Free worked solutions · 2026 Edition

IB Chemistry R3.2: Electron transfer reactions worked solutions

Worked solutions to every practice question in R3.2: Electron transfer reactions of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

  • SL & HL
  • 23 questions
  • Full worked solutions
  • Free, no sign-up

Basic practice

Question B1

[1 mark]
Which change is an oxidation? (A) \ce{MnO4- -> Mn^2+} (B) \ce{SO3^2- -> SO4^2-} (C) \ce{Cl2 -> 2Cl-} (D) \ce{Fe^3+ -> Fe^2+}
Show solutionHide solution
Sulfur goes from +4 to +6, an increase. In the other three the oxidation state falls (Mn +7\to+2, Cl 0\to-1, Fe +3\to+2): all reductions.

Final answer

B

Question C1

[1 mark]
A cell is written \ce{Fe(s)}\,|\,\ce{Fe^2+(aq)}\,\|\,\ce{Cu^2+(aq)}\,|\,\ce{Cu(s)}, with a potassium nitrate salt bridge. Which statement is correct while it supplies current? (A) Electrons pass through the salt bridge. (B) Nitrate ions move towards the iron half-cell. (C) Copper is the anode. (D) The iron electrode gains mass.
Show solutionHide solution
Iron is on the left, so it is the anode and oxidized: \ce{Fe -> Fe^2+ + 2e-} adds positive ions to that solution; nitrate ions enter to maintain electrical neutrality. Electrons travel in the wire, copper is the cathode, and the iron electrode loses mass.

Final answer

B

Question E1

[1 mark]
One mole of each compound is completely hydrogenated with a nickel catalyst. Which uses the most hydrogen? (A) hex-1-ene (B) hex-1-yne (C) benzene (D) cyclohexene
Show solutionHide solution
Hex-1-ene and cyclohexene use 1 mol of \ce{H2}, hex-1-yne uses 2, benzene uses 3 (three \pi bonds; the ring stays).

Final answer

C

Question F1

HL[1 mark]
The standard electrode potentials are \ce{Al^3+ + 3e- <=> Al}, -1.66\,\mathrm{V}, and \ce{Cu^2+ + 2e- <=> Cu}, +0.34\,\mathrm{V}. What is E^\ominus_\mathrm{cell} for an aluminium-copper cell? (A) +1.32\,\mathrm{V} (B) +2.00\,\mathrm{V} (C) +4.34\,\mathrm{V} (D) -2.00\,\mathrm{V}
Show solutionHide solution
E^\ominus_\mathrm{cell}=0.34-(-1.66)=+2.00\,\mathrm{V}. The values are not scaled by the electron numbers (3 and 2).

Final answer

B

Question B2

[1 mark]
Find the oxidation state of V in \ce{VO2+}. The ion contains one V and two O atoms. (a) State the value. [1]
Show solutionHide solution
Let V be x. Then x+2(-2)=+1, giving x=+5.

Final answer

+5

Question B3

[1 mark]
In \ce{2Al + 3Ni^2+ -> 2Al^3+ + 3Ni}, which species is the oxidizing agent? (A) \ce{Al} (B) \ce{Al^3+} (C) \ce{Ni^2+} (D) \ce{Ni}
Show solutionHide solution
\ce{Ni^2+} gains electrons and is reduced from +2 to 0, so it oxidizes Al. The agent is the whole ion.

Final answer

C

Question B4

[2 marks]
A little bromine water is mixed separately with aqueous potassium iodide and potassium chloride. (a) Identify the solution in which a displacement occurs and name the halogen formed. [2]
Show solutionHide solution
Bromine oxidizes iodide because oxidizing power falls down Group 17: \ce{Br2 + 2I- -> 2Br- + I2}. It cannot oxidize chloride; that tube has no displacement.

Final answer

KI; iodine

Question B5

[2 marks]
A clean iron nail reacts with dilute sulfuric acid. (a) Write the ionic equation, including the formula and state of the gas. [2]
Show solutionHide solution
Iron loses two electrons while two protons gain them: \ce{Fe(s) + 2H+(aq) -> Fe^2+(aq) + H2(g)}. Dilute non-oxidizing acid produces iron(II).

Final answer

Fe + 2H+ -> Fe2+ + H2(g)

Tutorial practice

Question G5

[5 marks]
Acidified permanganate oxidizes ethanedioate ions, \ce{C2O4^2-}, to carbon dioxide. (a) Write the half-equation for the oxidation of ethanedioate. [1] (b) Combine it with \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} to give the overall ionic equation. [2] (c) A 25.0\,\mathrm{cm^3} sample needs 20.00\,\mathrm{cm^3} of 0.0200\,\mathrm{mol\,dm^{-3}} permanganate. Calculate the ethanedioate concentration. [2]
Show solutionHide solution
(a) \ce{C2O4^2- -> 2CO2 + 2e-}. (b) Reduction \times2 (10e^-) and oxidation \times5 (10e^-): \ce{2MnO4- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}. Charge: left -2+16-10=+4, right +4. (c) n(\ce{MnO4-})=0.0200\times0.02000=4.00\times10^{-4}\,\mathrm{mol}; n(\ce{C2O4^2-})=4.00\times10^{-4}\times5/2=1.00\times10^{-3}\,\mathrm{mol}; concentration =1.00\times10^{-3}/0.0250=0.0400\,\mathrm{mol\,dm^{-3}}.

Final answer

(a) C2O4^2- -> 2CO2 + 2e-; (b) 2MnO4- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2; (c) 0.0400 mol dm^-3

Question H4

[4 marks]
A zinc half-cell and a nickel half-cell are joined by a potassium nitrate salt bridge. Electrons flow in the wire from zinc to nickel. (a) State which electrode is the anode and give its sign. [1] (b) Write the half-equation at each electrode. [2] (c) State which way nitrate ions move, and how the mass of the nickel electrode changes. [1]
Show solutionHide solution
(a) Electrons leave zinc, so zinc is oxidized: it is the anode and negative. (b) \ce{Zn -> Zn^2+ + 2e-} at the anode; \ce{Ni^2+ + 2e- -> Ni} at the cathode. (c) Anions move towards the anode compartment, which is gaining \ce{Zn^2+}. Nickel is deposited, so the nickel electrode gains mass.

Final answer

(a) zinc, negative; (b) Zn -> Zn^2+ + 2e-; Ni^2+ + 2e- -> Ni; (c) towards the zinc half-cell; nickel gains mass

Question J4

[4 marks]
Pentan-1-ol is warmed with acidified dichromate(VI). (a) Name the product obtained if it is distilled out as it forms. [1] (b) Heated under reflux with excess oxidant instead, it gives a different product. Name it and write the equation using [\ce{O}]. [2] (c) State the colour change of the oxidizing agent. [1]
Show solutionHide solution
(a) Pentanal. (b) Pentanoic acid: \ce{CH3(CH2)3CH2OH + 2[O] -> CH3(CH2)3COOH + H2O} (aldehyde stage then acid). (c) Orange to green, as \ce{Cr2O7^2-} is reduced to \ce{Cr^3+}.

Final answer

(a) pentanal; (b) pentanoic acid; CH3(CH2)3CH2OH + 2[O] -> CH3(CH2)3COOH + H2O; (c) orange to green

Question T1

[3 marks]
Metals P, Q and R form 2+ ions. P deposits Q from \ce{Q^2+} solution; R deposits P from \ce{P^2+} solution. (a) Arrange the metals in increasing ease of oxidation. [1] (b) Write the ionic equation when R is added to \ce{Q^2+} solution and identify the reducing agent. [2]
Show solutionHide solution
P donates electrons more readily than Q, while R donates more readily than P, so increasing ease is Q, P, R. Therefore \ce{R + Q^2+ -> R^2+ + Q}. R is the electron donor and reducing agent.

Final answer

Q < P < R; R + Q2+ -> R2+ + Q; R

Question T2

[3 marks]
Molten magnesium chloride is electrolysed using inert electrodes. (a) Write both electrode half-equations, naming the electrodes. [2] (b) State the energy conversion and the charge carrier in the melt. [1]
Show solutionHide solution
At the negative cathode, \ce{Mg^2+ + 2e- -> Mg}. At the positive anode, \ce{2Cl- -> Cl2 + 2e-}. The supply drives a non-spontaneous conversion of electrical into chemical energy. Moving ions, not electrons, conduct in the melt.

Final answer

Cathode Mg2+ + 2e- -> Mg; anode 2Cl- -> Cl2 + 2e-; electrical to chemical; ions

Question T3

[3 marks]
On discharge, a simplified secondary cell has the overall reaction \ce{A + B^2+ -> A^2+ + B}. (a) Write the half-equation at the A electrode on discharge and again on recharge. [2] (b) Explain why the A electrode changes from anode to cathode. [1]
Show solutionHide solution
Discharge: \ce{A -> A^2+ + 2e-}. Recharging reverses the chemistry: \ce{A^2+ + 2e- -> A}. Electrode names follow the process; A is oxidized during discharge and reduced during recharge.

Final answer

Discharge A -> A2+ + 2e-; recharge A2+ + 2e- -> A; oxidation changes to reduction

Question T4

[3 marks]
A remote sensor can use either a primary cell with 80 h of service or a secondary cell with 50 h per charge. A visit to replace or recharge the cell is expensive. (a) Give one operational advantage of the primary cell in this setting. [1] (b) Give one resource advantage of the secondary cell, and one condition needed for that advantage to be realized. [2]
Show solutionHide solution
The primary cell operates 30 h longer per visit. The secondary cell can be reused over many cycles, reducing discarded cells and demand for replacement materials. That benefit requires a practical charging source and repeated use; a single abandoned cycle gives little reuse benefit.

Final answer

Longer interval between visits; reuse reduces replacement-cell waste if recharging and collection are practical

Question T5

[3 marks]
A sample of pentan-2-one is treated with sodium borohydride and then aqueous acid. (a) Name the organic product. [1] (b) Write an equation using [\ce{H}] and state the change in oxidation state of the carbonyl carbon. [2]
Show solutionHide solution
The ketone is reduced to pentan-2-ol: \ce{CH3COCH2CH2CH3 + 2[H] -> CH3CH(OH)CH2CH2CH3}. The carbonyl carbon initially has +2 from the C=O bond; in the alcohol the C–O contribution is +1 and the C–H contribution is -1, giving 0.

Final answer

Pentan-2-ol; CH3COCH2CH2CH3 + 2[H] -> CH3CH(OH)CH2CH2CH3; +2 to 0

Advanced practice

Question L5

HL[5 marks]
Supplied reduction potentials: For \ce{2Fe^3+ + Sn^2+ -> 2Fe^2+ + Sn^4+}, use E^\ominus(\ce{Fe^3+/Fe^2+})=+0.77\,\mathrm V and E^\ominus(\ce{Sn^4+/Sn^2+})=+0.15\,\mathrm V. (a) Calculate E^\ominus_\mathrm{cell}. [1] (b) Calculate \Delta G^\ominus in \mathrm{kJ\,mol^{-1}}. [2] (c) A student writes E^\ominus_\mathrm{cell}=2(0.77)-0.15. State the mistake. [1] (d) Explain why potentials are measured against the standard hydrogen electrode, and state the hydrogen pressure. [1]
Show solutionHide solution
(a) 0.77-0.15=+0.62\,\mathrm{V}. (b) n=2: \Delta G^\ominus=-2\times96\,500\times0.62=-1.20\times10^{5}\,\mathrm{J\,mol^{-1}}=-120\,\mathrm{kJ\,mol^{-1}}. (c) Electrode potentials are not multiplied by the coefficients in the equation; the 2 belongs in n only. (d) A voltmeter measures only a difference, so every half-cell is compared with one agreed reference; hydrogen at 100\,\mathrm{kPa}.

Final answer

(a) +0.62 V; (b) -120 kJ/mol; (c) potential not multiplied by the coefficient; (d) only differences are measurable, 100 kPa

Question M5

HL[7 marks]
Aqueous solutions are electrolysed; electrodes are inert unless stated. (a) Concentrated sodium chloride solution: name the product at each electrode. [2] (b) Copper(II) sulfate solution, copper electrodes: state what happens at the anode and to the blue colour. [2] (c) A steel key is plated with copper here. Which terminal of the supply is the key joined to? [1] (d) In the copper(II) sulfate cell, replace the copper anode with graphite while maintaining the current. Deduce the changes in blue colour and acidity, with an anode equation. [2]
Show solutionHide solution
(a) Cathode: hydrogen (water is reduced, because \ce{Na+} is harder to reduce). Anode: chlorine. (b) The copper anode dissolves, \ce{Cu -> Cu^2+ + 2e-}, replacing the \ce{Cu^2+} removed at the cathode, so the blue colour does not fade. (c) The key is the cathode, so it is joined to the negative terminal. (d) Graphite does not replace copper ions removed at the cathode, so the blue colour fades. Water is oxidized, \ce{2H2O -> O2 + 4H+ + 4e-}, increasing acidity.

Final answer

(a) cathode H2, anode Cl2; (b) anode dissolves, colour unchanged; (c) negative terminal; (d) blue fades, acidity increases; 2H2O -> O2 + 4H+ + 4e-

Question A1

HL[5 marks]
A proposed recovery process passes \ce{Fe^2+} solution over silver metal. Use E^\ominus(\ce{Fe^2+/Fe})=-0.45\,\mathrm V and E^\ominus(\ce{Ag+/Ag})=+0.80\,\mathrm V. (a) Write the proposed overall equation and calculate its standard cell potential. [2] (b) Assess whether this process deposits iron under standard conditions, and explain what reversing the reaction would achieve. [2] (c) Explain why a favorable potential alone does not specify a reaction rate. [1]
Show solutionHide solution
The proposed reduction is \ce{Fe^2+ + 2e- -> Fe} and oxidation is \ce{2Ag -> 2Ag+ + 2e-}. Thus \ce{Fe^2+ + 2Ag -> Fe + 2Ag+} and E^\ominus=-0.45-0.80=-1.25\,\mathrm V. Iron deposition is not spontaneous under standard conditions. Reversing gives +1.25\,\mathrm V and deposits silver on iron. Potential describes thermodynamic driving force; activation barriers and surface effects control rate.

Final answer

Fe2+ + 2Ag -> Fe + 2Ag+; -1.25 V; no; reverse deposits Ag; kinetics controls rate

Question A2

HL[6 marks]
A copper coating is deposited on a steel part using aqueous copper(II) sulfate and a copper anode. The current is 0.800\,\mathrm A for 2412.5\,\mathrm s; the measured coating mass is 0.572\,\mathrm g. Use F=96500\,\mathrm{C\,mol^{-1}} and M(\ce{Cu})=63.55\,\mathrm{g\,mol^{-1}}. (a) Write both electrode equations and state the terminal connected to the steel. [2] (b) Calculate the theoretical coating mass. [2] (c) Calculate the current efficiency and suggest one process that could make it less than 100 percent. [2]
Show solutionHide solution
Copper dissolves at the anode, \ce{Cu -> Cu^2+ + 2e-}; copper ions are reduced on the negative steel cathode, \ce{Cu^2+ + 2e- -> Cu}. Q=It=1930\,\mathrm C, so n(e^-)=1930/96500=0.0200\,\mathrm{mol} and n(\ce{Cu})=0.0100\,\mathrm{mol}. Theoretical mass =0.0100(63.55)=0.6355\,\mathrm g. Efficiency =0.572/0.6355\times100=90.0\%. A competing cathodic reduction producing hydrogen consumes charge without depositing copper.

Final answer

Anode Cu -> Cu2+ + 2e-; cathode Cu2+ + 2e- -> Cu; negative; 0.6355 g; 90.0%; hydrogen evolution

Connected questions

Question X1

Connected questions[6 marks]
R3.2 + R2.1 + S1.4. A 1.20\,\mathrm g iron-containing solid is dissolved to make 100.0\,\mathrm{cm^3} of solution containing only iron(II) as the reducing solute. A 20.00\,\mathrm{cm^3} portion requires 18.40\,\mathrm{cm^3} of 0.0100\,\mathrm{mol\,dm^{-3}} acidified permanganate. (a) Combine \ce{Fe^2+ -> Fe^3+ + e-} with the permanganate reduction half-equation to obtain the ionic equation. [2] (b) Calculate the mass percentage of iron in the solid. Use M(\ce{Fe})=55.85\,\mathrm{g\,mol^{-1}}. [3] (c) Explain the effect on the calculated percentage if some iron(II) is oxidized by air before titration. [1]
Show solutionHide solution
The permanganate half-equation gains five electrons, so multiply the iron half-equation by five: \ce{MnO4- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+}. The aliquot consumes 0.0100(0.01840)=1.84\times10^{-4}\,\mathrm{mol} permanganate and contains 9.20\times10^{-4}\,\mathrm{mol} Fe. The full flask contains five times this, 0.00460\,\mathrm{mol} Fe, mass 0.25691\,\mathrm g. Percentage =0.25691/1.20\times100=21.4\%. Air oxidation removes titratable iron(II), reducing the titre and underestimating iron.

Final answer

MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+; 21.4%; too low

Question X2

Connected questions · HL[6 marks]
R3.2 + R1.4. For \ce{Zn + Cu^2+ -> Zn^2+ + Cu}, take E^\ominus(\ce{Cu^2+/Cu})=+0.34\,\mathrm V, E^\ominus(\ce{Zn^2+/Zn})=-0.76\,\mathrm V, F=96500\,\mathrm{C\,mol^{-1}}, and supplied \Delta H^\ominus=-218\,\mathrm{kJ\,mol^{-1}}. (a) Calculate E^\ominus_\mathrm{cell} and \Delta G^\ominus at 298\,\mathrm K. [2] (b) Calculate \Delta S^\ominus in \mathrm{J\,K^{-1}\,mol^{-1}}. [2] (c) Assuming constant \Delta H^\ominus and \Delta S^\ominus, predict the effect of a temperature rise on \Delta G^\ominus and on E^\ominus_\mathrm{cell}. [2]
Show solutionHide solution
E^\ominus=0.34-(-0.76)=1.10\,\mathrm V. Two electrons transfer, so \Delta G^\ominus=-2(96500)(1.10)=-212300\,\mathrm{J\,mol^{-1}}. From \Delta G^\ominus=\Delta H^\ominus-T\Delta S^\ominus, \Delta S^\ominus=(-218000+212300)/298=-19.1\,\mathrm{J\,K^{-1}\,mol^{-1}}. Since entropy change is negative, increasing T increases \Delta G^\ominus, making it less negative. With E^\ominus=-\Delta G^\ominus/(nF), the positive cell potential decreases.

Final answer

+1.10 V; -212.3 kJ/mol; -19.1 J/K/mol; Gibbs energy less negative and E decreases

Question X3

Connected questions[6 marks]
R3.2 + S3.2 + R3.1. An unknown alcohol has formula \ce{C4H10O}. Oxidation without C–C bond cleavage gives an organic acid; a separate sample of that acid reacts with aqueous sodium carbonate. (a) State the class of alcohol and give two possible alcohol names. [2] (b) Choose one candidate. Name its aldehyde and acid, and state how the apparatus is changed to favor the acid. [2] (c) Write the acid–carbonate equation using RCOOH and explain why this final reaction is not redox. [2]
Show solutionHide solution
Formation of an acid without carbon-chain cleavage identifies a primary alcohol. The two candidates are butan-1-ol and 2-methylpropan-1-ol. For butan-1-ol the aldehyde is butanal and the acid is butanoic acid. Distil to remove aldehyde; use reflux with excess oxidant to retain it for further oxidation to acid. For the branched candidate the corresponding names are 2-methylpropanal and 2-methylpropanoic acid. \ce{2RCOOH + Na2CO3 -> 2RCOONa + CO2 + H2O}. This is proton transfer: carbonate carbon remains +4, and the organic carbon oxidation states do not change.

Final answer

Primary; butan-1-ol or 2-methylpropan-1-ol; butanal/butanoic acid (or 2-methylpropanal/2-methylpropanoic acid); reflux; 2RCOOH + Na2CO3 -> 2RCOONa + CO2 + H2O; no oxidation-state changes

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Electron transfer reactions from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book