Electrochemistry
Welcome to HSLC Guru. This English-medium ASSEB study material on Class 12 Chemistry Chapter 3 — Electrochemistry — gives you a complete revision pack: a focused summary, all important short and long answer questions, numericals based on the Nernst equation and Faraday’s laws, MCQs, fill-in-the-blanks, true/false, a glossary and a formula table. Everything is aligned to the ASSEB Class 12 Chemistry syllabus and HS final examination pattern.
Chapter Summary
Electrochemistry is the branch of chemistry that deals with the relationship between chemical energy and electrical energy and the inter-conversion of one form into the other. An electrochemical cell is a device that converts chemical energy into electrical energy (galvanic / voltaic cell) or uses electrical energy to drive a non-spontaneous chemical reaction (electrolytic cell). In a galvanic cell such as the Daniell cell, oxidation takes place at the anode (negative electrode) and reduction at the cathode (positive electrode); the two half-cells are connected by a salt bridge that completes the inner circuit and maintains electrical neutrality. In an electrolytic cell the polarity is reversed — the anode is positive and the cathode is negative — because an external source forces the electrons.
The tendency of an electrode to lose or gain electrons is measured by its electrode potential. Since absolute electrode potentials cannot be measured, the Standard Hydrogen Electrode (SHE) is taken as the reference and is assigned a potential of exactly zero volts at 298 K, 1 atm H2 and 1 M H+. Standard electrode potentials of various electrodes, arranged in increasing order of reduction potential, give the electrochemical series. This series predicts the feasibility of redox reactions, the strength of oxidising and reducing agents, and the direction of cell reaction. The standard EMF of a cell is given by
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}$
The variation of cell potential with concentration is described by the Nernst equation:
$E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{RT}{nF}\ln Q$
At 298 K, putting in numerical values of R, T and F and changing natural log to common log, the equation becomes:
$E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n}\log Q$
The Gibbs energy change of a cell reaction is related to its EMF by $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ and to the equilibrium constant by $\Delta G^\circ = -RT\ln K$. Combining these two we get $\log K = \frac{nE^\circ_{\text{cell}}}{0.0591}$, which lets us evaluate K from a measured cell potential.
For solutions, electrolytic conductance is studied through resistance R, conductance G = 1/R, specific (or electrolytic) conductance $\kappa = \frac{1}{\rho}$ and molar conductance $\Lambda_m = \frac{\kappa \times 1000}{c}$, where c is the molar concentration in mol L-1. Specific conductance decreases on dilution, while molar conductance increases on dilution and approaches a limiting value $\Lambda_m^\circ$ at infinite dilution. Kohlrausch’s law of independent migration of ions states that at infinite dilution each ion contributes a definite, constant share to the molar conductance regardless of its partner ion, so $\Lambda_m^\circ = \nu_+\lambda_+^\circ + \nu_-\lambda_-^\circ$. The degree of dissociation of a weak electrolyte is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
Electrolysis is the decomposition of an electrolyte by passage of electric current. The amount deposited or liberated obeys Faraday’s laws: (i) mass deposited is proportional to the quantity of electricity, $w = Z \cdot Q = Z \cdot I \cdot t$; (ii) when the same quantity of electricity passes through different electrolytes, the masses of the substances deposited are in the ratio of their equivalent weights, $\frac{w_1}{w_2} = \frac{E_1}{E_2}$. One faraday equals 96500 C and deposits one gram-equivalent of any substance. Primary cells like the dry (Leclanché) cell and the mercury cell deliver current until the chemicals are exhausted and cannot be recharged. Secondary cells such as the lead-storage battery and the nickel-cadmium cell can be recharged by reversing the current. Fuel cells (e.g. H2–O2 fuel cell) convert the energy of combustion of a fuel directly into electrical energy with high efficiency. Corrosion of iron is an electrochemical process in which the metal acts as the anode (oxidised to Fe2+) and another spot acts as the cathode where oxygen is reduced; rust (hydrated ferric oxide) is the final product. Prevention is by barrier coatings, galvanising, sacrificial protection, and cathodic protection.
1-Mark Questions
Q1. What is an electrochemical cell?
Answer: An electrochemical cell is a device that either converts chemical energy of a spontaneous redox reaction into electrical energy (galvanic cell) or uses electrical energy to drive a non-spontaneous reaction (electrolytic cell).
Q2. Why is a salt bridge used in a galvanic cell?
Answer: A salt bridge completes the inner circuit, maintains electrical neutrality of the two half-cells by allowing migration of ions, and prevents the build-up of charge that would otherwise stop the cell reaction.
Q3. What is the standard electrode potential of SHE?
Answer: The standard electrode potential of the Standard Hydrogen Electrode is exactly zero volts (0.00 V) at 298 K, 1 atm H2 pressure and 1 M H+ concentration.
Q4. Define molar conductance.
Answer: Molar conductance ($\Lambda_m$) is the conductance of all the ions produced by one mole of an electrolyte in solution. It is given by $\Lambda_m = \frac{\kappa \times 1000}{c}$, where c is concentration in mol L-1.
Q5. State Faraday’s first law of electrolysis.
Answer: The mass (w) of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte: $w = Z \cdot I \cdot t$, where Z is the electrochemical equivalent.
Q6. What is the value of one faraday?
Answer: One faraday is the charge carried by one mole of electrons and is equal to 96500 coulombs (more precisely 96485 C mol-1).
Q7. Write the cell representation of the Daniell cell.
Answer: Zn(s) | Zn2+(aq, 1 M) || Cu2+(aq, 1 M) | Cu(s).
Q8. Why does specific conductance of a solution decrease on dilution?
Answer: Because dilution reduces the number of current-carrying ions per unit volume of the solution, so the conductance of a unit volume (specific conductance) falls.
Q9. Give one example each of a primary and a secondary cell.
Answer: Primary cell — dry (Leclanché) cell; Secondary cell — lead-storage battery.
Q10. What is corrosion?
Answer: Corrosion is the slow electrochemical deterioration of a metal due to its reaction with the surroundings, e.g. rusting of iron in moist air to give hydrated iron(III) oxide.
2 / 3-Mark Questions
Q11. Distinguish between a galvanic cell and an electrolytic cell.
Answer: (i) A galvanic cell converts chemical energy into electrical energy through a spontaneous redox reaction, while an electrolytic cell uses electrical energy to drive a non-spontaneous reaction. (ii) In a galvanic cell, the anode is negative and the cathode is positive; in an electrolytic cell the anode is positive and the cathode is negative. (iii) The two electrodes of a galvanic cell are placed in separate compartments connected by a salt bridge, whereas both electrodes of an electrolytic cell dip into the same electrolyte.
Q12. State and explain Kohlrausch’s law of independent migration of ions.
Answer: Kohlrausch’s law states that at infinite dilution, where ionic interactions vanish, each ion makes a definite contribution to the molar conductance of an electrolyte irrespective of the nature of the other ion with which it is associated. Mathematically, $\Lambda_m^\circ = \nu_+\lambda_+^\circ + \nu_-\lambda_-^\circ$. The law is used to (i) calculate $\Lambda_m^\circ$ for weak electrolytes, (ii) determine degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$ and (iii) evaluate dissociation constants of weak acids and bases.
Q13. Why is the standard hydrogen electrode chosen as the reference electrode?
Answer: Absolute electrode potentials cannot be measured experimentally; only potential differences between two electrodes can be measured. The SHE is reproducible, gives a stable potential under defined conditions (1 M H+, 1 atm H2, 298 K), is reversible, and by international convention is assigned the value 0.00 V. All other electrode potentials are reported relative to it.
Q14. What are the applications of the electrochemical series?
Answer: (i) Predicting the relative strength of oxidising and reducing agents — strong reducing agents lie at the top, strong oxidising agents at the bottom. (ii) Predicting the feasibility of a redox reaction (positive E°cell means feasible). (iii) Predicting whether a metal will displace hydrogen from dilute acids. (iv) Calculating standard EMF of a cell. (v) Predicting products of electrolysis at the electrodes.
Q15. Explain why molar conductance increases on dilution while specific conductance decreases.
Answer: Specific conductance is the conductance per unit volume; on dilution the number of ions per unit volume decreases, so $\kappa$ decreases. Molar conductance is the conductance of all ions from one mole of electrolyte; on dilution the volume containing one mole increases (V = 1000/c), and although ions per unit volume fall, the total number of ions in that volume rises (better dissociation, less inter-ionic attraction), so $\Lambda_m$ increases and finally reaches a limiting value $\Lambda_m^\circ$.
Q16. Describe the construction and working of the H2–O2 fuel cell.
Answer: A H2–O2 fuel cell consists of two porous carbon electrodes containing finely divided platinum or palladium catalyst, dipped in concentrated KOH solution. Hydrogen is bubbled through one electrode (anode) and oxygen through the other (cathode). At the anode: 2H2 + 4OH– → 4H2O + 4e–; at the cathode: O2 + 2H2O + 4e– → 4OH–. Net reaction: 2H2 + O2 → 2H2O. The cell delivers 0.9 V, has high efficiency, is pollution-free and is used in space vehicles.
5 / 7-Mark Questions (with Numericals)
Q17. Derive the Nernst equation for a single electrode and for a cell. Hence write its form at 298 K.
Answer: Consider the half-reaction Mn+ + ne– → M. From thermodynamics, $\Delta G = \Delta G^\circ + RT\ln Q$. Substituting $\Delta G = -nFE$ and $\Delta G^\circ = -nFE^\circ$, we get $-nFE = -nFE^\circ + RT\ln Q$, which on rearrangement gives the Nernst equation for a single electrode: $E = E^\circ – \frac{RT}{nF}\ln \frac{1}{[M^{n+}]}$. For a complete cell with cell reaction having reaction quotient Q: $E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{RT}{nF}\ln Q$. Substituting R = 8.314 J K-1 mol-1, T = 298 K, F = 96500 C, and converting natural log to log10: $E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n}\log Q$. The equation lets us calculate cell potential at any concentration and to relate $E^\circ_{\text{cell}}$ to the equilibrium constant through $\log K = \frac{nE^\circ_{\text{cell}}}{0.0591}$.
Q18. Calculate the EMF of the cell Zn | Zn2+ (0.01 M) || Cu2+ (0.1 M) | Cu at 298 K. Given E°(Cu2+/Cu) = +0.34 V, E°(Zn2+/Zn) = -0.76 V.
Answer: The cell reaction is Zn + Cu2+ → Zn2+ + Cu, n = 2. Standard EMF: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}} = 0.34 – (-0.76) = 1.10$ V. Reaction quotient: $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{0.01}{0.1} = 0.1$. Applying the Nernst equation at 298 K: $E_{\text{cell}} = 1.10 – \frac{0.0591}{2}\log(0.1) = 1.10 – \frac{0.0591}{2}(-1) = 1.10 + 0.02955 \approx 1.130$ V.
Q19. The standard EMF of the cell Zn | Zn2+ || Cu2+ | Cu is 1.10 V at 298 K. Calculate the equilibrium constant of the cell reaction.
Answer: n = 2, $E^\circ_{\text{cell}} = 1.10$ V. Using $\log K = \frac{nE^\circ_{\text{cell}}}{0.0591} = \frac{2 \times 1.10}{0.0591} = \frac{2.20}{0.0591} = 37.225$. Therefore $K = 10^{37.225} \approx 1.68 \times 10^{37}$, indicating an essentially complete reaction.
Q20. A current of 1.50 A is passed for 30 minutes through a CuSO4 solution. Calculate the mass of copper deposited (atomic mass of Cu = 63.5).
Answer: Quantity of electricity Q = I × t = 1.50 × 30 × 60 = 2700 C. Equivalent mass of Cu (Cu2+ + 2e– → Cu) = 63.5 / 2 = 31.75 g eq-1. By Faraday’s first law, $w = \frac{E \times I \times t}{96500} = \frac{31.75 \times 2700}{96500} = \frac{85725}{96500} = 0.888$ g. Hence about 0.89 g of copper is deposited.
Q21. The same quantity of electricity is passed through solutions of AgNO3 and CuSO4. If 1.08 g of silver is deposited, what mass of copper is deposited? (Ag = 108, Cu = 63.5)
Answer: By Faraday’s second law, $\frac{w_1}{w_2} = \frac{E_1}{E_2}$. E(Ag) = 108/1 = 108, E(Cu) = 63.5/2 = 31.75. So $\frac{1.08}{w_{Cu}} = \frac{108}{31.75}$, giving $w_{Cu} = \frac{1.08 \times 31.75}{108} = 0.3175$ g. About 0.318 g of copper is deposited.
Q21A. Calculate the molar conductance of 0.025 M aqueous KCl solution if the specific conductance is 0.0035 S cm-1.
Answer: Using $\Lambda_m = \frac{\kappa \times 1000}{c}$, we substitute $\kappa = 0.0035$ S cm-1 and c = 0.025 mol L-1: $\Lambda_m = \frac{0.0035 \times 1000}{0.025} = \frac{3.5}{0.025} = 140$ S cm2 mol-1. Hence the molar conductance of 0.025 M KCl is 140 S cm2 mol-1.
Q21B. Describe the construction, working and reactions of the lead-storage battery.
Answer: A lead-storage battery has a number of cells connected in series. Each cell has a lead anode and a grid of lead packed with lead dioxide (PbO2) as cathode, immersed in 38% sulphuric acid. During discharge: at the anode Pb + SO42- → PbSO4 + 2e–; at the cathode PbO2 + 4H+ + SO42- + 2e– → PbSO4 + 2H2O. Net: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O. The cell delivers about 2 V per cell. On charging, the reactions are reversed and PbSO4 is converted back to Pb at the cathode and PbO2 at the anode while H2SO4 is regenerated. It is widely used in automobiles and inverters because of high power output and easy recharge.
Q21C. Explain the mechanism of corrosion of iron and mention any two methods of prevention.
Answer: In the electrochemical theory, an iron surface in moist air develops tiny anodic and cathodic regions. At the anodic region, iron is oxidised: Fe → Fe2+ + 2e–. The electrons flow through the metal to the cathodic region where, in presence of moisture and dissolved oxygen, reduction occurs: O2 + 4H+ + 4e– → 2H2O. The Fe2+ ions then move out and are further oxidised by atmospheric oxygen to Fe3+, finally giving hydrated ferric oxide (Fe2O3·xH2O) — rust. Prevention: (i) Barrier protection — painting, oiling, electroplating with chromium or nickel, or galvanising with zinc to keep moisture away. (ii) Cathodic / sacrificial protection — connecting the iron object to a more active metal such as Mg or Zn so that the active metal corrodes preferentially and protects the iron.
Multiple Choice Questions (MCQ)
Q22. The standard electrode potential of SHE is:
(a) 1.0 V (b) 0.0 V (c) -1.0 V (d) 0.59 V
Answer: (b) 0.0 V.
Q23. The unit of specific conductance in SI is:
(a) S m-1 (b) S m2 mol-1 (c) ohm m (d) S
Answer: (a) S m-1.
Q24. One faraday of electricity is equal to:
(a) 9650 C (b) 96500 C (c) 1 C (d) 6.022 × 1023 C
Answer: (b) 96500 C.
Q25. Which of the following is a secondary cell?
(a) Dry cell (b) Mercury cell (c) Lead-storage cell (d) Daniell cell
Answer: (c) Lead-storage cell.
Q26. In a galvanic cell:
(a) Anode is positive (b) Cathode is negative (c) Oxidation at anode, reduction at cathode (d) Reduction at anode
Answer: (c).
Q27. The Nernst equation at 298 K for n = 2 has slope:
(a) 0.0591 V (b) 0.0295 V (c) 0.118 V (d) 0.59 V
Answer: (b) 0.0295 V.
Q28. Molar conductance increases on dilution because:
(a) Number of ions per unit volume increases (b) Degree of dissociation increases (c) Viscosity increases (d) Resistance increases
Answer: (b).
Q29. Rusting of iron is fastest in:
(a) Dry air (b) Moist air (c) Vacuum (d) Pure oxygen at low temperature
Answer: (b) Moist air.
Q30. The relation between $\Delta G^\circ$ and $E^\circ_{\text{cell}}$ is:
(a) $\Delta G^\circ = nFE^\circ$ (b) $\Delta G^\circ = -nFE^\circ$ (c) $\Delta G^\circ = -RT\ln E^\circ$ (d) $\Delta G^\circ = nF/E^\circ$
Answer: (b).
Q31. The degree of dissociation of a weak electrolyte is given by:
(a) $\alpha = \Lambda_m^\circ / \Lambda_m$ (b) $\alpha = \Lambda_m / \Lambda_m^\circ$ (c) $\alpha = \kappa / c$ (d) $\alpha = c / \kappa$
Answer: (b) $\alpha = \Lambda_m / \Lambda_m^\circ$.
Fill in the Blanks
Q32. The electrode at which oxidation takes place in a galvanic cell is the __________.
Answer: anode.
Q33. The unit of molar conductance in SI is __________.
Answer: S m2 mol-1.
Q34. The standard EMF of a Daniell cell is approximately __________ V.
Answer: 1.10.
Q35. The electrolyte in a lead-storage battery is __________.
Answer: dilute sulphuric acid (H2SO4).
Q36. A H2–O2 fuel cell uses __________ as electrolyte.
Answer: aqueous KOH.
True / False
Q37. In an electrolytic cell, the cathode is positive. (False)
Q38. Specific conductance increases on dilution. (False)
Q39. One faraday equals 96500 coulombs. (True)
Q40. The dry cell is rechargeable. (False)
Q41. Rusting of iron involves both oxidation of Fe and reduction of O2. (True)
Glossary
| Term | Meaning |
|---|---|
| Galvanic cell | Electrochemical cell that converts chemical energy to electrical energy spontaneously. |
| Electrolytic cell | Cell in which electrical energy drives a non-spontaneous reaction. |
| SHE | Standard Hydrogen Electrode, reference electrode with E° = 0.00 V. |
| Electrochemical series | Arrangement of elements in increasing order of standard reduction potential. |
| Specific conductance ($\kappa$) | Conductance of unit volume of an electrolyte solution. |
| Molar conductance ($\Lambda_m$) | Conductance of all ions from one mole of electrolyte at concentration c. |
| Kohlrausch’s law | At infinite dilution, each ion contributes its own constant share to $\Lambda_m^\circ$. |
| Faraday | Charge of one mole of electrons; 1 F = 96500 C. |
| Primary cell | Non-rechargeable galvanic cell, e.g. dry cell, mercury cell. |
| Secondary cell | Rechargeable cell, e.g. lead-storage, NiCd. |
| Fuel cell | Cell that converts the energy of combustion of a fuel directly to electricity. |
| Corrosion | Electrochemical deterioration of a metal by its environment. |
Formula Table
| Quantity | Formula |
|---|---|
| Standard cell EMF | $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}$ |
| Nernst equation (general) | $E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{RT}{nF}\ln Q$ |
| Nernst equation at 298 K | $E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n}\log Q$ |
| Gibbs energy and EMF | $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ |
| Gibbs energy and K | $\Delta G^\circ = -RT\ln K$; $\log K = \frac{nE^\circ_{\text{cell}}}{0.0591}$ |
| Specific conductance | $\kappa = \frac{1}{\rho}$ |
| Molar conductance | $\Lambda_m = \frac{\kappa \times 1000}{c}$ |
| Faraday’s first law | $w = Z \cdot Q = Z \cdot I \cdot t$ |
| Faraday’s second law | $\frac{w_1}{w_2} = \frac{E_1}{E_2}$ |
| Degree of dissociation | $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$ |
Quick Revision Notes
1. In every electrochemical cell, oxidation occurs at the anode and reduction at the cathode, regardless of whether it is galvanic or electrolytic. The polarity sign convention, however, is reversed between the two types.
2. The standard cell potential $E^\circ_{\text{cell}}$ is positive for a spontaneous reaction. Since $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, a positive EMF gives a negative Gibbs energy change.
3. The Nernst equation $E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0591}{n}\log Q$ shows that the cell potential decreases as the reaction proceeds because Q rises until at equilibrium Q = K and $E_{\text{cell}} = 0$. At that point $\log K = \frac{nE^\circ_{\text{cell}}}{0.0591}$.
4. For an electrolyte solution, $\kappa = \frac{1}{\rho}$ and $\Lambda_m = \frac{\kappa \times 1000}{c}$. Use SI units (S m-1, S m2 mol-1) or cgs units (S cm-1, S cm2 mol-1) consistently in numerical problems.
5. For weak electrolytes, $\Lambda_m^\circ$ cannot be obtained by extrapolation; it is calculated using Kohlrausch’s law from the limiting molar conductances of strong electrolytes that share its ions. The degree of dissociation is $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
6. In numericals based on Faraday’s laws, always convert minutes to seconds, calculate Q = I·t, then use $w = \frac{E \cdot Q}{96500}$ where E is the equivalent mass.
This complete revision pack of HSLC Guru on ASSEB Class 12 Chemistry Chapter 3 — Electrochemistry — covers cell concepts, EMF, Nernst equation, conductance, Faraday’s laws, batteries, fuel cells and corrosion, with all examination-oriented questions, MCQs, fills, true/false, glossary and formulas needed for the HS final.