Redox Processes (SL) — IB Chemistry SL SL Study Guide

For: IB Chemistry SL candidates sitting IB Chemistry SL.

Covers: oxidation states and electron transfer principles, balancing redox reactions in acidic media, comparing voltaic and electrolytic cell structure/function, and using the reactivity series to predict reaction outcomes.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Redox Processes (SL)?

Redox (reduction-oxidation) reactions are chemical processes involving the net transfer of electrons between reactants, forming the basis of batteries, corrosion, metal extraction, and many biological metabolic pathways. For IB Chemistry SL, redox content makes up ~10% of your final exam marks across paper 1 multiple choice and paper 2 structured response sections, with frequent questions requiring you to apply rules to new, unfamiliar reaction contexts. Common synonyms for redox reactions include electron-transfer reactions and electrochemical reactions.

2. Oxidation states and electron transfer

The core principle of redox is electron transfer, which you can track using oxidation states (OS): the hypothetical charge an atom would carry if all bonds in a compound were fully ionic, with no covalent character. Use these standard rules to calculate oxidation states, listed in order of priority:

1. The oxidation state of any uncombined element is 0 (e.g. ${\text{O}}_2$, $\text{Fe}$, ${\text{Cl}}_2$ all have OS = 0)
2. The sum of oxidation states for a neutral compound is 0; for a polyatomic ion, it equals the net charge of the ion
3. Group 1 metals always have OS = +1, Group 2 metals = +2, fluorine always = -1
4. Hydrogen has OS = +1 except in metal hydrides (e.g. $\text{NaH}$) where it = -1
5. Oxygen has OS = -2 except in peroxides (e.g. ${\text{H}}_2{\text{O}}_2$ where OS = -1) and when bonded to fluorine (e.g. ${\text{OF}}_2$ where OS = +2)

Key definitions

• Oxidation: Loss of electrons, increase in oxidation state (OIL = Oxidation Is Loss)
• Reduction: Gain of electrons, decrease in oxidation state (RIG = Reduction Is Gain)
• Half-reaction: A balanced equation showing only the oxidation or reduction step of a full redox reaction, with electrons explicitly included. Electrons appear on the product side of oxidation half-reactions, and the reactant side of reduction half-reactions.

Worked example

Calculate the oxidation state of chlorine in the perchlorate ion ${\text{ClO}}_4^{-}$:

1. Oxygen has OS = -2, total ion charge = -1
2. Set up the equation: $\text{Cl}+4(-2)=-1$
3. Solve: $\text{Cl}-8=-1\to \text{Cl}=+7$

Exam tip: Always write oxidation states with the sign first (e.g. +7 not 7+) to avoid confusion with ion charge notation, which places the sign second (e.g. 2+ for a magnesium ion). Examiners deduct marks for mixed notation.

3. Balancing redox equations

IB SL only requires you to balance redox reactions occurring in acidic aqueous solutions. Follow this step-by-step method to avoid common errors:

1. Split the unbalanced full reaction into two separate half-reactions (oxidation and reduction, identified via oxidation state changes)
2. Balance all atoms except hydrogen and oxygen in each half-reaction
3. Balance oxygen atoms by adding ${\text{H}}_2\text{O}$ molecules to the side of the half-reaction with fewer oxygen atoms
4. Balance hydrogen atoms by adding ${\text{H}}^{+}$ ions to the side of the half-reaction with fewer hydrogen atoms
5. Balance net charge by adding electrons to the side with a more positive total charge
6. Multiply each half-reaction by a small integer so the total number of electrons lost in oxidation equals the total number gained in reduction
7. Add the two half-reactions together, cancel out identical species (including electrons) that appear on both sides, and verify that total charge and total atoms are balanced on both sides of the final equation.

Worked example

Balance the reaction of hydrogen peroxide (${\text{H}}_2{\text{O}}_2$) with iodide ions (${\text{I}}^{-}$) in acidic solution to form water and iodine solid:

1. Half-reactions:

• Oxidation (I goes from -1 to 0): ${\text{I}}^{-}\to {\text{I}}_2$
• Reduction (O goes from -1 to -2): ${\text{H}}_2{\text{O}}_2\to {\text{H}}_2\text{O}$

2. Balance non-H/O atoms: $2{\text{I}}^{-}\to {\text{I}}_2$, ${\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}$
3. Balance O: already balanced, no extra ${\text{H}}_2\text{O}$ needed
4. Balance H: add 2 ${\text{H}}^{+}$ to the left of the reduction half-reaction: ${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}\to 2{\text{H}}_2\text{O}$
5. Balance charge:

• Oxidation: left charge = -2, right = 0, add 2e⁻ to right: $2{\text{I}}^{-}\to {\text{I}}_2+2e^{-}$
• Reduction: left charge = +2, right = 0, add 2e⁻ to left: ${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2e^{-}\to 2{\text{H}}_2\text{O}$

6. Electron counts are equal (2 lost, 2 gained), no multiplication needed
7. Add and cancel electrons: final balanced equation: $${\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}+2{\text{I}}^{-}\to {\text{I}}_2+2{\text{H}}_2\text{O}$$ Verify charge balance: left = 0 + 2 - 2 = 0, right = 0 + 0 = 0, correct.

4. Voltaic and electrolytic cells

Electrochemical cells convert between chemical energy and electrical energy, with two core types assessed in SL:

General rules for all cells

• Anode: Site of oxidation (mnemonic: AN OX)
• Cathode: Site of reduction (mnemonic: RED CAT)
• Electrons flow from anode to cathode through the external wire in all cells
• Ions flow through electrolytes (or salt bridges) to balance charge buildup

Voltaic (galvanic) cells

Voltaic cells generate electrical energy from spontaneous redox reactions (ΔG < 0), and are the basis of household batteries. Key components:

1. Two half-cells: each consists of a solid metal electrode submerged in an aqueous solution of its own metal ions
2. Salt bridge: an inert gel/filter paper soaked in a soluble ionic compound (e.g. ${\text{KNO}}_3$) that allows mobile ions to flow between half-cells to balance charge, completing the circuit without mixing the two solutions
3. External wire connecting the two electrodes, often with a voltmeter to measure cell potential

• Charge labels: Anode = negative (spontaneously releases electrons), Cathode = positive (attracts electrons)

Electrolytic cells

Electrolytic cells use an external electrical power source to drive non-spontaneous redox reactions (ΔG > 0), used for electroplating, extracting reactive metals like aluminium, and splitting water into hydrogen and oxygen. Key components:

1. A single container with a molten or aqueous electrolyte (conductive ionic solution)
2. Two electrodes connected to an external DC power supply

• Charge labels: Anode = positive (connected to the positive terminal of the power supply, pulls electrons from the electrolyte), Cathode = negative (connected to the negative terminal of the power supply, pushes electrons into the electrolyte)

Worked comparative example

Exam tip: You will often be asked to label cell diagrams for 3-4 marks. Always include direction of electron flow, ion flow, anode/cathode labels, and charge signs to earn full marks.

5. Reactivity series

The reactivity (activity) series ranks metals (and hydrogen) by their tendency to be oxidized (lose electrons), i.e. their strength as reducing agents. The core SL series, ordered from most reactive to least reactive, is: $$\text{K}>\text{Na}>\text{Ca}>\text{Mg}>\text{Al}>\text{Zn}>\text{Fe}>\text{Pb}>(\text{H})>\text{Cu}>\text{Ag}>\text{Au}$$

Key predictive rules:

1. A more reactive (higher) elemental metal will displace the ion of a less reactive (lower) metal from aqueous solution. For example, Zn is above Cu, so ${\text{Zn}}_{(s)}+{\text{CuSO}}_{4(aq)}\to {\text{ZnSO}}_{4(aq)}+{\text{Cu}}_{(s)}$ is spontaneous, but the reverse reaction will not occur.
2. Metals above hydrogen in the series will displace ${\text{H}}_2$ gas from dilute non-oxidizing acids (e.g. HCl, ${\text{H}}_2{\text{SO}}_4$). Metals below hydrogen will not react with dilute acids.
3. The higher a metal is in the series, the more energy is required to extract it from its ore: very reactive metals like Al require electrolysis, while less reactive metals like Fe can be extracted via reduction with carbon.

Worked example

Predict if a reaction occurs when copper metal is added to silver nitrate (${\text{AgNO}}_3$) solution, and write the balanced full equation if it does:

1. Cu is above Ag in the reactivity series, so a reaction will occur
2. Oxidation half-reaction: $\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$
3. Reduction half-reaction: $2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag}$
4. Full balanced equation: ${\text{Cu}}_{(s)}+2{\text{AgNO}}_{3(aq)}\to {\text{Cu(NO}}_3{\text{)}}_{2(aq)}+2{\text{Ag}}_{(s)}$

6. Common Pitfalls (and how to avoid them)

• Wrong move: Writing oxidation states as 2+ instead of +2, mixing notation with ion charges. Why students do it: Confusion between oxidation state and ion charge conventions. Correct move: Explicitly label values, write oxidation states with sign first (+2) and ion charges with sign second (2+).
• Wrong move: Stating the salt bridge in a voltaic cell carries electrons. Why students do it: Confusing ion flow with electron flow. Correct move: Electrons only flow through the external wire; the salt bridge carries mobile ions to balance charge buildup in half-cells.
• Wrong move: Assigning the same charge signs to anode/cathode across voltaic and electrolytic cells. Why students do it: Memorizing "anode is positive" without context. Correct move: First apply AN OX/RED CAT, then remember spontaneous voltaic cells have negative anodes (electron source) while electrolytic cells have positive anodes connected to the power supply positive terminal.
• **Wrong move: Only balancing atoms, not charge, when balancing redox equations. Why students do it: Carrying over habits from standard chemical equation balancing. Correct move: Always verify total charge is equal on both sides of the final balanced equation, in addition to atom counts.
• Wrong move: Predicting displacement reactions using the position of the ion instead of the elemental metal in the reactivity series. Why students do it: Mixing up reducing agent (elemental metal) and oxidizing agent (metal ion) roles. Correct move: The elemental metal must be higher in the series than the metal in the ionic compound for a displacement reaction to proceed.

7. Practice Questions (IB Chemistry SL Style)

Question 1 (Paper 1, 1 mark)

What is the oxidation state of nitrogen in the ammonium ion ${\text{NH}}_4^{+}$? A) -3 B) +1 C) +3 D) +5

Worked solution: Correct answer A. Calculation: H has OS = +1, total ion charge = +1. $N+4(+1)=+1\to N=+1-4=-3$.

Question 2 (Paper 2, 4 marks)

Balance the following redox reaction in acidic solution: $${\text{MnO}}_4^{-}+{\text{SO}}_3^{2-}\to {\text{Mn}}^{2+}+{\text{SO}}_4^{2-}$$

Worked solution (mark scheme):

1. (1 mark) Correct half-reactions:

• Oxidation (S OS +4 → +6): ${\text{SO}}_3^{2-}+{\text{H}}_2\text{O}\to {\text{SO}}_4^{2-}+2{\text{H}}^{+}+2e^{-}$
• Reduction (Mn OS +7 → +2): ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5e^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

2. (1 mark) Equalize electron count: Multiply oxidation half-reaction by 5, reduction by 2 to get 10 electrons transferred total
3. (2 marks) Final balanced equation, with electrons and duplicate species cancelled: $$2{\text{MnO}}_4^{-}+6{\text{H}}^{+}+5{\text{SO}}_3^{2-}\to 2{\text{Mn}}^{2+}+5{\text{SO}}_4^{2-}+3{\text{H}}_2\text{O}$$ (1 mark for correct atom balance, 1 mark for correct charge balance: left charge = -2 +6 -10 = -6, right = +4 -10 = -6, equal)

Question 3 (Paper 2, 3 marks)

An electrolytic cell is used to electroplate silver onto a copper spoon. a) State if the spoon acts as the anode or cathode (1 mark) b) State the direction of flow of silver ions in the electrolyte (1 mark) c) Explain why a power supply is required for this process (1 mark)

Worked solution: a) (1 mark) Electroplating involves reduction of silver ions to solid silver metal, which occurs at the cathode, so the spoon is the cathode. b) (1 mark) Silver cations are positive, so they flow toward the negative cathode (spoon). c) (1 mark) Copper is less reactive than silver, so the reduction of silver ions onto copper is non-spontaneous, requiring an external power supply to drive the reaction.

8. Quick Reference Cheatsheet

9. What's Next

Redox processes are a foundational topic that connects to multiple other parts of the IB Chemistry SL syllabus. You will apply oxidation state rules to classify organic oxidation and reduction reactions in the Organic Chemistry topic, and redox titration calculations build directly on the stoichiometry skills you learned earlier in the course. If you progress to IB Chemistry HL, you will expand on this content to cover standard electrode potential calculations, electrolysis of aqueous solutions, and fuel cell technology, all of which build directly on the SL concepts covered in this guide. Redox is also a very popular topic for internal assessment (IA) investigations, including experiments measuring corrosion rates, voltaic cell potential changes, or redox titration of commercial products like vitamin C tablets.

If you are stuck on any redox concept, from balancing tricky equations to distinguishing between voltaic and electrolytic cell components, you can ask Ollie, your 24/7 AI tutor, for personalized explanations, extra practice questions, or step-by-step walkthroughs at any time on the homepage. You can also explore our other IB Chemistry SL study guides to reinforce related topics like stoichiometric relationships and chemical bonding before moving on to organic chemistry content.