IB Chemistry Quiz: Understand Electron Transfer Reactions
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Understand Electron Transfer ReactionsQuestion 1 of 20

A student performs three experiments with metals X, Y, and Z and their aqueous nitrate solutions.

Experiment 1: A piece of metal X is placed in a solution of Y(NO₃)₂. A solid deposit of metal Y forms. Experiment 2: A piece of metal Y is placed in a solution of Z(NO₃)₂. No reaction is observed. Experiment 3: A piece of metal X is placed in a solution of Z(NO₃)₂. A solid deposit of metal Z forms.

Based on these observations, what is the correct order of decreasing reactivity for these metals?

X > Y > Z
Y > Z > X
Z > X > Y
X > Z > Y
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Electron Transfer Reactions

Practice Understand Electron Transfer Reactions in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Electron Transfer Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Question 1

A student performs three experiments with metals X, Y, and Z and their aqueous nitrate solutions.

Experiment 1: A piece of metal X is placed in a solution of Y(NO₃)₂. A solid deposit of metal Y forms. Experiment 2: A piece of metal Y is placed in a solution of Z(NO₃)₂. No reaction is observed. Experiment 3: A piece of metal X is placed in a solution of Z(NO₃)₂. A solid deposit of metal Z forms.

Based on these observations, what is the correct order of decreasing reactivity for these metals?

  1. X > Y > Z
  2. Y > Z > X
  3. Z > X > Y
  4. X > Z > Y (correct answer)
Explanation: Experiment 1 shows that metal X can displace Y from its salt solution, so X is more reactive than Y (X > Y). Experiment 2 shows that metal Y cannot displace Z, so Z is more reactive than Y (Z > Y). Experiment 3 shows that metal X can displace Z, so X is more reactive than Z (X > Z). Combining these results gives the order of decreasing reactivity as X > Z > Y.

Question 2

Consider the following half-equations: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) C₂O₄²⁻(aq) → 2CO₂(g) + 2e⁻

What is the coefficient for H⁺(aq) in the overall balanced equation for the reaction between permanganate and oxalate ions, and on which side of the equation does it appear?

  1. 16, on the reactants side (correct answer)
  2. 8, on the reactants side
  3. 16, on the products side
  4. 8, on the products side
Explanation: To balance the electrons, the first half-equation must be multiplied by 2, and the second by 5. This gives: 2MnO₄⁻(aq) + 16H⁺(aq) + 10e⁻ → 2Mn²⁺(aq) + 8H₂O(l) 5C₂O₄²⁻(aq) → 10CO₂(g) + 10e⁻ Adding these two equations and cancelling the 10e⁻ from both sides gives the overall equation: 2MnO₄⁻(aq) + 16H⁺(aq) + 5C₂O₄²⁻(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g). The coefficient for H⁺(aq) is 16, and it appears on the reactants side.

Question 3

In a titration, 25.00 cm³ of hydrogen peroxide, H₂O₂, reacts completely with 30.00 cm³ of 0.0500 mol dm⁻³ acidified potassium permanganate, KMnO₄. The unbalanced equation is: ‾\underline{\hspace{2em}} MnO₄⁻(aq) + ‾\underline{\hspace{2em}} H₂O₂(aq) + ‾\underline{\hspace{2em}} H⁺(aq) → ‾\underline{\hspace{2em}} Mn²⁺(aq) + ‾\underline{\hspace{2em}} O₂(g) + ‾\underline{\hspace{2em}} H₂O(l) The mole ratio of MnO₄⁻ to H₂O₂ in the balanced equation is 2:5. What is the concentration of the H₂O₂ solution?

  1. 0.0750 mol dm⁻³
  2. 0.150 mol dm⁻³ (correct answer)
  3. 0.0240 mol dm⁻³
  4. 0.0600 mol dm⁻³
Explanation:
  1. Calculate moles of MnO₄⁻ used: n(MnO₄⁻) = C × V = 0.0500 mol dm⁻³ × (30.00/1000) dm³ = 0.00150 mol. 2. Use the mole ratio to find moles of H₂O₂: n(H₂O₂) = n(MnO₄⁻) × (5/2) = 0.00150 mol × 2.5 = 0.00375 mol. 3. Calculate the concentration of H₂O₂: C(H₂O₂) = n / V = 0.00375 mol / (25.00/1000) dm³ = 0.150 mol dm⁻³.

Question 4

What is the change in oxidation state of the carbonyl carbon atom when ethanal (CH₃CHO) is reduced to ethanol (CH₃CH₂OH)?

  1. From +1 to -1 (correct answer)
  2. From +2 to -2
  3. From -1 to +1
  4. From 0 to -2
Explanation: In ethanal (CH₃CHO), the carbonyl carbon is bonded to one H (+1), one C (0 contribution), and double-bonded to one O (-2). The oxidation state is +1. In ethanol (CH₃CH₂OH), this carbon is bonded to two H atoms (+1 each), one O (-2), and one C (0 contribution). The oxidation state is -1. The change is from +1 to -1, representing a reduction (gain of electrons).

Question 5

The reaction between ethanol and acidified dichromate(VI) ions produces ethanoic acid and chromium(III) ions. What is the total change in the oxidation state of chromium for one mole of dichromate(VI) ions, Cr₂O₇²⁻, that reacts?

  1. A decrease of 3
  2. A decrease of 6 (correct answer)
  3. An increase of 3
  4. An increase of 6
Explanation: First, determine the oxidation state of Cr in Cr₂O₇²⁻. Let the oxidation state be x. The oxidation state of O is -2. So, 2x + 7(-2) = -2, which gives 2x = +12, so x = +6. The final product is chromium(III) ion, Cr³⁺, where the oxidation state is +3. The change for one Cr atom is from +6 to +3, which is a change of -3. Since there are two Cr atoms in one Cr₂O₇²⁻ ion, the total change per ion is 2 × (-3) = -6. Therefore, the total change is a decrease of 6.

Question 6

When chlorine gas is bubbled through cold, dilute sodium hydroxide solution, the following reaction occurs: Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l) Which statement accurately describes the role of chlorine in this reaction?

  1. Chlorine is the oxidizing agent because it is converted to NaCl.
  2. Chlorine is the reducing agent because it is converted to NaClO.
  3. Chlorine is neither oxidized nor reduced as it reacts with itself.
  4. Chlorine is both oxidized to an oxidation state of +1 and reduced to an oxidation state of -1. (correct answer)
Explanation: In Cl₂, the oxidation state of chlorine is 0. In the product NaCl, the oxidation state of chlorine is -1. This is a reduction. In the product NaClO, the oxidation state of chlorine is +1 (since Na is +1 and O is -2). This is an oxidation. Because the same element, chlorine, has undergone both oxidation and reduction in the same reaction, it has been both oxidized and reduced. This type of reaction is known as disproportionation.

Question 7

A compound with the molecular formula C₃H₈O is oxidized by heating with acidified potassium dichromate(VI) solution to form a product, Y. Compound Y does not give a positive test with Tollens' reagent. What is the systematic name of the starting compound?

  1. Propan-1-ol
  2. Propan-2-ol (correct answer)
  3. Propanal
  4. Propanoic acid
Explanation: The starting compound C₃H₈O is an alcohol. The product Y does not react with Tollens' reagent, which means it is not an aldehyde. Therefore, Y must be a ketone. Ketones are formed from the oxidation of secondary alcohols. The secondary alcohol with the formula C₃H₈O is propan-2-ol. Its oxidation produces propanone, a ketone. Propan-1-ol is a primary alcohol and would be oxidized to propanal (an aldehyde) and then to propanoic acid.

Question 8

Equal molar quantities of four different powdered metals are added to separate beakers containing 1.0 mol dm⁻³ hydrochloric acid. The time taken for the metal to be completely consumed is recorded. Which metal would be expected to be consumed in the shortest time?

  1. Zinc
  2. Lead
  3. Iron
  4. Magnesium (correct answer)
Explanation: The rate of reaction between a metal and an acid is determined by the metal's reactivity. A more reactive metal will react more vigorously and thus be consumed faster. The reactivity series for these metals is Magnesium > Zinc > Iron > Lead. Magnesium is the most reactive metal listed, so it will react the fastest with HCl and be consumed in the shortest time.

Question 9

A voltaic cell is set up with two unknown metals, P and Q, and their corresponding 1.0 mol dm⁻³ nitrate solutions. It is observed that the mass of electrode P increases and the concentration of Qⁿ⁺(aq) ions increases. Which conclusion must be correct?

  1. Metal P is more reactive than metal Q, and electrons flow from P to Q.
  2. Metal Q is more reactive than metal P, and electrons flow from Q to P. (correct answer)
  3. Metal P is more reactive than metal Q, and electrons flow from Q to P.
  4. Metal Q is more reactive than metal P, and electrons flow from P to Q.
Explanation: An increase in the mass of electrode P means that reduction is occurring there (Pⁿ⁺(aq) + ne⁻ → P(s)), so P is the cathode. An increase in the concentration of Qⁿ⁺(aq) ions means that oxidation is occurring at electrode Q (Q(s) → Qⁿ⁺(aq) + ne⁻), so Q is the anode. In a voltaic cell, the more reactive metal is oxidized at the anode. Therefore, metal Q is more reactive than metal P. Electrons always flow from the anode (Q) to the cathode (P) through the external wire.

Question 10

The following reactions are observed to occur spontaneously:

I. Zn(s) + Pb²⁺(aq) → Zn²⁺(aq) + Pb(s)

II. Fe(s) + Zn²⁺(aq) → No reaction Based on this information, what is the correct order of increasing strength as reducing agents?

  1. Pb < Fe < Zn (correct answer)
  2. Pb < Zn < Fe
  3. Fe < Zn < Pb
  4. Zn < Fe < Pb
Explanation: A stronger reducing agent is more easily oxidized (more reactive). From reaction I, Zn reduces Pb²⁺, so Zn is a stronger reducing agent than Pb. From reaction II, Fe does not reduce Zn²⁺, so Zn is a stronger reducing agent than Fe. To compare Fe and Pb, we can infer the reactivity series: Zn > Fe and Zn > Pb. A more reactive metal is a stronger reducing agent. The general reactivity series places Zn > Fe > Pb. Therefore, the order of increasing strength as reducing agents is Pb < Fe < Zn.

Question 11

A voltaic cell is constructed using a magnesium electrode in a 1.0 mol dm⁻³ Mg(NO₃)₂ solution and a silver electrode in a 1.0 mol dm⁻³ AgNO₃ solution. Magnesium is more reactive than silver. Which statement is correct as the cell operates?

  1. The concentration of Ag⁺(aq) ions increases in the silver half-cell.
  2. Electrons flow from the silver electrode to the magnesium electrode via the external wire.
  3. The magnesium electrode is the negative electrode, and its mass decreases over time. (correct answer)
  4. Cations from the salt bridge migrate towards the magnesium half-cell to maintain charge balance.
Explanation: Since magnesium is more reactive, it will be oxidized at the anode: Mg(s) → Mg²⁺(aq) + 2e⁻. The anode is the negative electrode in a voltaic cell. As solid Mg is converted to Mg²⁺ ions, the mass of the magnesium electrode decreases. A: Ag⁺ is reduced to Ag, so its concentration decreases. B: Electrons flow from the anode (Mg) to the cathode (Ag). D: The magnesium half-cell is producing positive Mg²⁺ ions, so anions from the salt bridge will migrate there to balance the charge.

Question 12

In a redox titration, 25.0 mL of 0.150 M Fe²⁺ solution is titrated with 0.0200 M MnO₄⁻ in acidic conditions. The reaction is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. What volume of MnO₄⁻ is required to reach the equivalence point?

  1. 18.8 mL because each MnO₄⁻ accepts 5 electrons
  2. 37.5 mL because electron transfer follows 5:5 stoichiometry
  3. 18.8 mL because the molar ratio is 1:5 MnO₄⁻ to Fe²⁺
  4. 37.5 mL because 5 moles Fe²⁺ react per mole MnO₄⁻ (correct answer)
Explanation: Moles of Fe²⁺ = 0.025 L × 0.150 M = 0.00375 mol. From the balanced equation, 5 mol Fe²⁺ react with 1 mol MnO₄⁻, so moles of MnO₄⁻ needed = 0.00375 mol Fe²⁺ × (1 mol MnO₄⁻/5 mol Fe²⁺) = 0.000750 mol. Volume of MnO₄⁻ = 0.000750 mol ÷ 0.0200 M = 0.0375 L = 37.5 mL.

Question 13

A student constructs an electrochemical cell using zinc and copper electrodes in their respective 1.0 M sulfate solutions. After the cell operates for some time, the student observes that the zinc electrode has lost 0.65 g of mass. Assuming 100% efficiency, what mass change occurs at the copper electrode, and what is the total charge transferred?

  1. Copper electrode gains 0.63 g; total charge transferred is 1930 C because electron flow balances the mass changes (correct answer)
  2. Copper electrode gains 0.63 g; total charge transferred is 965 C because each zinc atom transfers two electrons to copper
  3. Copper electrode gains 1.26 g; total charge transferred is 1930 C because copper has twice the molar mass of zinc in this redox process
  4. Copper electrode gains 0.32 g; total charge transferred is 965 C because the electron transfer efficiency varies between electrodes
Explanation: The reactions are: Zn → Zn²⁺ + 2e⁻ (anode) and Cu²⁺ + 2e⁻ → Cu (cathode). Moles of Zn oxidized = 0.65 g ÷ 65.4 g/mol = 0.00994 mol. Since each Zn transfers 2e⁻, total electrons = 0.00994 × 2 = 0.0199 mol. These electrons reduce Cu²⁺: moles of Cu deposited = 0.0199 mol e⁻ ÷ 2 e⁻/Cu = 0.00994 mol. Mass of Cu deposited = 0.00994 mol × 63.5 g/mol = 0.63 g. Total charge = 0.0199 mol e⁻ × 96485 C/mol = 1920 C ≈ 1930 C. Choice B has correct mass but wrong charge calculation, C has wrong mass (doubled incorrectly), D has wrong mass and charge.

Question 14

In a fuel cell operating on the reaction H₂ + ½O₂ → H₂O, the theoretical cell voltage is 1.23 V under standard conditions. During operation at 0.8 V with a current density of 200 mA/cm², the cathode has an area of 50 cm². What is the power output per unit cathode area, and what is the primary cause of the voltage loss from theoretical?

  1. Power density is 160 mW/cm²; voltage loss is primarily due to ohmic resistance in the electrolyte and electrode materials limiting electron flow
  2. Power density is 160 mW/cm²; voltage loss is primarily due to activation overpotential for the oxygen reduction reaction at the cathode surface (correct answer)
  3. Power density is 246 mW/cm²; voltage loss is primarily due to ohmic resistance in the electrolyte and electrode materials limiting electron flow
  4. Power density is 246 mW/cm²; voltage loss is primarily due to activation overpotential for the oxygen reduction reaction at the cathode surface
Explanation: Power density = voltage × current density = 0.8 V × 200 mA/cm² = 160 mW/cm². The voltage loss is 1.23 - 0.8 = 0.43 V. In fuel cells, the largest voltage losses typically occur due to activation overpotential, especially for the oxygen reduction reaction (ORR) at the cathode, which is kinetically slow and requires significant overpotential to proceed at practical rates. At moderate current densities like 200 mA/cm², activation losses dominate over ohmic losses. The cathode area (50 cm²) affects total current but not current density or power density calculations. Choices C and D incorrectly calculate power density, while A and C incorrectly identify ohmic resistance as the primary loss mechanism.

Question 15

An electroplating process uses a current of 3.5 A to deposit silver from AgNO₃ solution onto a steel substrate. After 25 minutes, the actual mass of silver deposited is 4.1 g. What is the current efficiency, and what is the most likely cause of the efficiency loss?

  1. 70% efficiency; loss due to hydrogen gas evolution competing with silver reduction (correct answer)
  2. 85% efficiency; loss due to hydrogen gas evolution competing with silver reduction
  3. 70% efficiency; loss due to silver ion migration away from electrode
  4. 85% efficiency; loss due to silver ion migration away from electrode
Explanation: Theoretical calculation: Q = 3.5 A × 25 × 60 s = 5250 C. Moles of electrons = 5250/96485 = 0.0544 mol. Since Ag⁺ + e⁻ → Ag, theoretical mass = 0.0544 mol × 107.9 g/mol = 5.87 g. Current efficiency = (4.1/5.87) × 100% = 70%. The main cause of efficiency loss in silver electroplating is competing reduction of H⁺ or H₂O to form H₂ gas at the cathode.

Question 16

In the reaction 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O, the rate of electron transfer depends on the concentration of reactants. If the reaction rate is proportional to [MnO₄⁻]¹[C₂O₄²⁻]², what happens to the rate of electron transfer when [MnO₄⁻] is doubled and [C₂O₄²⁻] is halved?

  1. Rate decreases by factor of 2 because fewer reducing agent molecules are available despite increased oxidizing agent concentration
  2. Rate remains unchanged because the kinetic effects of concentration changes exactly balance each other in this electron transfer process
  3. Rate decreases by factor of 2 because the overall reaction order indicates that oxalate concentration has greater kinetic influence than permanganate (correct answer)
  4. Rate increases by factor of 2 because permanganate accepts more electrons per molecule than oxalate donates per collision event
Explanation: The rate law is rate = k[MnO₄⁻]¹[C₂O₄²⁻]². Initial rate = k[MnO₄⁻][C₂O₄²⁻]². New rate = k[2MnO₄⁻][C₂O₄²⁻/2]² = k[MnO₄⁻] × 2 × [C₂O₄²⁻]² × (1/4) = (1/2) × k[MnO₄⁻][C₂O₄²⁻]². The rate decreases by a factor of 2. This occurs because C₂O₄²⁻ appears squared in the rate law, so halving its concentration reduces the rate by (1/2)² = 1/4, while doubling [MnO₄⁻] increases rate by factor of 2. Net effect: 2 × 1/4 = 1/2. Choice A gives correct conclusion but oversimplified reasoning, B is incorrect about the balance, D incorrectly suggests rate increases.

Question 17

Which of the following chemical equations represents a reaction that is not a redox reaction?

  1. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
  2. CuO(s) + H₂(g) → Cu(s) + H₂O(l)
  3. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) (correct answer)
  4. Cl₂(g) + 2KI(aq) → 2KCl(aq) + I₂(aq)
Explanation: A redox reaction involves a change in oxidation states. In reaction C, a precipitation reaction, the oxidation states of all elements remain constant: Ag is +1, N is +5, O is -2, Na is +1, and Cl is -1 on both sides of the equation. In A, Na changes from 0 to +1 and H from +1 to 0. In B, Cu changes from +2 to 0 and H from 0 to +1. In D, Cl changes from 0 to -1 and I from -1 to 0.

Question 18

Consider the reaction: 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l). Which statement is correct?

  1. Hydrogen sulfide (H₂S) is the oxidizing agent because its sulfur atom is oxidized.
  2. Sulfur dioxide (SO₂) is the oxidizing agent because its sulfur atom is reduced. (correct answer)
  3. This is a disproportionation reaction because sulfur is both oxidized and reduced.
  4. The oxidation state of oxygen changes from -2 in SO₂ to -1 in H₂O.
Explanation: In H₂S, the oxidation state of S is -2. In SO₂, the oxidation state of S is +4. In the product, elemental sulfur (S), the oxidation state is 0. The sulfur in H₂S is oxidized (-2 to 0), making H₂S the reducing agent. The sulfur in SO₂ is reduced (+4 to 0), making SO₂ the oxidizing agent. A is incorrect because the species that is oxidized is the reducing agent. C is incorrect as this is comproportionation, not disproportionation. D is incorrect as the oxidation state of oxygen remains -2.

Question 19

But-2-ene reacts with hydrogen gas in the presence of a platinum catalyst. Which statement best describes this reaction?

  1. It is an oxidation reaction because hydrogen atoms are added.
  2. It is an addition reaction where the carbon atoms are reduced. (correct answer)
  3. It is a substitution reaction where hydrogen replaces the double bond.
  4. It is a reduction reaction where the carbon atoms are oxidized.
Explanation: The reaction is the hydrogenation of an alkene (but-2-ene + H₂ → butane), which is a type of addition reaction. During this reaction, the C=C double bond is broken and C-H bonds are formed. The oxidation state of the carbon atoms involved in the double bond decreases (from -1 to -2), so the carbon atoms are reduced. Therefore, it is both an addition and a reduction reaction. Option B correctly identifies both aspects.

Question 20

Propanal is treated with a reducing agent such as lithium aluminium hydride (LiAlH₄), followed by aqueous acid work-up. What is the major organic product?

  1. Propan-1-ol (correct answer)
  2. Propan-2-ol
  3. Propane
  4. Propanoic acid
Explanation: The reduction of an aldehyde yields a primary alcohol. Propanal is a three-carbon aldehyde. Its reduction will add two hydrogen atoms across the C=O double bond, converting the aldehyde functional group (-CHO) into a primary alcohol functional group (-CH₂OH). The product is therefore propan-1-ol. Propan-2-ol would be formed from the reduction of a ketone (propanone). Propanoic acid is an oxidation product of propanal.