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Apply Electron Transfer Reactions - IB Chemistry

  1. Consider the disproportionation reaction of chlorine in hot, concentrated sodium hydroxide solution: 3Cl₂(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l). Which statement correctly describes the process?
    Answer: Chlorine is reduced to NaCl and oxidized to NaClO₃. In Cl₂, the oxidation number of chlorine is 0. In NaCl, the oxidation number of chlorine is -1 (a reduction). In NaClO₃, the oxidation number of chlorine is +5 (Na is +1, each O is -2, so Cl + 3(-2) = -1, Cl = +5), which is an oxidation. Since the same element (chlorine) is both oxidized and reduced, it is a disproportionation reaction. Choice A incorrectly states the oxidation state in NaClO₃ is +3. Choice C is incorrect because redox reactions involve changes in oxidation state. Choice D is incorrect because Cl₂ is both the oxidizing and reducing agent.
  2. A current of 0.50 A is passed through a solution of silver nitrate, AgNO₃(aq), for 30 minutes. Which expression gives the mass of silver, in g, deposited at the cathode? (Molar mass of Ag = 107.87 g mol⁻¹; Faraday constant, F = 96500 C mol⁻¹)
    Answer: (0.50 × 30 × 60 × 107.87) / 96500 First, calculate the total charge (Q) passed in Coulombs: Q = current (I) × time (t). Time must be in seconds, so t = 30 min × 60 s/min = 1800 s. Q = 0.50 A × 1800 s = 900 C. Next, calculate the moles of electrons (nen_e) using the Faraday constant: nen_e = Q / F = 900 C / 96500 C mol⁻¹. The reduction of silver is Ag⁺ + e⁻ → Ag, so the mole ratio of electrons to silver is 1:1. Thus, moles of Ag = nen_e. Finally, calculate the mass of silver: mass = moles × molar mass = (900 / 96500) mol × 107.87 g mol⁻¹. Combining these steps into one expression: mass = (I × t × M) / F = (0.50 × (30 × 60) × 107.87) / 96500. Choice A matches this expression.
  3. Consider the following half-reactions and their standard reduction potentials: Br2+2e−→2Br−Br_2 + 2e^- \rightarrow 2Br^- (E° = +1.07 V) and I2+2e−→2I−I_2 + 2e^- \rightarrow 2I^- (E° = +0.54 V). When Br2Br_2 is added to a solution containing both Cl−Cl^- and I−I^- ions, which statement best describes the expected reaction, given that Cl2+2e−→2Cl−Cl_2 + 2e^- \rightarrow 2Cl^- has E° = +1.36 V?
    Answer: Br2Br_2 will oxidize only I−I^- because Br2Br_2 is a stronger oxidizing agent than I2I_2 When you encounter redox questions with multiple half-reactions, you need to determine which species can act as oxidizing agents by comparing their reduction potentials. A stronger oxidizing agent (higher E°) can oxidize species that correspond to weaker oxidizing agents (lower E°). To determine what Br2Br_2 can oxidize, compare its reduction potential (+1.07 V) with the other halogens. Since Cl2Cl_2 has a higher reduction potential (+1.36 V), Br2Br_2 cannot oxidize Cl−Cl^- to Cl2Cl_2 - this would be thermodynamically unfavorable. However, Br2Br_2 has a higher reduction potential than I2I_2 (+0.54 V), so it can oxidize I−I^- to I2I_2. The reaction Br2+2I−→2Br−+I2Br_2 + 2I^- \rightarrow 2Br^- + I_2 has a positive cell potential of 1.07 - 0.54 = +0.53 V, making it spontaneous. Choice A incorrectly suggests Br2Br_2 can oxidize both ions - while it's true that Br2Br_2 has an intermediate potential, this doesn't mean it can oxidize all species. Choice B is wrong because Br2Br_2 can definitely oxidize I−I^-, as shown by the favorable thermodynamics. Choice C reverses the logic - Br2Br_2 cannot oxidize Cl−Cl^- because chlorine is more electronegative and has a higher reduction potential. Remember the key rule: in redox chemistry, you can only go "downhill" in reduction potential. A species can only oxidize others that have lower (less positive) reduction potentials than itself. Always arrange the potentials in order to visualize possible reactions.
  4. In the context of redox reactions, what is the primary role of a salt bridge in a voltaic cell?
    Answer: To allow the migration of ions between the half-cells to maintain electrical neutrality. The salt bridge contains an electrolyte and allows ions to move between the two half-cells. As oxidation occurs at the anode, positive ions are produced, creating a net positive charge. At the cathode, positive ions are consumed, creating a net negative charge. The salt bridge allows anions to flow into the anode compartment and cations to flow into the cathode compartment to neutralize this charge buildup, which would otherwise stop the flow of electrons and halt the cell's operation. Electrons flow through the external wire, not the salt bridge.
  5. In the redox titration of Fe2+Fe^{2+} with MnO4−MnO_4^- in acidic solution, 25.0 mL of 0.020 M MnO4−MnO_4^- is required to reach the endpoint when titrating an unknown Fe2+Fe^{2+} solution. The balanced equation is: 5Fe2++MnO4−+8H+→5Fe3++Mn2++4H2O5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O. What mass of iron was present in the original solution?
    Answer: 0.14 g Moles of MnO₄⁻ used = 0.025 L × 0.020 M = 5.0 × 10⁻⁴ mol. From the balanced equation, the molar ratio is 5 Fe²⁺ : 1 MnO₄⁻. Therefore, moles of Fe²⁺ = 5 × 5.0 × 10⁻⁴ = 2.5 × 10⁻³ mol. Mass of Fe = 2.5 × 10⁻³ mol × 55.85 g/mol = 0.140 g. The key is recognizing that the stoichiometry shows 5 electrons transferred from Fe²⁺ ions to each MnO₄⁻ ion (which goes from +7 to +2 oxidation state, gaining 5 electrons).
  6. What is the standard cell potential (E⦵_cell) and the standard Gibbs free energy change (ΔG⦵) for the spontaneous reaction between these two half-cells? (Faraday constant, F ≈ 96500 C mol⁻¹)
    Answer: E⦵_cell = +0.23 V, ΔG⦵ = -44 kJ mol⁻¹ For a spontaneous reaction, E⦵_cell must be positive. This occurs when the half-reaction with the higher E⦵ value is the cathode (reduction) and the one with the lower E⦵ is the anode (oxidation). So, Fe³⁺ is reduced and I⁻ is oxidized. The anode reaction is 2I⁻ → I₂ + 2e⁻. The cathode reaction is Fe³⁺ + e⁻ → Fe²⁺. E⦵_cell = E⦵_cathode - E⦵_anode = (+0.77 V) - (+0.54 V) = +0.23 V. The overall reaction is 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. The number of moles of electrons transferred, n, is 2. ΔG⦵ = -nFE⦵_cell = -2 mol 96500 C mol⁻¹ 0.23 V = -44390 J mol⁻¹ ≈ -44 kJ mol⁻¹. Choice C incorrectly uses n=1.
  7. The reaction of the thiosulfate ion, S₂O₃²⁻, with iodine, I₂, produces the tetrathionate ion, S₄O₆²⁻, and iodide ions, I⁻. What is the change in the average oxidation state of a sulfur atom in this reaction?
    Answer: From +2 to +2.5 First, determine the average oxidation state of sulfur in the thiosulfate ion, S₂O₃²⁻. Let the oxidation state of S be x. The oxidation state of O is -2. The overall charge is -2. So, 2x + 3(-2) = -2, which gives 2x - 6 = -2, so 2x = +4, and x = +2. Next, determine the average oxidation state of sulfur in the tetrathionate ion, S₄O₆²⁻. Let the oxidation state of S be y. 4y + 6(-2) = -2, which gives 4y - 12 = -2, so 4y = +10, and y = +2.5. The change in the average oxidation state is from +2 to +2.5.
  8. Consider the reaction: 2MnO4−+5H2O2+6H+→2Mn2++5O2+8H2O2MnO_4^- + 5H_2O_2 + 6H^+ \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O. In this reaction, what is the total number of electrons transferred when 0.15 mol of MnO4−MnO_4^- reacts completely?
    Answer: 0.75 mol electrons First, determine the oxidation state changes. Mn in MnO₄⁻ has oxidation state +7, and in Mn²⁺ it's +2, so each Mn gains 5 electrons. O in H₂O₂ has oxidation state -1, and in O₂ it's 0, so each O loses 1 electron (each H₂O₂ loses 2 electrons total). For 0.15 mol MnO₄⁻: electrons gained = 0.15 mol × 5 = 0.75 mol electrons. To verify: 0.15 mol MnO₄⁻ requires (5/2) × 0.15 = 0.375 mol H₂O₂, which loses 0.375 × 2 = 0.75 mol electrons. The electron transfer balances at 0.75 mol electrons total.
  9. What is the change in the oxidation number of the specified carbon atom when propan-1-ol is oxidized to propanoic acid?
    Answer: From -1 to +3 To find the oxidation number of a specific carbon, consider the bonds to it. In propan-1-ol (CH₃CH₂CH₂OH), focus on the carbon bonded to the -OH group. It is bonded to two H atoms, one C atom, and one O atom. Assigning oxidation states: H is +1, O is -2. Let the carbon's oxidation state be x. The CH₂OH fragment is not neutral. Let's use the method of assigning electrons based on electronegativity. For the C in CH₂OH: It is bonded to 2 H's, 1 C, and 1 O. C is more electronegative than H; split C-C bonds; O is more electronegative than C. So, this C gets electrons from two C-H bonds (+2e) and one C-C bond (+1e), for a total of 3 electrons gained relative to its neutral state if protons are considered. Formal method: Oxidation state for C in CH₂OH: x + 2(+1) + (-2) + (+1) = 0 for the molecule? No. Simpler: For the CH₂OH carbon, the sum of oxidation states must balance the fragment. Let's consider the C atom itself. C bonded to 2 H's (-2), 1 C (0), 1 O (+1). Sum = -1. So the oxidation state is -1. In propanoic acid (CH₃CH₂COOH), the carboxyl carbon is bonded to one C, one O (in OH), and double-bonded to another O. Ignore C-C. C-O gives +1, C=O gives +2. Total = +3. The change is from -1 to +3.
  10. When ethene (C₂H₄) is hydrogenated to form ethane (C₂H₆), the reaction can be classified as a reduction. What is the justification for this classification based on oxidation states?
    Answer: Each carbon atom's oxidation state changes from -2 to -3. In ethene (H₂C=CH₂), each carbon is bonded to two hydrogen atoms. Let the oxidation state of C be x. Using the rule that H is +1, we have 2x + 4(+1) = 0, so 2x = -4 and x = -2 for each carbon. In ethane (H₃C-CH₃), each carbon is bonded to three hydrogen atoms. Let the oxidation state of C be y. 2y + 6(+1) = 0, so 2y = -6 and y = -3 for each carbon. The oxidation state of each carbon atom decreases from -2 to -3. A decrease in oxidation state is reduction. While adding hydrogen is often a reduction, the reason is the change in oxidation states.
  11. During the electroplating of an object with silver from a silver nitrate solution, AgNO₃(aq), a constant current is applied for a fixed amount of time. If the experiment were repeated using a tin(II) nitrate solution, Sn(NO₃)₂(aq), with the same current and time, how would the moles of metal plated compare?
    Answer: Half as many moles of tin would be plated compared to silver. The amount of charge passed is the same in both experiments (same current and time). The charge is proportional to the moles of electrons. Let the moles of electrons be 'n'. For silver, the reduction is Ag⁺(aq) + e⁻ → Ag(s). So, 'n' moles of electrons will produce 'n' moles of Ag. For tin(II), the reduction is Sn²⁺(aq) + 2e⁻ → Sn(s). So, 'n' moles of electrons will produce 'n/2' moles of Sn. Therefore, half as many moles of tin would be plated compared to silver.
  12. In the electrolysis of molten NaClNaCl, a current of 5.0 A is passed for 45 minutes. Simultaneously, the same current electrolyzes an aqueous solution of CuSO4CuSO_4. What is the ratio of moles of NaNa produced to moles of CuCu produced?
    Answer: 2:1 Both cells receive the same charge: Q = 5.0 A × 45 min × 60 s/min = 13,500 C. Moles of electrons = 13,500 C ÷ 96,485 C/mol = 0.140 mol e⁻. For Na production: Na⁺ + e⁻ → Na, so moles of Na = 0.140 mol. For Cu production: Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.140 ÷ 2 = 0.070 mol. Ratio of Na:Cu = 0.140:0.070 = 2:1. The key insight is that Na requires 1 electron per atom while Cu requires 2 electrons per atom, so with the same total electron flow, twice as many Na atoms are produced.
  13. In a redox reaction, Cr2O72−Cr_2O_7^{2-} oxidizes Fe2+Fe^{2+} to Fe3+Fe^{3+} while being reduced to Cr3+Cr^{3+} in acidic solution. If 2.40 g of FeSO4⋅7H2OFeSO_4 \cdot 7H_2O (molar mass = 278 g/mol) reacts completely, how many moles of Cr2O72−Cr_2O_7^{2-} are required?
    Answer: 0.00144 mol When you encounter redox stoichiometry problems, you need to balance the half-reactions first, then use mole ratios to find the required amounts. This question tests your ability to work with balanced redox equations and perform stoichiometric calculations. Start by writing the balanced equation. The half-reactions are: - Oxidation: Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^- - Reduction: Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O To balance electrons, multiply the iron half-reaction by 6: Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2OCr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O This shows that 1 mole of Cr2O72−Cr_2O_7^{2-} reacts with 6 moles of Fe2+Fe^{2+}. Next, find moles of FeSO4⋅7H2OFeSO_4 \cdot 7H_2O: 2.40 g278 g/mol=0.00863 mol\frac{2.40 \text{ g}}{278 \text{ g/mol}} = 0.00863 \text{ mol} Since each formula unit contains one Fe2+Fe^{2+}, you have 0.00863 mol of Fe2+Fe^{2+}. Using the 6:1 ratio: 0.00863 mol Fe2+6=0.00144 mol Cr2O72−\frac{0.00863 \text{ mol } Fe^{2+}}{6} = 0.00144 \text{ mol } Cr_2O_7^{2-} Answer D (0.00144 mol) is correct. Answer B (0.00863 mol) assumes a 1:1 ratio, ignoring the balanced equation. Answer A (0.00431 mol) uses an incorrect 2:1 ratio. Answer C (0.00288 mol) uses a 3:1 ratio, perhaps confusing electron counts. Always balance the complete redox equation before attempting stoichiometric calculations. The electron balance determines the mole ratios, which are crucial for getting the right answer.
  14. A student constructs an electrochemical cell where the anode reaction is Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^- and the cathode reaction is 2Ag++2e−→2Ag2Ag^+ + 2e^- \rightarrow 2Ag. If the initial concentrations are [Zn2+]=0.01M[Zn^{2+}] = 0.01 M and [Ag+]=1.0M[Ag^+] = 1.0 M, and after some time the [Zn2+]=0.05M[Zn^{2+}] = 0.05 M, what is the new concentration of Ag+Ag^+?
    Answer: 0.92 M The overall reaction is Zn + 2Ag⁺ → Zn²⁺ + 2Ag. The change in [Zn²⁺] = 0.05 - 0.01 = 0.04 M increase. According to stoichiometry, for every 1 mol of Zn²⁺ produced, 2 mol of Ag⁺ are consumed. Therefore, Δ[Ag⁺] = -2 × 0.04 = -0.08 M. New [Ag⁺] = 1.0 - 0.08 = 0.92 M. This demonstrates the importance of using reaction stoichiometry to relate concentration changes of different species in electron transfer reactions.
  15. A voltaic cell is constructed using two half-cells: Ni²⁺(aq) / Ni(s) and Al³⁺(aq) / Al(s). Given the standard electrode potentials: Ni²⁺(aq) + 2e⁻ ⇌ Ni(s) E⦵ = -0.26 V Al³⁺(aq) + 3e⁻ ⇌ Al(s) E⦵ = -1.66 V Which change would cause the greatest increase in the initial cell potential (E_cell)?
    Answer: Increasing the concentration of Ni²⁺(aq). The half-reaction with the higher (less negative) E⦵ is the cathode (reduction): Ni²⁺ + 2e⁻ → Ni. The other is the anode (oxidation): Al → Al³⁺ + 3e⁻. The overall reaction is 3Ni²⁺(aq) + 2Al(s) → 3Ni(s) + 2Al³⁺(aq). According to Le Châtelier's principle, to increase the cell potential (make the forward reaction more favorable), the equilibrium should be shifted to the right. This is achieved by increasing the concentration of a reactant (Ni²⁺) or decreasing the concentration of a product (Al³⁺). Therefore, increasing [Ni²⁺] will increase the cell potential. Increasing the surface area of an electrode affects the rate but not the potential. Increasing [Al³⁺] would shift the equilibrium to the left, decreasing the cell potential.
  16. A piece of zinc metal is placed in a solution of copper(II) sulfate. Which of the following statements is NOT correct?
    Answer: The concentration of Zn²⁺ ions in the solution decreases. Zinc is more reactive than copper, so it will displace copper from the solution. The reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). In this reaction, Zn is oxidized (loses electrons) so it is the reducing agent (A is correct). The blue color of the solution is due to the hydrated Cu²⁺ ions. As they are consumed, the color fades (B is correct). Zinc metal is converted into Zn²⁺ ions, so the concentration of Zn²⁺ ions increases, not decreases (C is incorrect). Because zinc is higher in the reactivity series than copper (or because E⦵(Zn²⁺/Zn) is more negative than E⦵(Cu²⁺/Cu)), the reaction is spontaneous (D is correct).
  17. The reaction 2Cr(s) + 3Fe²⁺(aq) → 2Cr³⁺(aq) + 3Fe(s) has a standard cell potential, E⦵_cell, of +0.30 V. What is the standard Gibbs free energy change, ΔG⦵, for this reaction in kJ? (Faraday constant, F = 96500 C mol⁻¹)
    Answer: -174 kJ The relationship between ΔG⦵ and E⦵_cell is given by ΔG⦵ = -nFE⦵_cell. First, determine n, the number of moles of electrons transferred. The half-reactions are: Cr → Cr³⁺ + 3e⁻ and Fe²⁺ + 2e⁻ → Fe. To balance the electrons for the overall reaction, the chromium half-reaction is multiplied by 2 and the iron half-reaction by 3, resulting in 6 moles of electrons being transferred. So, n=6. ΔG⦵ = -6 96500 C mol⁻¹ 0.30 V = -173700 J = -173.7 kJ ≈ -174 kJ. Distractors are based on incorrect values of n (n=3 gives -87 kJ, n=2 gives -58 kJ) or a sign error.
  18. Consider the redox reaction: 3Cl2+6OH−→ClO3−+5Cl−+3H2O3Cl_2 + 6OH^- \rightarrow ClO_3^- + 5Cl^- + 3H_2O. Which statement correctly describes the electron transfer processes occurring in this disproportionation reaction?
    Answer: One chlorine atom loses 5 electrons to form ClO3−ClO_3^- while five chlorine atoms each gain 1 electron to form Cl−Cl^- In this disproportionation, Cl₂ (oxidation state 0) forms both ClO₃⁻ (Cl has +5 oxidation state) and Cl⁻ (oxidation state -1). One Cl atom goes from 0 to +5 (loses 5 electrons), while five Cl atoms go from 0 to -1 (each gains 1 electron). This balances the electron transfer: 5 electrons lost = 5 electrons gained. Choice A incorrectly suggests the same atom both loses and gains electrons. Choice B has the electron transfer directions reversed. Choice D incorrectly describes the stoichiometry and mechanism.
  19. The overall reaction in a hydrogen fuel cell is 2H₂(g) + O₂(g) → 2H₂O(l). Which statement correctly describes the processes occurring at the electrodes?
    Answer: At the negative electrode (anode), hydrogen is oxidized, and at the positive electrode (cathode), oxygen is reduced. In a fuel cell, as in any voltaic cell, oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). Hydrogen is the fuel, so it is oxidized: H₂ → 2H⁺ + 2e⁻ (in acidic medium). Oxygen is the oxidant, so it is reduced: O₂ + 4H⁺ + 4e⁻ → 2H₂O (in acidic medium). Therefore, hydrogen is oxidized at the anode, and oxygen is reduced at the cathode.
  20. An aqueous solution of copper(II) sulfate, CuSO₄(aq), is electrolyzed using inert graphite electrodes. What are the products at the anode and cathode?
    Answer: Anode: Oxygen gas; Cathode: Copper metal At the negative electrode (cathode), reduction occurs. Possible species to be reduced are Cu²⁺(aq) and H₂O(l). The reduction potential of Cu²⁺ (+0.34 V) is higher than that of water (-0.83 V at pH 7), so Cu²⁺ is preferentially reduced: Cu²⁺(aq) + 2e⁻ → Cu(s). At the positive electrode (anode), oxidation occurs. Possible species to be oxidized are SO₄²⁻(aq) and H₂O(l). The sulfate ion is very difficult to oxidize. Water is oxidized instead: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻. Therefore, copper metal is produced at the cathode and oxygen gas at the anode.

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