Board Previous Year Questions

How to Score Full Marks in the Board Exam

This section is a complete bank of board-style questions with model answers for Electrochemistry. Write each answer the way an examiner wants: state the formula or definition first (e.g. $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$), substitute with units, and box the answer. In Nernst and $\Delta G$/$K_c$ numericals, always identify $n$ from the balanced equation before substituting — method marks follow.

1-Mark Questions (Definitions & Direct)

Q1. Define molar conductivity. Answer: Molar conductivity is the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is given by $\Lambda_m = \dfrac{\kappa \times 1000}{c}$, in S cm$^2$ mol$^{-1}$.

Q2. Why is the standard hydrogen electrode called a reference electrode? Answer: Its standard electrode potential is defined as exactly 0.00 V, and the potentials of all other electrodes are measured relative to it.

Q3. Write the relationship between standard cell potential and Gibbs energy. Answer: $\Delta_r G^\circ = -nFE^\circ_{cell}$, where $n$ is the moles of electrons transferred and $F = 96500$ C mol$^{-1}$.

Q4. State Faraday's first law of electrolysis. Answer: The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed: $m = ZIt$.

1-Mark Questions (continued)

Q5. What is the SI unit of cell constant? Answer: The cell constant $G^* = l/A$ has the unit m$^{-1}$ (or cm$^{-1}$).

Q6. Write the Nernst equation for a general cell at 298 K. Answer: $E_{cell} = E^\circ_{cell} - \dfrac{0.059}{n}\log Q$, where $Q$ is the reaction quotient.

Q7. Define the term 'limiting molar conductivity'. Answer: It is the molar conductivity of an electrolyte at infinite dilution (when concentration approaches zero), denoted $\Lambda^\circ_m$.

Q8. What is one faraday of charge equal to? Answer: One faraday is the charge on one mole of electrons, approximately $96500$ C mol$^{-1}$.

2-Mark Reasoning Questions

Q9. Why does the conductivity of a solution decrease but its molar conductivity increase on dilution? Answer: On dilution, the number of ions per unit volume decreases, so the conductivity (per unit volume) falls. However, ions move more freely because interionic attractions decrease, and weak electrolytes dissociate more, so the molar conductivity (per mole of electrolyte) rises.

Q10. State Kohlrausch's law of independent migration of ions and give one application. Answer: At infinite dilution, the limiting molar conductivity of an electrolyte is the sum of the independent contributions of its cation and anion: $\Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$. Application: determining $\Lambda^\circ_m$ of a weak electrolyte (e.g. acetic acid), which cannot be obtained by extrapolation.

2-Mark Reasoning Questions (continued)

Q11. Why is alternating current (not direct current) used to measure the resistance of an electrolyte solution? Answer: AC is used to prevent electrolysis and the resulting concentration changes and polarisation at the electrodes, which would otherwise alter the resistance during the measurement and give an unreliable reading.

Q12. Why does the cell potential of a Daniell cell decrease as it discharges? Answer: As the cell runs, [Zn$^{2+}$] increases and [Cu$^{2+}$] decreases, so the reaction quotient $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ rises. By the Nernst equation $E_{cell} = E^\circ_{cell} - \dfrac{0.059}{2}\log Q$, a larger $Q$ lowers $E_{cell}$, until it reaches zero in the fully discharged cell.

Q13. Why is the mercury cell able to maintain a constant voltage during its life? Answer: Its overall reaction, Zn(Hg) + HgO $\rightarrow$ ZnO + Hg, involves only solids and liquids, so there is no concentration of ions in the usual sense changing significantly during discharge. As a result, the cell potential remains nearly constant at about 1.35 V.

2-3 Mark Reasoning Questions

Q14. Distinguish between a galvanic cell and an electrolytic cell (any three points). Answer: (i) A galvanic cell converts chemical energy to electrical energy; an electrolytic cell converts electrical energy to chemical energy. (ii) The reaction in a galvanic cell is spontaneous; in an electrolytic cell it is non-spontaneous. (iii) In a galvanic cell the anode is negative and the cathode positive; in an electrolytic cell the anode is positive and the cathode negative. (Oxidation occurs at the anode and reduction at the cathode in both.)

Q15. Explain why the electrolysis of aqueous NaCl gives hydrogen at the cathode rather than sodium. Answer: In aqueous solution, the reduction of water to H$_2$ is far easier than the reduction of Na$^+$ because Na$^+$/Na has a very negative reduction potential. So water is preferentially reduced at the cathode, liberating H$_2$ gas, while Na$^+$ remains in solution. (Molten NaCl, having no water, gives Na metal.)

Q16. Why does a lead storage battery's sulphuric acid become more dilute on discharge? Answer: The overall discharge reaction, Pb + PbO$_2$ + 2H$_2$SO$_4$ $\rightarrow$ 2PbSO$_4$ + 2H$_2$O, consumes H$_2$SO$_4$ and produces water. This dilutes the electrolyte, so the acid density falls — and measuring the density is a way to check the state of charge.

3-Mark Numericals

Q17. The conductivity of a 0.20 M KCl solution at 298 K is 0.0248 S cm$^{-1}$. Calculate its molar conductivity. Answer: $\Lambda_m = \dfrac{\kappa \times 1000}{c} = \dfrac{0.0248 \times 1000}{0.20} = \mathbf{124\ S\ cm^2\ mol^{-1}}$.

Q18. Calculate $\Delta_r G^\circ$ for the Daniell cell ($E^\circ_{cell} = 1.10$ V, $n = 2$). ($F = 96500$ C mol$^{-1}$) Answer: $\Delta_r G^\circ = -nFE^\circ_{cell} = -2 \times 96500 \times 1.10 = -212300$ J $= \mathbf{-212.3\ kJ\ mol^{-1}}$.

Q19. Calculate $E_{cell}$ for Zn | Zn$^{2+}$(0.1 M) || Cu$^{2+}$(0.01 M) | Cu, given $E^\circ_{cell} = 1.10$ V. Answer: $E_{cell} = 1.10 - \dfrac{0.059}{2}\log\dfrac{0.1}{0.01} = 1.10 - 0.0295\log 10 = 1.10 - 0.0295 = \mathbf{1.07\ V}$.

3-Mark Numericals (continued)

Q20. Calculate the molar conductivity of acetic acid using Kohlrausch's law, given $\Lambda^\circ_m$(CH$_3$COONa) = 91.0, $\Lambda^\circ_m$(HCl) = 425.9, and $\Lambda^\circ_m$(NaCl) = 126.4 S cm$^2$ mol$^{-1}$. Answer: By Kohlrausch's law, $\Lambda^\circ_m$(CH$_3$COOH) = $\Lambda^\circ_m$(CH$_3$COONa) + $\Lambda^\circ_m$(HCl) - $\Lambda^\circ_m$(NaCl) = 91.0 + 425.9 - 126.4 = $\mathbf{390.5\ S\ cm^2\ mol^{-1}}$.

Q21. How much charge is required to deposit 0.5 mol of copper from CuSO$_4$ solution (Cu$^{2+}$ + 2e$^-$ $\rightarrow$ Cu)? Answer: 0.5 mol Cu needs $0.5 \times 2 = 1$ mol of electrons $= 1 \times 96500 = \mathbf{96500\ C}$.

Q22. Calculate the standard cell potential $E^\circ_{cell}$ for a cell with $E^\circ_{cathode} = +0.34$ V and $E^\circ_{anode} = -0.76$ V. Answer: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = \mathbf{1.10\ V}$.

3-Mark Numericals (continued)

Q23. For a cell, $E^\circ_{cell} = 0.59$ V and $n = 2$. Calculate the equilibrium constant $K_c$ at 298 K. Answer: $\log K_c = \dfrac{nE^\circ_{cell}}{0.059} = \dfrac{2 \times 0.59}{0.059} = 20$. Therefore $K_c = \mathbf{1 \times 10^{20}}$.

Q24. A current of 5 A flows for 30 minutes through molten AlCl$_3$. What mass of Al (M = 27) is deposited? Answer: $Q = It = 5 \times 1800 = 9000$ C. Moles of electrons $= 9000/96500 = 0.0933$. Al$^{3+}$ needs 3 e$^-$, so moles of Al $= 0.0933/3 = 0.0311$. Mass $= 0.0311 \times 27 = \mathbf{0.84\ g}$.

Q25. The molar conductivity of 0.001 M acetic acid is 48.15 S cm$^2$ mol$^{-1}$ and $\Lambda^\circ_m = 390.5$. Calculate the degree of dissociation. Answer: $\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m} = \dfrac{48.15}{390.5} = \mathbf{0.123}$ (12.3% dissociated).

3-Mark Numericals & Reactions

Q26. A conductivity cell filled with 0.1 M KCl (conductivity 1.29 S m$^{-1}$) has a resistance of 100 $\Omega$. Find the cell constant. Answer: $\kappa = \dfrac{G^*}{R} \Rightarrow G^* = \kappa \times R = 1.29 \times 100 = \mathbf{129\ m^{-1}}$ (= 1.29 cm$^{-1}$).

Q27. Write the electrode reactions and the overall reaction of the lead storage battery during discharge. Answer: Anode: Pb + SO$_4^{2-}$ $\rightarrow$ PbSO$_4$ + 2e$^-$. Cathode: PbO$_2$ + SO$_4^{2-}$ + 4H$^+$ + 2e$^-$ $\rightarrow$ PbSO$_4$ + 2H$_2$O. Overall: Pb + PbO$_2$ + 2H$_2$SO$_4$ $\rightarrow$ 2PbSO$_4$ + 2H$_2$O.

Q28. Write the electrode reactions of the hydrogen-oxygen fuel cell (alkaline medium). Answer: Anode: 2H$_2$ + 4OH$^-$ $\rightarrow$ 4H$_2$O + 4e$^-$. Cathode: O$_2$ + 2H$_2$O + 4e$^-$ $\rightarrow$ 4OH$^-$. Overall: 2H$_2$ + O$_2$ $\rightarrow$ 2H$_2$O.

3-Mark Numericals (continued)

Q29. Calculate the time required to deposit 1.08 g of silver (M = 108, n = 1) using a current of 3 A. Answer: Moles of Ag $= 1.08/108 = 0.01$ mol = 0.01 mol electrons. $Q = 0.01 \times 96500 = 965$ C. $t = Q/I = 965/3 = \mathbf{321.7\ s}$ (~5.4 min).

Q30. The potential of a hydrogen electrode in a solution of pH 5 (H$_2$ at 1 bar) is required. Answer: For 2H$^+$ + 2e$^-$ $\rightarrow$ H$_2$, $E = -0.059 \times \text{pH} = -0.059 \times 5 = \mathbf{-0.295\ V}$.

Q31. Calculate $E_{cell}$ for a concentration cell Cu | Cu$^{2+}$(0.001 M) || Cu$^{2+}$(0.1 M) | Cu. Answer: For a concentration cell $E^\circ_{cell} = 0$. $E_{cell} = -\dfrac{0.059}{2}\log\dfrac{0.001}{0.1} = -0.0295 \times (-2) = \mathbf{0.059\ V}$.

5-Mark / Long-Answer Questions

Q32. Derive the relationship $E^\circ_{cell} = \dfrac{0.059}{n}\log K_c$. Answer: At equilibrium the cell does no net work, so $E_{cell} = 0$ and the reaction quotient $Q$ equals the equilibrium constant $K_c$. Substituting into the Nernst equation $E_{cell} = E^\circ_{cell} - \dfrac{0.059}{n}\log Q$ gives $0 = E^\circ_{cell} - \dfrac{0.059}{n}\log K_c$. Rearranging, $E^\circ_{cell} = \dfrac{0.059}{n}\log K_c$. (This can also be obtained by combining $\Delta_r G^\circ = -nFE^\circ_{cell}$ with $\Delta_r G^\circ = -RT\ln K_c$.)

Q33. Explain the construction and working of the standard hydrogen electrode. Answer: The standard hydrogen electrode (SHE) consists of a platinum electrode coated with platinum black, dipped in a 1 M H$^+$ solution, over which pure hydrogen gas at 1 bar is bubbled at 298 K. The half-reaction is 2H$^+$(aq, 1 M) + 2e$^-$ $\rightleftharpoons$ H$_2$(g, 1 bar). Its electrode potential is defined as exactly 0.00 V, and it serves as the reference against which all other electrode potentials are measured by constructing a cell with the SHE.

5-Mark / Long-Answer Questions (continued)

Q34. Describe how corrosion (rusting of iron) occurs as an electrochemical process, and give two methods of prevention. Answer: Rusting is essentially a tiny galvanic cell on the iron surface in the presence of water and air. At an anodic spot, iron is oxidised: Fe $\rightarrow$ Fe$^{2+}$ + 2e$^-$ ($E^\circ = -0.44$ V). The electrons travel through the metal to a cathodic spot where oxygen is reduced: O$_2$ + 2H$_2$O + 4e$^-$ $\rightarrow$ 4OH$^-$. The Fe$^{2+}$ ions are further oxidised by atmospheric oxygen and water to hydrated ferric oxide (Fe$_2$O$_3\cdot x$H$_2$O), which is rust. Prevention: (i) barrier protection by painting/greasing or galvanisation (coating with zinc); (ii) sacrificial (cathodic) protection by connecting the iron to a more reactive metal such as magnesium, which corrodes instead.

Q35. The conductivity of 0.001 M acetic acid is 4.95 × 10$^{-5}$ S cm$^{-1}$. Calculate its molar conductivity, and given $\Lambda^\circ_m = 390.5$, find the degree of dissociation. Answer: $\Lambda_m = \dfrac{\kappa \times 1000}{c} = \dfrac{4.95 \times 10^{-5} \times 1000}{0.001} = 49.5$ S cm$^2$ mol$^{-1}$. Degree of dissociation $\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m} = \dfrac{49.5}{390.5} = \mathbf{0.127}$.

Q36. State and explain the two Faraday's laws of electrolysis. Answer: First law: the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed, $m = ZIt$ ($Z$ = electrochemical equivalent). Second law: when the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent masses. Together they show that depositing one mole of a metal M$^{n+}$ requires $n$ faradays ($n \times 96500$ C).