1500 \ \Omega$, $\kappa = 0.146 \times 10^{-3} \text{ S cm}^{-1}$
Formula: $\text{Cell Constant} (b) = \kappa \times R$
$b = (0.146 \times 10^{-3}) \times 1500$
$b = 0.219 \text{ cm}^{-1}$
46. How much charge is required for the reduction of 1 mole of $Al^{3+}$ to $Al$?
Reaction: $Al^{3+} + 3e^- \rightarrow Al$
To reduce 1 mole of $Al^{3+}$, 3 moles of electrons are required.
Charge of 1 mole of electrons = 1 Faraday (F) = 96500 C.
Total charge required = $3 \text{ F} = 3 \times 96500 \text{ C} = 289,500 \text{ C}$.
47. Given $E^\circ_{Zn^{2+}/Zn} = -0.76 \text{ V}$ and $E^\circ_{Cu^{2+}/Cu} = +0.34 \text{ V}$. Calculate standard cell potential ($E^\circ_{cell}$) for the Daniell cell.
The lower reduction potential acts as the anode (oxidation). So, Zinc is anode and Copper is cathode.
Formula: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
$E^\circ_{cell} = (+0.34 \text{ V}) - (-0.76 \text{ V})$
$E^\circ_{cell} = 0.34 + 0.76 = 1.10 \text{ V}$
48. Can you store copper sulfate ($CuSO_4$) solution in a zinc pot? Give reason.
No. We cannot store $CuSO_4$ in a zinc pot.
Reason: Zinc is placed above copper in the electrochemical series (it has a lower standard reduction potential). Therefore, zinc is a stronger reducing agent than copper. Zinc will displace copper from the copper sulfate solution ($Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$), causing the zinc pot to dissolve/corrode.
49. Why is AC (Alternating Current) used instead of DC (Direct Current) while measuring the resistance of an electrolytic solution?
If DC current is passed through an electrolytic solution, it causes electrolysis (chemical decomposition) which alters the concentration of the solution near the electrodes and causes polarization. This results in inaccurate resistance measurements. Using high-frequency AC prevents electrolysis and polarization, allowing accurate measurement.
50. Calculate the molar conductivity ($\Lambda_m$) if the conductivity of a 0.05 M solution is $0.0025 \text{ S cm}^{-1}$.
Given: $\kappa = 0.0025 \text{ S cm}^{-1}$, $C = 0.05 \text{ mol L}^{-1}$
Formula: $\Lambda_m = \frac{1000 \times \kappa}{C}$
$\Lambda_m = \frac{1000 \times 0.0025}{0.05}$
$\Lambda_m = \frac{2.5}{0.05} = 50 \text{ S cm}^2 \text{ mol}^{-1}$.
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