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RBSE SS Chemistry study materials
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RBSE SS Chemistry practice questions

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Question 1Solid State and Surface Chemistry

How many atoms are effectively present in one simple cubic unit cell in which atoms occupy only the eight corners?

  1. 2
  2. 1 Correct answer.
  3. 4
  4. 8

Explanation

Each corner atom is shared by eight adjacent unit cells, so its contribution to one cell is 1/8. For eight corners the total is 8 × 1/8 = 1 atom per simple cubic unit cell.

Question 2Solid State and Surface Chemistry

In the sodium chloride crystal lattice, what is the coordination number of each Na+ ion?

  1. 8
  2. 4
  3. 12
  4. 6 Correct answer.

Explanation

In NaCl each Na+ ion is surrounded by six Cl- ions arranged octahedrally, and each Cl- ion is likewise surrounded by six Na+ ions. The lattice therefore has 6:6 coordination.

Question 3Solid State and Surface Chemistry

The yellow colour produced when NaCl crystals are heated in sodium vapour is due to which species?

  1. Schottky defects
  2. Interstitial chloride ions
  3. F-centres Correct answer.
  4. Frenkel defects

Explanation

When NaCl is heated in sodium vapour, sodium atoms deposit and chloride-ion vacancies are created. Electrons trapped in these anion vacancies are called F-centres, and they absorb visible light to give the yellow colour.

Question 4Solid State and Surface Chemistry

A metal crystallises in a face-centred cubic unit cell with edge length 400 pm and molar mass 60 g mol^-1. Using N_A = 6.022 × 10^23 mol^-1, what is its density?

  1. 1.56 g cm^-3
  2. 3.11 g cm^-3
  3. 12.5 g cm^-3
  4. 6.23 g cm^-3 Correct answer.

Explanation

For an fcc cell Z = 4, and the edge length 400 pm must first be converted to 4.00 × 10^-8 cm. Density d = ZM/(N_A a^3) = (4 × 60)/(6.022 × 10^23 × (4.00 × 10^-8)^3) = 6.23 g cm^-3.

Question 5Solid State and Surface Chemistry

How many tetrahedral voids are present per unit cell in a cubic close-packed (ccp) structure?

  1. 8 Correct answer.
  2. 4
  3. 12
  4. 2

Explanation

In ccp the number of atoms per unit cell is four. Tetrahedral voids are twice the number of atoms, so the cell contains 2 × 4 = 8 tetrahedral voids.

Question 6Solid State and Surface Chemistry

Which statement correctly describes a Frenkel defect in an ionic crystal?

  1. It produces equal numbers of cation and anion vacancies throughout the lattice.
  2. A cation moves to an interstitial site, leaving its lattice site vacant. Correct answer.
  3. It is the main defect in sodium chloride and is responsible for its colour.
  4. It decreases the density of the crystal because lattice sites are vacated.

Explanation

A Frenkel defect is formed when an ion, usually a cation, leaves its normal lattice position and occupies an interstitial site. Because the ion remains in the crystal, the density is almost unchanged, and the defect is favoured when the ions differ considerably in size.

Question 7Solid State and Surface Chemistry

For the Freundlich adsorption isotherm, what does the slope of the plot of log(x/m) against log p represent?

  1. n
  2. log k
  3. k
  4. 1/n Correct answer.

Explanation

The Freundlich equation is x/m = k p^(1/n). Taking logarithms gives log(x/m) = log k + (1/n) log p, so a plot of log(x/m) against log p has slope 1/n and intercept log k.

Question 8Solid State and Surface Chemistry

Which property is shown by a colloid but not by a true solution?

  1. It scatters a beam of light, producing a visible Tyndall cone. Correct answer.
  2. It cannot be filtered through ordinary filter paper but can be filtered through a semipermeable membrane.
  3. It settles completely under gravity within a few minutes.
  4. It is optically homogeneous and shows no scattering of light.

Explanation

Colloidal particles are large enough to scatter light, giving the Tyndall effect, which true solutions do not show. Colloids pass through ordinary filter paper but are retained by a semipermeable membrane.

Question 9Solid State and Surface Chemistry

Silicon doped with boron behaves as a p-type semiconductor. What is the correct reason?

  1. Boron has one more valence electron than silicon, donating free electrons.
  2. Boron atoms occupy interstitial sites and donate electrons.
  3. Boron removes silicon atoms, leaving Schottky defects.
  4. Boron has one fewer valence electron than silicon, creating electron holes. Correct answer.

Explanation

Boron is a group 13 element and has three valence electrons, one fewer than silicon. When it replaces a silicon atom in the lattice, an electron-deficient site or hole is created, so the doped crystal conducts by holes and is p-type.

Question 10Solid State and Surface Chemistry

X-rays of wavelength 154 pm are diffracted by a crystal at a glancing angle of 30.0° in the first order. What is the interplanar spacing d?

  1. 77 pm
  2. 154 pm Correct answer.
  3. 267 pm
  4. 308 pm

Explanation

Bragg's law is nλ = 2d sin θ. For first order n = 1 and sin 30.0° = 0.500, so d = λ/(2 sin θ) = 154 pm/(2 × 0.500) = 154 pm.

Question 11Solid State and Surface Chemistry

In a body-centred cubic (bcc) arrangement of identical spheres, what percentage of the unit-cell volume is occupied by the spheres?

  1. 68% Correct answer.
  2. 90.4%
  3. 52.4%
  4. 74%

Explanation

In a bcc unit cell the spheres touch along the body diagonal, giving a packing efficiency of 68%. This value follows from the two atoms per cell and the relation r = √3a/4; simple cubic gives 52.4% and ccp/fcc gives 74%.

Question 12Solid State and Surface Chemistry

Which pair of features is characteristic of physisorption on a solid surface?

  1. Low enthalpy of adsorption and reversibility Correct answer.
  2. High specificity and very low activation energy
  3. High enthalpy of adsorption and irreversibility
  4. Formation of covalent bonds and multilayer coverage

Explanation

Physisorption arises from weak van der Waals forces, so it has a low enthalpy of adsorption, roughly 20–40 kJ mol^-1, and is reversible at moderate conditions. Chemisorption involves bond formation and is typically high in enthalpy and irreversible.

Question 13Solid State and Surface Chemistry

A metal crystallises in a body-centred cubic unit cell with edge length 300 pm. If its molar mass is 54 g mol^-1, what is the density of the metal? Use N_A = 6.022 × 10^23 mol^-1.

  1. 3.32 g cm^-3
  2. 6.64 g cm^-3 Correct answer.
  3. 13.3 g cm^-3
  4. 1.66 g cm^-3

Explanation

For a bcc cell, Z = 2. The cell volume is (300 pm)^3 = 2.70 × 10^-23 cm^3, and the mass per cell is 2 × 54 / (6.022 × 10^23) = 1.79 × 10^-22 g. Dividing mass by volume gives a density of about 6.64 g cm^-3.

Question 14Solid State and Surface Chemistry

Which defect is produced when equal numbers of cations and anions are missing from an ionic lattice, keeping the crystal electrically neutral?

  1. Metal-excess defect
  2. Schottky defect Correct answer.
  3. Frenkel defect
  4. Interstitial defect

Explanation

A Schottky defect consists of a missing cation and a missing anion in equal numbers, so electrical neutrality is preserved. It is common in highly ionic solids with comparable cation and anion sizes, such as NaCl.

Question 15Solid State and Surface Chemistry

According to the Langmuir adsorption isotherm, which plot should be linear for adsorption of a gas on a solid?

  1. x/m against p
  2. log p against log(x/m)
  3. p/(x/m) against p Correct answer.
  4. log(x/m) against log p

Explanation

The Langmuir isotherm is x/m = ap/(1 + bp). Rearranging gives p/(x/m) = 1/a + (b/a)p, so a plot of p/(x/m) against p is linear with slope b/a and intercept 1/a.

Question 16Solid State and Surface Chemistry

An ionic solid AB has a CsCl-type structure with unit-cell edge length 400 pm. If the radius of A+ is 170 pm, what is the radius of B−, assuming the ions touch along the body diagonal?

  1. 400 pm
  2. 346 pm
  3. 176 pm Correct answer.
  4. 230 pm

Explanation

In the CsCl-type structure the ions touch along the body diagonal, so 2(r+ + r−) = √3a. With a = 400 pm, r+ + r− = (√3 × 400)/2 = 346.4 pm; hence r− = 346.4 − 170 = 176.4 pm, about 176 pm.

Question 17Solutions

Which concentration term expresses the number of moles of solute dissolved in one kilogram of solvent and is independent of temperature?

  1. Normality
  2. Mass percentage
  3. Molality Correct answer.
  4. Molarity

Explanation

Molality is defined as the moles of solute present per kilogram of solvent, so it is built only from mass and moles. Because no volume measurement appears in the definition, a change in temperature that expands or contracts the solution does not change the molality. Molarity and normality, by contrast, involve the volume of solution and therefore vary with temperature.

Question 18Solutions

At constant temperature, the solubility of a sparingly soluble gas in a liquid, according to Henry's law, is directly proportional to which quantity?

  1. the total volume of the solution prepared
  2. the boiling point of the solvent
  3. its partial pressure above the solvent Correct answer.
  4. the molar mass of the gas

Explanation

Henry's law states that the mole fraction of a gas dissolved in a liquid is proportional to the partial pressure of that gas in the vapour phase above the liquid, the constant of proportionality being the Henry's law constant. Raising the pressure therefore pushes more gas into solution, which is why soft drinks are bottled under carbon dioxide pressure.

Question 19Solutions

The osmotic pressure of a dilute solution is called a colligative property because it depends on which of the following?

  1. the colour and density of the solution
  2. the molar mass of the solvent only
  3. the nature of the solute particles only
  4. the number of solute particles and not their nature Correct answer.

Explanation

Colligative properties are those that depend only on the number of solute particles present in a given amount of solvent, not on their chemical identity. Osmotic pressure, like relative lowering of vapour pressure, elevation of boiling point and depression of freezing point, changes in proportion to the particle count. This is why 1 mol of glucose and 0.5 mol of sodium chloride produce similar effects when ionisation is ignored.

Question 20Solutions

5.85 g of sodium chloride (molar mass 58.5 g mol⁻¹) is dissolved in water and the solution is made up to 500 mL. What is the molarity of the solution?

  1. 0.05 M
  2. 0.50 M
  3. 0.20 M Correct answer.
  4. 0.10 M

Explanation

The amount of sodium chloride is 5.85 g ÷ 58.5 g mol⁻¹ = 0.10 mol. Molarity is moles of solute divided by the volume of solution in litres, so 0.10 mol ÷ 0.500 L = 0.20 M. Note that the solution volume, not the volume of water added, must be used.

Question 21Solutions

For a binary solution containing a non-volatile solute in a volatile solvent, the relative lowering of vapour pressure is equal to which quantity?

  1. the mole fraction of the solvent in the solution
  2. the molality of the solute multiplied by 1000
  3. the ratio of the mass of solute to the mass of solvent
  4. the mole fraction of the solute in the solution Correct answer.

Explanation

Raoult's law for a non-volatile solute gives (p° − p)/p° = x₂, where x₂ is the mole fraction of the solute. The relative lowering of vapour pressure therefore depends only on the mole fraction of solute, not on its identity. This relation is the basis for determining molar masses of non-volatile solutes from vapour-pressure measurements.

Question 22Solutions

When 1.8 g of glucose (molar mass 180 g mol⁻¹) is dissolved in 100 g of water, what is the elevation in boiling point? (Kb for water = 0.512 K kg mol⁻¹)

  1. 0.0051 K
  2. 1.02 K
  3. 0.051 K Correct answer.
  4. 0.512 K

Explanation

The amount of glucose is 1.8 g ÷ 180 g mol⁻¹ = 0.010 mol, and the mass of solvent is 0.100 kg, so the molality is 0.010 ÷ 0.100 = 0.10 mol kg⁻¹. Applying ΔTb = Kb × m gives 0.512 × 0.10 = 0.0512 K, approximately 0.051 K. Glucose does not dissociate, so no van't Hoff factor is needed.

Question 23Solutions

An aqueous solution contains 0.1 mol of sodium chloride per litre at 300 K. Assuming complete dissociation, what is its osmotic pressure? (R = 0.083 L bar K⁻¹ mol⁻¹)

  1. 9.96 bar
  2. 2.49 bar
  3. 4.98 bar Correct answer.
  4. 0.83 bar

Explanation

Sodium chloride dissociates completely into two ions, so the van't Hoff factor i = 2. Using π = iCRT = 2 × 0.1 mol L⁻¹ × 0.083 L bar K⁻¹ mol⁻¹ × 300 K gives about 4.98 bar. The factor of two must be included because osmotic pressure counts particles, not formula units.

Question 24Solutions

A mixture of ethanol and water shows a large positive deviation from Raoult's law. Which statement about this mixture is correct?

  1. Its components can be separated completely by fractional distillation at all compositions.
  2. It forms a minimum-boiling azeotrope. Correct answer.
  3. It boils at a temperature higher than both pure components at every composition.
  4. The ethanol–water attractions are stronger than in the pure liquids.

Explanation

A positive deviation means the new interactions between unlike molecules are weaker than those in the pure liquids, so escaping tendency and vapour pressure are higher than Raoult's law predicts. Higher vapour pressure implies a lower boiling point, and at one composition such a system gives a minimum-boiling azeotrope. Ethanol–water is the standard example, which is why absolute ethanol cannot be obtained by ordinary fractional distillation.

Question 25Solutions

Acetic acid dimerises in benzene. If 60% of the acid is dimerised in a 0.1 molal solution, what is the van't Hoff factor of the acid?

  1. 1.60
  2. 0.40
  3. 0.70 Correct answer.
  4. 1.30

Explanation

When two molecules associate to form one dimer, the number of particles decreases, so i = 1 − α/2, where α is the degree of association. With α = 0.60, i = 1 − 0.30 = 0.70. Because i is less than one, the observed molar mass of the acid in benzene is greater than its formula mass, being equal to the normal value divided by 0.70.

Question 26Solutions

A 0.1 molal aqueous solution of calcium chloride is assumed to dissociate completely. What is the depression in freezing point? (Kf for water = 1.86 K kg mol⁻¹)

  1. 1.116 K
  2. 0.558 K Correct answer.
  3. 0.372 K
  4. 0.186 K

Explanation

Calcium chloride gives three ions per formula unit, so i = 3 for complete dissociation. Applying ΔTf = i × Kf × m = 3 × 1.86 × 0.1 gives 0.558 K. The observed depression is therefore three times that of a non-electrolyte solution of the same molality.

Question 27Solutions

Which statement correctly describes an ideal solution formed by two volatile liquids?

  1. It obeys Raoult's law at every composition, with ΔmixH = 0. Correct answer.
  2. It forms a minimum-boiling azeotrope at one composition.
  3. It shows a large negative deviation from Raoult's law.
  4. It has ΔmixV > 0 and a strong solute–solvent interaction.

Explanation

An ideal solution obeys Raoult's law over the entire composition range because the intermolecular forces between unlike molecules are essentially the same as those between like molecules. Consequently no heat is absorbed or released on mixing (ΔmixH = 0) and there is no volume change (ΔmixV = 0). Ideal solutions also never form azeotropes, since each component's partial pressure stays proportional to its mole fraction throughout.

Question 28Solutions

Dissolution of potassium nitrate in water is endothermic (ΔsolH > 0). How does its solubility in water change when the temperature is raised?

  1. Solubility increases as the temperature rises. Correct answer.
  2. Solubility decreases because the process is endothermic.
  3. Solubility stays exactly constant at every temperature.
  4. Solubility rises and then falls sharply above 313 K.

Explanation

For an endothermic dissolution, heat behaves like a reactant, so raising the temperature shifts the dissolution equilibrium towards the dissolved state (Le Chatelier's principle). The extra thermal energy also helps break the solute–solute and solvent–solvent attractions that must be overcome. Hence the solubility of solids such as KNO₃ increases with temperature, which is why their crystals are purified by recrystallisation from hot water.

Question 29Solutions

18 g of glucose (molar mass 180 g mol⁻¹) is dissolved in 90 g of water (molar mass 18 g mol⁻¹). What is the mole fraction of glucose in the solution?

  1. 0.20
  2. 0.017
  3. 0.98
  4. 0.020 Correct answer.

Explanation

Moles of glucose = 18/180 = 0.10 mol and moles of water = 90/18 = 5.0 mol, so the total is 5.10 mol. The mole fraction of glucose is therefore 0.10/5.10 = 0.0196, which rounds to 0.020. Mole fraction is a ratio of moles and must be calculated after converting both masses with their own molar masses.

Question 30Solutions

Four aqueous solutions of equal molality, each 0.1 mol kg⁻¹, are prepared. Assuming complete dissociation of the electrolytes, which solution has the lowest freezing point?

  1. 0.1 molal aluminium sulphate Correct answer.
  2. 0.1 molal urea
  3. 0.1 molal potassium chloride
  4. 0.1 molal glucose

Explanation

Depression of freezing point is proportional to the total number of particles, i.e. ΔTf = i·Kf·m, so the solute with the largest van't Hoff factor gives the lowest freezing point. Aluminium sulphate dissociates as Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻, giving i = 5, the highest value among the four solutes. Glucose and urea are non-electrolytes (i = 1) and KCl gives i = 2, so all three depress the freezing point less than aluminium sulphate does.

Question 31Solutions

The vapour pressure of pure water at 298 K is 23.8 mm Hg. What is the vapour pressure of a solution prepared by dissolving 3.0 g of urea (molar mass 60 g mol⁻¹) in 90 g of water?

  1. 23.8 mm Hg
  2. 24.9 mm Hg
  3. 22.5 mm Hg
  4. 23.6 mm Hg Correct answer.

Explanation

Moles of urea = 3.0/60 = 0.050 mol and moles of water = 90/18 = 5.0 mol, so the mole fraction of water is 5.0/5.05 = 0.990. Urea is non-volatile, so only water contributes to the vapour pressure and Raoult's law gives p = 0.990 × 23.8 = 23.6 mm Hg. The small lowering from 23.8 mm Hg is consistent with the low mole fraction of solute present.

Question 32Solutions

A 0.10 molal aqueous solution of a weak monobasic acid freezes at –0.2046 °C. Taking Kf for water as 1.86 K kg mol⁻¹, what is the percentage degree of dissociation of the acid?

  1. 1%
  2. 10% Correct answer.
  3. 20%
  4. 5%

Explanation

The observed depression is 0.2046 K, so the van't Hoff factor is i = ΔTf/(Kf·m) = 0.2046/(1.86 × 0.10) = 1.10. For a weak monobasic acid HA ⇌ H⁺ + A⁻, each molecule that ionises increases the particle count by one, so i = 1 + α and α = 1.10 − 1 = 0.10. The percentage degree of dissociation is therefore 10%.

Question 33Electrochemistry

In a Daniell cell, what is the main function of the salt bridge?

  1. It prevents any movement of charge between the two compartments.
  2. It completes the circuit by allowing ions to migrate between the two half-cell solutions. Correct answer.
  3. It acts as a catalyst and increases the rate of the electrode reactions.
  4. It supplies electrons to the cathode through the external circuit.

Explanation

The salt bridge contains a concentrated inert electrolyte, usually KCl in agar, whose ions migrate into the two half-cells. This migration maintains electrical neutrality so that the external flow of electrons can continue. The bridge also keeps the two solutions separate and prevents them from mixing directly.

Question 34Electrochemistry

The molar conductivity of an electrolyte solution is expressed in which unit?

  1. S cm⁻¹
  2. S cm² mol⁻¹ Correct answer.
  3. Ω cm
  4. mol cm⁻²

Explanation

Molar conductivity is defined as Λm = κ/c, where κ is conductivity in S cm⁻¹ and c is concentration in mol cm⁻³. Substituting these units gives S cm² mol⁻¹, whose SI equivalent is S m² mol⁻¹.

Question 35Electrochemistry

In the galvanic cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), which statement about the electrodes is correct?

  1. Zinc is oxidised at the anode and copper ions are reduced at the cathode. Correct answer.
  2. Copper is oxidised at the anode and zinc ions are reduced at the cathode.
  3. Reduction occurs at the zinc electrode while oxidation occurs at the copper electrode.
  4. Both zinc and copper are oxidised, one at each electrode.

Explanation

Zinc has the more negative standard reduction potential (−0.76 V) than copper (+0.34 V), so zinc is the stronger reducing agent. Zinc metal therefore loses electrons and is oxidised at the anode, while Cu²⁺ ions gain those electrons and are reduced at the cathode.

Question 36Electrochemistry

For the cell Zn(s) | Zn²⁺(0.10 M) || Cu²⁺(0.010 M) | Cu(s) at 298 K, the standard cell potential is 1.10 V. Using the Nernst equation, what is the cell potential?

  1. 1.04 V
  2. 1.13 V
  3. 1.10 V
  4. 1.07 V Correct answer.

Explanation

The cell reaction transfers two electrons, so Ecell = E°cell − (0.0591/2) log([Zn²⁺]/[Cu²⁺]). The concentration ratio is 0.10/0.010 = 10, and log 10 = 1. Hence Ecell = 1.10 − 0.0296 = 1.07 V.

Question 37Electrochemistry

A cell reaction involves two electrons and has E°cell = +0.46 V at 298 K. What is ΔG° for the reaction? (F = 96485 C mol⁻¹)

  1. −88.8 kJ mol⁻¹ Correct answer.
  2. −177.6 kJ mol⁻¹
  3. −44.4 kJ mol⁻¹
  4. +88.8 kJ mol⁻¹

Explanation

The relation between standard free energy and cell potential is ΔG° = −nFE°cell. Substituting n = 2, F = 96485 C mol⁻¹ and E° = 0.46 V gives −88766 J mol⁻¹, approximately −88.8 kJ mol⁻¹. The negative sign shows that the cell reaction is spontaneous under standard conditions.

Question 38Electrochemistry

At 298 K, the limiting molar conductivities are Λ°m(NaCl) = 126.4, Λ°m(HCl) = 426.2 and Λ°m(CH₃COONa) = 91.0 S cm² mol⁻¹. Using Kohlrausch's law, what is Λ°m(CH₃COOH)?

  1. 390.8 S cm² mol⁻¹ Correct answer.
  2. 260.8 S cm² mol⁻¹
  3. 643.6 S cm² mol⁻¹
  4. 517.2 S cm² mol⁻¹

Explanation

Kohlrausch's law allows the limiting molar conductivity of a weak electrolyte to be built from strong electrolytes: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl). Substituting gives 91.0 + 426.2 − 126.4 = 390.8 S cm² mol⁻¹. The Na⁺ and Cl⁻ contributions cancel, leaving λ°(H⁺) + λ°(CH₃COO⁻).

Question 39Electrochemistry

A current of 2.0 A is passed through molten AlCl₃ for 30 minutes. What mass of aluminium is deposited at the cathode? (Molar mass of Al = 27 g mol⁻¹, F = 96500 C mol⁻¹)

  1. 0.672 g
  2. 0.336 g Correct answer.
  3. 3.36 g
  4. 1.01 g

Explanation

The charge passed is Q = It = 2.0 × 1800 = 3600 C, so the moles of electrons are 3600/96500 = 0.0373 mol. Because Al³⁺ requires three electrons per atom, the moles of aluminium are 0.0373/3 = 0.0124 mol. Multiplying by 27 g mol⁻¹ gives about 0.336 g of aluminium.

Question 40Electrochemistry

During the discharge of a lead-acid storage battery, which change occurs at the anode?

  1. PbO₂ is reduced to PbSO₄.
  2. Pb is oxidised to PbSO₄. Correct answer.
  3. Pb is reduced to PbSO₄.
  4. PbSO₄ is oxidised to PbO₂.

Explanation

On discharge, lead metal at the anode loses electrons and is oxidised to sparingly soluble PbSO₄, while PbO₂ at the cathode is reduced to PbSO₄. The overall process consumes sulphuric acid and produces water, lowering the density of the electrolyte.

Question 41Electrochemistry

For a two-electron cell reaction at 298 K, the standard cell potential is 0.295 V. What is the equilibrium constant Kc of the cell reaction?

  1. 1 × 10⁵
  2. 1 × 10²⁰
  3. 2 × 10¹⁰
  4. 1 × 10¹⁰ Correct answer.

Explanation

At 298 K the equilibrium constant is related to the standard cell potential by log Kc = nE°cell/0.0591. Substituting n = 2 and E°cell = 0.295 V gives log Kc = 0.590/0.0591 ≈ 10. Therefore Kc is approximately 1 × 10¹⁰.

Question 42Electrochemistry

Which statement best explains why iron rusts rapidly in neutral, aerated water containing dissolved oxygen?

  1. Oxygen is reduced at cathodic regions while iron is oxidised at anodic regions, and the Fe²⁺ formed is further oxidised to hydrated Fe(III) oxide. Correct answer.
  2. Iron acts as the cathode and is reduced to Fe²⁺ at cathodic regions.
  3. Hydrogen gas is oxidised at the anode to produce water and oxide ions.
  4. Rusting is a purely chemical process in which no electrons are transferred.

Explanation

Rusting is electrochemical corrosion. At anodic regions iron loses electrons to form Fe²⁺, and at cathodic regions dissolved oxygen gains electrons in the presence of water to form OH⁻. The Fe²⁺ is then oxidised by oxygen to hydrated iron(III) oxide, the familiar rust.

Question 43Electrochemistry

In the standard hydrogen electrode, the platinum foil is coated with platinum black. What is the main purpose of this coating?

  1. It prevents hydrogen gas from escaping from the electrode compartment.
  2. It increases the effective surface area and catalyses the H₂/H⁺ interconversion. Correct answer.
  3. It allows the platinum metal to dissolve slowly and keep the ion concentration constant.
  4. It fixes the electrode potential at exactly zero volt under all conditions.

Explanation

Platinum black is finely divided platinum that greatly increases the surface area of the electrode and acts as a catalyst for the oxidation of H₂ to H⁺ or the reduction of H⁺ to H₂. The zero-volt value is a convention fixed by definition, not produced by the coating. Good catalysis ensures the electrode reaction remains reversible and the potential quickly becomes steady.

Question 44Electrochemistry

A conductivity cell has two parallel platinum electrodes, each of area 1.5 cm², separated by a distance of 0.75 cm. What is the cell constant of this cell?

  1. 0.75 cm⁻¹
  2. 2.00 cm⁻¹
  3. 0.50 cm⁻¹ Correct answer.
  4. 1.125 cm⁻¹

Explanation

The cell constant is defined as l/A, the distance between the electrodes divided by their area. Here l/A = 0.75 cm ÷ 1.5 cm² = 0.50 cm⁻¹. Its unit is the reciprocal of length, and it is determined by the geometry of the cell, not by the electrolyte used.

Question 45Electrochemistry

For the half-cell Zn²⁺(aq) + 2e⁻ → Zn(s), the standard electrode potential is −0.76 V at 298 K. What is the electrode potential when [Zn²⁺] = 1.0 × 10⁻³ M?

  1. −0.67 V
  2. −0.94 V
  3. −0.76 V
  4. −0.85 V Correct answer.

Explanation

Apply the Nernst equation to the reduction half-reaction: E = E° − (0.059/2) log(1/[Zn²⁺]). With [Zn²⁺] = 10⁻³ M, log(10³) = 3, so E = −0.76 − (0.0295 × 3) = −0.85 V. Diluting the ion makes reduction less favourable, so the potential becomes more negative than the standard value.

Question 46Electrochemistry

A conductometric titration of hydrochloric acid with sodium hydroxide solution is carried out. How does the conductivity of the solution change as the base is added?

  1. It falls to a minimum at the equivalence point and then rises again. Correct answer.
  2. It falls continuously throughout the titration until excess base is present.
  3. It remains almost unchanged up to the equivalence point and then falls sharply.
  4. It rises steadily until the equivalence point and then stays constant.

Explanation

Before the equivalence point, fast-moving H⁺ ions are replaced by slower Na⁺ ions, so the conductivity decreases steadily. At the equivalence point only NaCl remains and the conductivity reaches its lowest value. Adding excess NaOH introduces highly mobile OH⁻ ions, so the conductivity rises again, giving the familiar V-shaped curve.

Question 47Electrochemistry

In an alkaline hydrogen–oxygen fuel cell, which half-reaction occurs at the anode?

  1. H₂(g) → 2H⁺(aq) + 2e⁻
  2. H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻ Correct answer.
  3. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
  4. O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

Explanation

In an alkaline fuel cell the electrolyte supplies OH⁻ ions, so hydrogen is oxidised by reaction with hydroxide to give water and electrons. The oxygen reduction that uses up water and produces OH⁻ takes place at the cathode. Writing H₂ → 2H⁺ + 2e⁻ would describe an acidic medium, not the alkaline electrolyte used here.

Question 48Electrochemistry

Copper is deposited from aqueous CuSO₄ solution at a cathode. What charge is needed to deposit 0.635 g of copper? (M(Cu) = 63.5 g mol⁻¹, F = 96485 C mol⁻¹)

  1. 483 C
  2. 1930 C Correct answer.
  3. 965 C
  4. 3860 C

Explanation

Moles of copper = 0.635 g ÷ 63.5 g mol⁻¹ = 0.010 mol. The cathode reaction Cu²⁺ + 2e⁻ → Cu requires two moles of electrons per mole of copper, so 0.020 mol of electrons are needed. Multiplying by Faraday's constant gives 0.020 × 96485 ≈ 1930 C.

Question 49Electrochemistry

For a cell reaction involving two electrons at 298 K, the cell emf changes with temperature at the rate dE/dT = −1.50 × 10⁻⁴ V K⁻¹. What is the standard entropy change of the cell reaction? (F = 96485 C mol⁻¹)

  1. +29.0 J K⁻¹ mol⁻¹
  2. −29.0 J K⁻¹ mol⁻¹ Correct answer.
  3. −14.5 J K⁻¹ mol⁻¹
  4. −58.0 J K⁻¹ mol⁻¹

Explanation

The relation ΔS° = nF(dE/dT) links the entropy change to the temperature coefficient of the emf. Substituting n = 2 gives ΔS° = 2 × 96485 × (−1.50 × 10⁻⁴) ≈ −29.0 J K⁻¹ mol⁻¹. The negative sign shows that the cell reaction produces order, and the magnitude follows directly from the given slope.

Question 50Electrochemistry

At 298 K, a 0.10 M solution of a weak monobasic acid has molar conductivity 20.0 S cm² mol⁻¹ and limiting molar conductivity 400 S cm² mol⁻¹. What is the acid dissociation constant?

  1. 5.0 × 10⁻³
  2. 2.6 × 10⁻⁴ Correct answer.
  3. 2.5 × 10⁻⁴
  4. 5.0 × 10⁻²

Explanation

The degree of dissociation is α = Λm/Λ°m = 20.0/400 = 0.050. Ostwald's dilution law then gives Ka = Cα²/(1 − α) = (0.10 × 0.0025)/0.95 ≈ 2.6 × 10⁻⁴. The denominator is close to but not exactly one, so retaining it is necessary for the accurate value.