IB DP Chemistry Reactivity 3.2: Electron transfer: standard electrode potentials and Gibbs energy – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set tests understandings Reactivity 3.2.12 to 3.2.16 of the IB’s Diploma Programme Chemistry guide (first assessment 2025, published February 2023). All of this is additional higher level content, so every question is HL only, for exam sessions in 2025 and later years.

Before starting, read the Reactivity 3.2.12-3.2.16 study guide or skim the condensed notes. Useful too: the DP Chemistry hub, the DP Chemistry checklist and our free diagnostic quizzes.

Data for every question. Standard electrode potentials, E⦵ / V: K⁺/K −2.93; Al³⁺/Al −1.66; H₂O/H₂, OH⁻ −0.83; Fe²⁺/Fe −0.45; Co²⁺/Co −0.28; Ni²⁺/Ni −0.26; Sn²⁺/Sn −0.14; H⁺/H₂ 0.00; Cu²⁺/Cu +0.34; I₂/I⁻ +0.54; Ag⁺/Ag +0.80; Br₂/Br⁻ +1.09; O₂, H⁺/H₂O +1.23; Cl₂/Cl⁻ +1.36; H₂O₂, H⁺/H₂O +1.78. F = 96 500 C mol⁻¹; R = 8.31 J K⁻¹ mol⁻¹. Final answers: 3 significant figures unless a question asks for something else.

Questions

1. (a) State the standard electrode potential assigned to the half-cell H⁺(aq) + e⁻ ⇌ ½H₂(g). [1] (b) Describe the standard hydrogen electrode, including its conditions. [2]

2. Use the data.

(a) Identify the strongest oxidizing agent among Al³⁺, Sn²⁺, Ag⁺ and Br₂. [1] (b) Identify the strongest reducing agent among Al, Fe, Cu and I⁻. [1] (c) Deduce which of the metals iron, copper and silver can reduce Sn²⁺(aq) to tin under standard conditions. [1]

3. A voltaic cell is made from an Al³⁺/Al half-cell and a Ni²⁺/Ni half-cell.

(a) Identify the anode and state its polarity. [1] (b) Calculate E⦵cell. [1] (c) Deduce the overall equation for the cell reaction. [2] (d) Calculate ΔG⦵ for the reaction in (c), in kJ mol⁻¹. [2]

4. Use E⦵cell to predict whether each reaction is spontaneous as written.

(a) Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g) [2] (b) I₂(aq) + 2Br⁻(aq) → 2I⁻(aq) + Br₂(aq) [2] (c) Ni(s) + Sn²⁺(aq) → Ni²⁺(aq) + Sn(s) [2]

5. Acidified hydrogen peroxide oxidizes iodide ions: H₂O₂(aq) + 2H⁺(aq) + 2I⁻(aq) → 2H₂O(l) + I₂(aq).

(a) Calculate E⦵cell. [1] (b) Calculate ΔG⦵ in kJ mol⁻¹. [2] (c) The equation is multiplied by 2. State the new values of E⦵cell and ΔG⦵. [2]

6. (a) A cell is built from a standard X²⁺/X half-cell and a standard Cu²⁺/Cu half-cell. E⦵cell = 0.62 V and X is the negative electrode. Determine E⦵(X²⁺/X) and identify X from the data. [2] (b) In a methanol fuel cell, 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l), ΔG⦵ = −1404 kJ mol⁻¹ and 12 mol of electrons are transferred. Calculate E⦵cell. [2]

7. Aqueous solutions are electrolysed with inert electrodes.

(a) For potassium iodide solution, deduce the product at each electrode, with half-equations, and explain the cathode product. [3] (b) For potassium sulfate solution, state the product at each electrode and explain the anode product. [2]

8. (a) Concentrated NaCl(aq) is electrolysed with inert electrodes. Write both half-equations and explain why the anode product differs from the E⦵ prediction. [3] (b) State and explain the main anode product when the solution is very dilute. [2] (c) CuSO₄(aq) is electrolysed with graphite electrodes, then with copper electrodes. Write each anode half-equation. [2]

9. Water containing a little dilute sulfuric acid is electrolysed with platinum electrodes.

(a) Write the half-equation at each electrode. [2] (b) Calculate E⦵cell for the overall reaction and state what it means for the supply voltage. [1] (c) 42 cm³ of hydrogen is collected. State the volume of oxygen collected under the same conditions. [1]

10. A brass brooch is to be silver-plated.

(a) State what is used as the anode, the cathode and the electrolyte. [2] (b) Give the cathode and anode half-equations. [2] (c) Explain why the concentration of the electrolyte stays almost constant. [1]

11. Steel for food cans is tin-plated using a tin anode and tin(II) sulfate solution.

(a) Write the cathode half-equation and explain why tin, not hydrogen, forms there. [2] (b) Explain why the tin anode is oxidized in preference to water. [1] (c) The tin anode is replaced by a graphite anode. State the new anode product and two changes to the electrolyte as electrolysis continues. [3] (d) With the power off, steel in the bath can react: Fe(s) + Sn²⁺(aq) → Fe²⁺(aq) + Sn(s). Calculate E⦵cell and ΔG⦵. [3]

12. (Links to Reactivity 1.4.) For H₂(g) + ½O₂(g) → H₂O(l), ΔH⦵ = −285.8 kJ mol⁻¹. Standard entropies, S⦵ / J K⁻¹ mol⁻¹: H₂(g) 130.7, O₂(g) 205.2, H₂O(l) 70.0.

(a) Calculate ΔS⦵. [1] (b) Calculate ΔG⦵ at 298 K. [2] (c) Use (b) to calculate E⦵cell for a hydrogen–oxygen fuel cell, and compare it with the E⦵ data. [3] (d) Estimate the cell potential at 353 K, assuming ΔH⦵ and ΔS⦵ do not change. [2]

Answers

1. (a) 0.00 V, by convention [1]. [1] (b) Platinum electrode in [H⁺] = 1.00 mol dm⁻³ [1]; hydrogen at 100 kPa, 298 K [1]. [2] Examiner insight: Name the platinum and all three conditions; “standard conditions” alone is incomplete.

2. (a) Br₂ (+1.09 V is the most positive) [1]. [1] (b) Al (−1.66 V is the most negative) [1]. [1] (c) Iron only (−0.45 V is below −0.14 V) [1]. [1] Examiner insight: Oxidizing agents sit on the left of the half-equations and reducing agents on the right; check the side before ranking.

3. (a) Aluminium is the anode and it is negative [1]. [1] (b) E⦵cell = −0.26 − (−1.66) = +1.40 V [1]. [1] (c) Al(s) → Al³⁺(aq) + 3e⁻ and Ni²⁺(aq) + 2e⁻ → Ni(s), so 6 electrons are needed [1]. 2Al(s) + 3Ni²⁺(aq) → 2Al³⁺(aq) + 3Ni(s) [1]. [2] (d) n = 6 [1]. ΔG⦵ = −6 × 96 500 × 1.40 = −810 600 J mol⁻¹ = −811 kJ mol⁻¹ [1]. [2] Examiner insight: E⦵cell stays at 1.40 V although the half-equations are multiplied; only n carries the scaling into ΔG⦵.

4. (a) E⦵cell = 0.00 − 0.34 = −0.34 V [1]; negative, so not spontaneous [1]. [2] (b) E⦵cell = 0.54 − 1.09 = −0.55 V [1]; not spontaneous; the reverse, bromine oxidizing iodide, is [1]. [2] (c) E⦵cell = −0.14 − (−0.26) = +0.12 V [1]; positive, so spontaneous [1]. [2] Examiner insight: Show the signed E⦵cell before the verdict; a bare “yes” or “no” gives a marker nothing to check.

5. (a) E⦵cell = 1.78 − 0.54 = +1.24 V [1]. [1] (b) n = 2 [1]; ΔG⦵ = −2 × 96 500 × 1.24 = −239 320 J mol⁻¹ = −239 kJ mol⁻¹ [1]. [2] (c) E⦵cell is unchanged at +1.24 V [1]; n = 4, so ΔG⦵ = −479 kJ mol⁻¹ [1]. [2] Examiner insight: In (c), state that E⦵cell does not depend on the amounts written, while ΔG⦵ does.

6. (a) Copper is the cathode, so 0.62 = 0.34 − E⦵(X²⁺/X) and E⦵(X²⁺/X) = −0.28 V [1]. X is cobalt [1]. [2] (b) E⦵cell = −ΔG⦵ ÷ nF = 1 404 000 ÷ (12 × 96 500) [1] = +1.21 V [1]. [2] Examiner insight: In (b), convert kJ to J before dividing, or the answer is a thousand times too small.

7. (a) Cathode: hydrogen, 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) [1]. K⁺ (−2.93 V) is far more negative than water (−0.83 V), so water is reduced instead [1]. Anode: iodine, 2I⁻(aq) → I₂(aq) + 2e⁻ [1]. [3] (b) Hydrogen at the cathode and oxygen at the anode [1]. Sulfate ions are not oxidized, so water (+1.23 V) is oxidized: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ [1]. [2] Examiner insight: Quote the two competing E⦵ values; “potassium is reactive” is weaker than the numbers.

8. (a) Cathode: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) [1]. Anode: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ [1]. E⦵ alone favours oxygen (+1.23 V against +1.36 V), but the values are close and high [Cl⁻] makes chloride oxidation win [1]. [3] (b) Mainly oxygen [1]; with few chloride ions, water is oxidized as the E⦵ values predict [1]. [2] (c) Graphite: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ [1]. Copper: Cu(s) → Cu²⁺(aq) + 2e⁻ [1]. [2] Examiner insight: In (c), writing the oxygen equation twice misses that a copper anode is itself oxidized.

9. (a) Cathode: 2H⁺(aq) + 2e⁻ → H₂(g) [1]. Anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ [1]. [2] (b) E⦵cell = 0.00 − 1.23 = −1.23 V, so the supply must provide more than 1.23 V [1]. [1] (c) 21 cm³ (2 : 1 ratio) [1]. [1] Examiner insight: A negative E⦵cell is expected for electrolysis; say what it means for the supply.

10. (a) Cathode: the brooch; anode: pure silver [1]. Electrolyte: a solution containing Ag⁺ ions, such as silver nitrate [1]. [2] (b) Cathode: Ag⁺(aq) + e⁻ → Ag(s) [1]. Anode: Ag(s) → Ag⁺(aq) + e⁻ [1]. [2] (c) Ag⁺ forms at the anode as fast as it is removed at the cathode [1]. [1] Examiner insight: Name the brooch as the cathode; reversed electrodes would strip metal from it instead.

11. (a) Sn²⁺(aq) + 2e⁻ → Sn(s) [1]. Sn²⁺/Sn (−0.14 V) is more positive than water (−0.83 V) [1]. [2] (b) Tin (−0.14 V) is much easier to oxidize than water (+1.23 V) [1]. [1] (c) Oxygen forms at the anode [1]. The Sn²⁺ concentration falls [1], and the solution becomes more acidic as H⁺ is produced [1]. [3] (d) E⦵cell = −0.14 − (−0.45) = +0.31 V [1]. n = 2 [1]; ΔG⦵ = −2 × 96 500 × 0.31 = −59 830 J mol⁻¹ = −59.8 kJ mol⁻¹ [1]. [3] Examiner insight: For (c), the anode half-equation shows where the H⁺ comes from.

12. (a) ΔS⦵ = 70.0 − (130.7 + ½ × 205.2) = −163.3 J K⁻¹ mol⁻¹ [1]. [1] (b) ΔG⦵ = −285.8 − 298 × (−0.1633) [1] = −237.1 kJ mol⁻¹ [1]. [2] (c) E⦵cell = 237 100 ÷ (2 × 96 500) [1] = +1.23 V [1]. From the data, E⦵cell = 1.23 − 0.00 = 1.23 V, so the two routes agree [1]. [3] (d) ΔG = −285.8 − 353 × (−0.1633) = −228.2 kJ mol⁻¹ [1]; E = 228 200 ÷ (2 × 96 500) = 1.18 V [1]. [2] Examiner insight: Use ΔS⦵ in kJ K⁻¹ mol⁻¹ in ΔH⦵ − TΔS⦵, then ΔG⦵ in J before dividing by nF.

Where marks are usually lost

• Subtracting E⦵ values the wrong way round, or reversing a sign and then subtracting.
• Multiplying an E⦵ value by a balancing coefficient.
• Using the ion charge for n instead of the total electrons in the overall equation.
• Mixing J and kJ in ΔG⦵ = −nFE⦵cell.
• Writing a reactive metal such as potassium as a cathode product from aqueous solution.
• Ignoring concentration (NaCl) or electrode material (CuSO₄).
• Making the object the anode in electroplating.

Next steps

• Condensed notes: electrode potentials and electrolysis
• Full study guide: Reactivity 3.2.12 to 3.2.16
• DP Chemistry course hub
• Printable DP Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Source document: the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 3.2, understandings 3.2.12 to 3.2.16 (additional higher level).