Introduction: Mastering Redox and Iodine–Thiosulfate Titrations

Welcome to one of the most practical and calculation-rich areas of A Level Chemistry! While the name iodine–thiosulfate titration might sound a little intimidating at first, the underlying chemistry follows a logical, step-by-step process. In this chapter, we will break down the rules of oxidation states, learn how to balance redox equations from half-equations, explore the exact practical steps of an iodine–thiosulfate titration, and conquer the multi-step quantitative calculations that regularly appear in exam papers.

Note: For foundational work on the mole concept and standard solution preparation, see the chapter "Formulae, equations and the mole".

1. The Fundamentals of Redox Chemistry

Oxidation and Reduction in Terms of Electrons

You have likely met the mnemonic OIL RIG before:

• Oxidation Is Loss of electrons.
• Reduction Is Gain of electrons.

Because electrons cannot simply appear or vanish on their own, oxidation and reduction always occur simultaneously. A reaction in which electrons are transferred from one species to another is called a redox reaction.

• An oxidising agent accepts electrons from another species (and is itself reduced).
• A reducing agent donates electrons to another species (and is itself oxidised).

Oxidation Numbers (Oxidation States)

An oxidation number (or oxidation state) is a formal number assigned to an atom in a compound or ion that reflects its apparent degree of oxidation. We represent oxidation numbers with Roman numerals (or standard signed integers, e.g., \(+2\), \(-1\)).

Rules for Assigning Oxidation Numbers:

1. Any uncombined element has an oxidation number of \(0\). For example, in \(\text{Mg}\), \(\text{O}_2\), \(\text{Cl}_2\), and \(\text{I}_2\), the oxidation number is \(0\).
2. In a simple monatomic ion, the oxidation number is equal to the charge on the ion. For example, \(\text{Na}^+\) is \(+1\), \(\text{Mg}^{2+}\) is \(+2\), \(\text{Cl}^-\) is \(-1\), and \(\text{O}^{2-}\) is \(-2\).
3. The sum of all oxidation numbers in a neutral compound is always \(0\).
4. The sum of all oxidation numbers in a polyatomic ion equals the overall charge on the ion. For example, in \(\text{SO}_4^{2-}\), the sum is \(-2\).
5. Certain elements almost always have fixed oxidation numbers in their compounds:
    • Group 1 metals: always \(+1\).
    • Group 2 metals: always \(+2\).
    • Fluorine: always \(-1\) (it is the most electronegative element).
    • Oxygen: almost always \(-2\) (except in peroxides such as \(\text{H}_2\text{O}_2\) where it is \(-1\), or bonded to fluorine where it can be positive, such as \(+2\) in \(\text{OF}_2\)).
    • Hydrogen: almost always \(+1\) (except in metal hydrides like \(\text{NaH}\), where it is \(-1\)).
    • Halogens (chlorine, bromine, iodine): usually \(-1\), unless bonded to a more electronegative element (such as oxygen or fluorine).

Redox in Terms of Oxidation Numbers:
• Oxidation is an increase in oxidation number.
• Reduction is a decrease in oxidation number.

Worked Example: Finding an Oxidation Number

Question: Find the oxidation number of sulfur in the thiosulfate ion, \(\text{S}_2\text{O}_3^{2-}\).
Solution:
Let the oxidation state of sulfur be \(x\).
Oxygen is \(-2\). There are \(3\) oxygen atoms, giving a total of \(3 \times (-2) = -6\).
The overall charge of the ion is \(-2\).
\(2x + (-6) = -2\)
\(2x = +4\)
\(x = +2\)
The average oxidation number of sulfur in \(\text{S}_2\text{O}_3^{2-}\) is \(+2\).

Balancing Redox Equations Using Half-Equations

To combine two half-equations into a balanced overall redox equation, follow this systematic method:

1. Identify the oxidation half-equation and the reduction half-equation.
2. Balance any atoms other than oxygen and hydrogen first.
3. In acidic solution, balance oxygen atoms by adding water molecules, \(\text{H}_2\text{O}\).
4. Balance hydrogen atoms by adding hydrogen ions, \(\text{H}^+\).
5. Balance the electrical charges by adding electrons, \(\text{e}^-\).
6. Multiply one or both half-equations by integers so that both equations contain the same number of electrons.
7. Add the two equations together and cancel out the electrons, along with any common \(\text{H}^+\) ions or \(\text{H}_2\text{O}\) molecules appearing on both sides.

Key Takeaway: Oxidation is a loss of electrons and an increase in oxidation number; reduction is a gain of electrons and a decrease in oxidation number. Oxidation numbers must sum to zero in neutral compounds and to the total ionic charge in polyatomic ions.

2. The Iodine–Thiosulfate Reaction

The Core Reaction

The heart of all iodine–thiosulfate volumetric analysis is the rapid and quantitative reaction between aqueous iodine, \(\text{I}_2\text{(aq)}\), and aqueous thiosulfate ions, \(\text{S}_2\text{O}_3^{2-}\text{(aq)}\):

\(2\text{S}_2\text{O}_3^{2-}\text{(aq)} + \text{I}_2\text{(aq)} \rightarrow \text{S}_4\text{O}_6^{2-}\text{(aq)} + 2\text{I}^-\text{(aq)}\)

Let's look at the two half-equations that make up this process:
• Reduction half-equation:
\(\text{I}_2\text{(aq)} + 2\text{e}^- \rightarrow 2\text{I}^-\text{(aq)}\)
Iodine is reduced from oxidation state \(0\) in \(\text{I}_2\) to \(-1\) in \(\text{I}^-\).
• Oxidation half-equation:
\(2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow \text{S}_4\text{O}_6^{2-}\text{(aq)} + 2\text{e}^-\)
Sulfur is oxidised from an average oxidation state of \(+2\) in thiosulfate (\(\text{S}_2\text{O}_3^{2-}\)) to \(+2.5\) in the tetrathionate ion (\(\text{S}_4\text{O}_6^{2-}\)).

Crucial Stoichiometric Ratio:
From the balanced equation, notice that:

\(1\text{ mol of }\text{I}_2 \equiv 2\text{ mol of }\text{S}_2\text{O}_3^{2-}\)

Whenever you titrate iodine with sodium thiosulfate solution, the number of moles of thiosulfate used is always twice the number of moles of iodine present in the flask.

Why Use an Indirect Titration?

Many common oxidising agents (such as chlorate(I) ions in household bleach, iodate(V) ions, copper(II) ions, or iron(III) ions) cannot be titrated directly with high precision because they do not have an obvious, sharp colour change at their end point.
Instead, we carry out a two-step indirect titration:

1. Step 1 (Liberation of iodine): A measured volume or mass of the oxidising agent is reacted with an excess of potassium iodide, \(\text{KI}\), in acidic solution. The oxidising agent oxidises the colourless iodide ions (\(\text{I}^-\)) into iodine (\(\text{I}_2\)). The solution turns a deep brown/red-brown colour.
2. Step 2 (Titration of iodine): The liberated iodine is then titrated directly with a standard solution of sodium thiosulfate of known concentration until all the iodine has reacted.

Because the amount of iodine produced is directly proportional to the amount of original oxidising agent, measuring the iodine tells us exactly how much oxidising agent was present!

Common Reactions Used to Generate Iodine

1. Reaction with Iodate(V) ions, \(\text{IO}_3^-\):
Potassium iodate(V) can act as a primary standard. In acid solution, iodate(V) oxidises iodide ions:
\(\text{IO}_3^-\text{(aq)} + 5\text{I}^-\text{(aq)} + 6\text{H}^+\text{(aq)} \rightarrow 3\text{I}_2\text{(aq)} + 3\text{H}_2\text{O}\text{(l)}\)
Stoichiometry: \(1\text{ mol of }\text{IO}_3^- \equiv 3\text{ mol of }\text{I}_2 \equiv 6\text{ mol of }\text{S}_2\text{O}_3^{2-}\).

2. Reaction with Chlorate(I) ions, \(\text{ClO}^-\) (Bleach analysis):
\(\text{ClO}^-\text{(aq)} + 2\text{I}^-\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)} + \text{H}_2\text{O}\text{(l)}\)
Stoichiometry: \(1\text{ mol of }\text{ClO}^- \equiv 1\text{ mol of }\text{I}_2 \equiv 2\text{ mol of }\text{S}_2\text{O}_3^{2-}\).

3. Reaction with Copper(II) ions, \(\text{Cu}^{2+}\):
\(2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI}\text{(s)} + \text{I}_2\text{(aq)}\)
Here, copper(II) is reduced to an off-white precipitate of copper(I) iodide (\(\text{CuI}\)), while iodide is oxidised to iodine.
Stoichiometry: \(2\text{ mol of }\text{Cu}^{2+} \equiv 1\text{ mol of }\text{I}_2 \equiv 2\text{ mol of }\text{S}_2\text{O}_3^{2-}\), which simplifies to \(1\text{ mol of }\text{Cu}^{2+} \equiv 1\text{ mol of }\text{S}_2\text{O}_3^{2-}\).

Key Takeaway: In an indirect titration, an unknown oxidising agent liberates an equivalent amount of \(\text{I}_2\) from excess \(\text{I}^-\). The liberated \(\text{I}_2\) is then titrated with \(\text{S}_2\text{O}_3^{2-}\) in a \(1 : 2\) mole ratio.

3. Practical Procedure and the Starch Indicator

The Titration Step-by-Step

Carrying out an iodine–thiosulfate titration accurately requires precise experimental technique:

1. Rinse a burette with distilled water and then with the standard sodium thiosulfate solution. Fill the burette and record the initial reading to the nearest \(0.05\text{ cm}^3\).
2. Pipette a known volume of the oxidising agent into a conical flask.
3. Add an excess of potassium iodide solution (often together with dilute acid if required). The mixture turns dark brown due to the formed iodine dissolved in aqueous iodide.
4. Place the conical flask on a white tile under the burette and begin titrating with sodium thiosulfate while swirling continuously.
5. As the titration proceeds, the brown colour fades to yellow, and then to a pale straw colour.
6. Do not add starch yet! Wait until the solution reaches this pale straw colour, then add about \(1\text{ cm}^3\) to \(2\text{ cm}^3\) of fresh starch indicator solution. The solution immediately turns deep blue-black.
7. Continue adding sodium thiosulfate dropwise until the blue-black colour vanishes completely, leaving a sharp, colourless end point (or a white/cream suspension if \(\text{CuI}\) is present).
8. Repeat until you achieve concordant titres (titres within \(0.10\text{ cm}^3\) of each other), and calculate the mean of these concordant values.

Why Must Starch Be Added Near the End Point?

This is a favorite question in OCR examinations!

• If starch is added right at the start when the iodine concentration is high, iodine binds very strongly to the amylose helices in the starch to form an insoluble blue-black starch–iodine complex.
• This complex breaks apart very slowly, preventing the iodine from reacting readily with the thiosulfate ions.
• This leads to an inaccurate, delayed end point and an over-estimation of the titre.
• Adding starch only when the mixture is pale straw ensures that the remaining concentration of iodine is low enough for the complex to dissociate rapidly and give a sharp, instantaneous colour change from blue-black to colourless.

Experimental Technique and Uncertainties

To maximise experimental accuracy in the laboratory and score full marks in paper 03 practical assessments:

• Reading the burette: Burettes have graduations every \(0.10\text{ cm}^3\). Measurements must be recorded to two decimal places ending in either \(.00\) or \(.05\text{ cm}^3\).
• Percentage error in a reading:
\(\text{percentage uncertainty} = \frac{\text{total apparatus uncertainty}}{\text{measurement value}} \times 100\)
Because a titre involves two burette readings (initial and final), the absolute uncertainty is doubled: \(\pm 0.05\text{ cm}^3 \times 2 = \pm 0.10\text{ cm}^3\).
• Concordancy: Only use titres that agree within \(\pm 0.10\text{ cm}^3\). The rough/trial titre is always excluded from the mean unless it happens to be concordant with subsequent runs.

Key Takeaway: Starch indicator turns blue-black in the presence of iodine and goes colourless at the end point. It must only be added when the solution fades to pale straw, avoiding the formation of an insoluble starch–iodine complex.

4. Step-by-Step Quantitative Calculations

Don't worry if redox titration calculations seem long. They are just a chain of simple calculations connected together. If you follow this four-step roadmap, you will never get lost:

• Step 1: Calculate the moles of thiosulfate used in the average concordant titre: \(n = c \times V\).
• Step 2: Use the stoichiometric ratio (\(2\text{S}_2\text{O}_3^{2-} : 1\text{I}_2\)) to find the moles of \(\text{I}_2\) titrated.
• Step 3: Relate the moles of \(\text{I}_2\) back to the moles of the original oxidising agent in the reaction flask using the first chemical equation.
• Step 4: Scale up if a volumetric aliquot (dilution) was used, and calculate the final concentration, mass, or percentage purity as requested.

Worked Example 1: Analysis of Bleach

Question: A student wants to determine the concentration of chlorate(I) ions, \(\text{ClO}^-\), in household bleach. A \(25.0\text{ cm}^3\) sample of bleach is diluted with distilled water to \(250.0\text{ cm}^3\) in a volumetric flask. A \(25.0\text{ cm}^3\) portion of this diluted solution is acidified and treated with an excess of \(\text{KI}\). The liberated iodine required an average concordant titre of \(21.40\text{ cm}^3\) of \(0.0500\text{ mol dm}^{-3}\) \(\text{Na}_2\text{S}_2\text{O}_3\). Calculate the concentration of \(\text{ClO}^-\) in the original bleach.

Step 1: Find moles of thiosulfate ions used
\(V = \frac{21.40}{1000} = 0.02140\text{ dm}^3\)
\(n(\text{S}_2\text{O}_3^{2-}) = c \times V = 0.0500\text{ mol dm}^{-3} \times 0.02140\text{ dm}^3 = 1.07 \times 10^{-3}\text{ mol}\)

Step 2: Find moles of iodine titrated
From the equation: \(2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-\)
\(n(\text{I}_2) = \frac{n(\text{S}_2\text{O}_3^{2-})}{2} = \frac{1.07 \times 10^{-3}}{2} = 5.35 \times 10^{-4}\text{ mol}\)

Step 3: Find moles of \(\text{ClO}^-\) in the \(25.0\text{ cm}^3\) sample
From the equation: \(\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}\)
The ratio of \(\text{ClO}^- : \text{I}_2\) is \(1 : 1\).
\(n(\text{ClO}^-\text{ in }25.0\text{ cm}^3) = 5.35 \times 10^{-4}\text{ mol}\)

Step 4: Scale up to the original sample
The \(25.0\text{ cm}^3\) titrated portion came from a \(250.0\text{ cm}^3\) volumetric flask:
\(n(\text{ClO}^-\text{ in }250.0\text{ cm}^3) = 5.35 \times 10^{-4}\text{ mol} \times \frac{250.0}{25.0} = 5.35 \times 10^{-3}\text{ mol}\)
These moles were originally present in the \(25.0\text{ cm}^3\) undiluted bleach sample:
\(c(\text{ClO}^-\text{ in original bleach}) = \frac{n}{V} = \frac{5.35 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.214\text{ mol dm}^{-3}\)

Worked Example 2: Determining Percentage of Copper in an Alloy

Question: A brass screw of mass \(2.80\text{ g}\) was dissolved in concentrated nitric acid and made up to \(250.0\text{ cm}^3\) with distilled water. A \(25.0\text{ cm}^3\) portion was neutralised, treated with excess potassium iodide, and titrated with \(0.100\text{ mol dm}^{-3}\) sodium thiosulfate. The mean titre was \(29.80\text{ cm}^3\). Calculate the percentage by mass of copper in the brass screw (\(A_r\text{ of }\text{Cu} = 63.5\)).

Equations provided:
\(2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI}\text{(s)} + \text{I}_2\text{(aq)}\)
\(2\text{S}_2\text{O}_3^{2-}\text{(aq)} + \text{I}_2\text{(aq)} \rightarrow \text{S}_4\text{O}_6^{2-}\text{(aq)} + 2\text{I}^-\text{(aq)}\)

Solution:
1. Calculate moles of thiosulfate:
\(n(\text{S}_2\text{O}_3^{2-}) = 0.100\text{ mol dm}^{-3} \times \frac{29.80}{1000}\text{ dm}^3 = 2.980 \times 10^{-3}\text{ mol}\)
2. Calculate moles of \(\text{I}_2\):
\(n(\text{I}_2) = \frac{2.980 \times 10^{-3}}{2} = 1.490 \times 10^{-3}\text{ mol}\)
3. Relate to moles of \(\text{Cu}^{2+}\) in the \(25.0\text{ cm}^3\) sample:
From the equation, \(1\text{ mol }\text{I}_2\) is produced by \(2\text{ mol }\text{Cu}^{2+}\).
\(n(\text{Cu}^{2+}) = 2 \times 1.490 \times 10^{-3} = 2.980 \times 10^{-3}\text{ mol}\)
(Notice that \(n(\text{Cu}^{2+}) = n(\text{S}_2\text{O}_3^{2-})\) overall!)
4. Scale up to the total \(250.0\text{ cm}^3\) solution:
\(n(\text{Cu}^{2+}\text{ total}) = 2.980 \times 10^{-3}\text{ mol} \times 10 = 2.980 \times 10^{-2}\text{ mol}\)
5. Calculate the mass of copper:
\(\text{mass} = n \times A_r = 2.980 \times 10^{-2}\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.8923\text{ g}\)
6. Calculate percentage by mass in the screw:
\(\% \text{ Cu} = \frac{1.8923\text{ g}}{2.80\text{ g}} \times 100 = 67.6\%\) (to \(3\) significant figures).

5. Other Redox Titrations: Manganate(VII) Titrations

In addition to iodine–thiosulfate systems, the syllabus covers manganate(VII) redox titrations (encountered in the Developing metals module). These follow the exact same stoichiometric principles!

• Potassium manganate(VII), \(\text{KMnO}_4\), acts as a powerful oxidising agent in acidic solution.
• The half-equation for the reduction of the deep purple manganate(VII) ion, \(\text{MnO}_4^-\), is:
\(\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5\text{e}^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O}\text{(l)}\)
• Manganese is reduced from \(+7\) (deep purple) to \(+2\) (virtually colourless / very pale pink).
• Self-indicating: Because \(\text{MnO}_4^-\) is intensely purple and \(\text{Mn}^{2+}\) is almost colourless, no external indicator is needed! The end point occurs when the first drop of excess \(\text{MnO}_4^-\) gives the solution a permanent pale pink colour.
• A classic application is titrating iron(II) ions, \(\text{Fe}^{2+}\):
\(\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + \text{e}^-\)
Overall equation:
\(\text{MnO}_4^-\text{(aq)} + 5\text{Fe}^{2+}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow \text{Mn}^{2+}\text{(aq)} + 5\text{Fe}^{3+}\text{(aq)} + 4\text{H}_2\text{O}\text{(l)}\)
Here, \(1\text{ mol of }\text{MnO}_4^- \equiv 5\text{ mol of }\text{Fe}^{2+}\).

Common Pitfalls to Avoid in Exams

• Adding starch too early: Remember, never add starch at the beginning of an iodine–thiosulfate titration. Only add it when the brown colour fades to pale straw/yellow.
• Forgetting the \(2 : 1\) thiosulfate-to-iodine ratio: Always write out the stoichiometric conversion explicitly: \(n(\text{I}_2) = \frac{1}{2} n(\text{S}_2\text{O}_3^{2-})\).
• Missing the dilution factor: Check whether the question used a sample directly or diluted it in a volumetric flask (e.g. taking \(25.0\text{ cm}^3\) from a \(250.0\text{ cm}^3\) flask requires multiplying by \(10\)).
• Significant figures: Match your final answer's significant figures to the data with the fewest significant figures in the question (usually \(3\) significant figures).
• Confusing the indicator change: In an iodine titration with thiosulfate in the burette, the end point is blue-black to colourless, not colourless to blue-black!

Chapter Summary

• Oxidation is electron loss and an increase in oxidation number; reduction is electron gain and a decrease in oxidation number.
• Iodine is titrated with thiosulfate: \(2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-\). The mole ratio of \(\text{S}_2\text{O}_3^{2-}\) to \(\text{I}_2\) is always \(2 : 1\).
• Starch indicator produces an intense blue-black colour with iodine and must be added near the end point (pale straw colour) to ensure sharp reversibility to colourless.
• Indirect titration allows quantitative analysis of oxidising agents such as \(\text{ClO}^-\), \(\text{IO}_3^-\), and \(\text{Cu}^{2+}\) by liberating an equivalent amount of \(\text{I}_2\).
• In \(\text{MnO}_4^-\) titrations, \(\text{MnO}_4^-\) is reduced to \(\text{Mn}^{2+}\) in a self-indicating reaction with an end point of first permanent pale pink.