Calculate quantities of substances in redox titrations

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 – Unit B1, Outcome 8k

You add a deep purple solution, drop by drop, into a pale green solution. Suddenly, one drop turns the entire flask a permanent faint pink — and you stop. That colour change is the endpoint of a redox titration, and it tells you that the reaction between the two substances is complete. From the volume and concentration of the solution you added, you can calculate exactly how much of the unknown substance was present in the flask. This lesson teaches you how to set up, balance, and solve those calculations step by step.

What you will learn

Identify the oxidizing agent and reducing agent in a redox titration reaction.
Write and balance a net ionic redox equation using half-reactions.
Use mole ratios from a balanced redox equation to relate moles of titrant to moles of analyte.
Calculate the concentration or volume of an unknown substance from redox titration data.
Apply correct significant digits and units throughout a multi-step redox titration calculation.

Prerequisite Review: Titration and Stoichiometry Basics

A titration is a laboratory technique in which a solution of known concentration, called the standard solution or titrant, is added carefully to a solution of unknown concentration, called the analyte, until the reaction is exactly complete. The point at which the reaction is complete is called the equivalence point. An indicator — a substance that changes colour — signals when you are at or very near the equivalence point; this signal point is called the endpoint.

From Chemistry 20 you know that concentration is defined as moles of solute divided by volume of solution in litres. In equation form:

c = \frac{n}{V}

, where

c

is concentration in mol/L,

n

is amount in mol, and

V

is volume in L. Rearranging gives

n = cV

. This relationship is the bridge between what you measure in the lab (volume and concentration) and what you calculate (moles).

You also know from stoichiometry that once you have moles of one substance, you use the mole ratio from the balanced equation to find moles of any other substance in the reaction. In a redox titration the balanced equation must account for both mass (atoms) and charge, so you must balance it using half-reactions before you can read off the correct mole ratio.

n = cV

Titrant: solution of known concentration added from the burette.
Analyte: solution of unknown concentration in the flask.
Equivalence point: moles of oxidizing agent exactly react with moles of reducing agent according to the balanced equation.
$n = cV$ converts measured volume and concentration into moles.
The balanced redox equation provides the mole ratio needed for stoichiometry.

What Makes a Redox Titration Different

In an acid-base titration, H⁺ transfers between species. In a redox titration, electrons transfer between species. The substance that loses electrons is the reducing agent; it is oxidized. The substance that gains electrons is the oxidizing agent; it is reduced. Both must appear correctly in the balanced equation before you can do any calculation.

A common and important redox titration in Chemistry 30 uses potassium permanganate,

\text{KMnO}_4

, as the titrant. Permanganate ion,

\text{MnO}_4^-

, is the oxidizing agent. In acidic solution it is reduced to the nearly colourless

\text{Mn}^{2+}

ion. Because permanganate is intensely purple and

\text{Mn}^{2+}

is nearly colourless, permanganate acts as its own indicator: the solution stays colourless (or the colour of the analyte) until one extra drop of

\text{MnO}_4^-

is added past the equivalence point, turning the solution a faint permanent pink. No separate indicator is needed.

Another common titration uses potassium dichromate,

\text{K}_2\text{Cr}_2\text{O}_7

, as the oxidizing agent, with a separate indicator such as diphenylamine. Whether the titrant or a separate indicator signals the endpoint, the calculation method is identical: find moles of titrant used, apply the mole ratio from the balanced equation, and solve for the unknown.

\text{MnO}_4^-(aq) + 8\,\text{H}^+(aq) + 5\,e^- \rightarrow \text{Mn}^{2+}(aq) + 4\,\text{H}_2\text{O}(l)

Redox titrations involve electron transfer, not proton transfer.
Oxidizing agent gains electrons (is reduced); reducing agent loses electrons (is oxidized).
$\text{MnO}_4^-$ in acidic solution is reduced to $\text{Mn}^{2+}$ ; the colour change from purple to colourless/pink signals the endpoint.
The balanced half-reaction method ensures both atoms and charge are conserved.
Mole ratio from the balanced overall equation is essential for stoichiometry.

Balancing the Redox Equation Using Half-Reactions

Before calculating, you must have a correctly balanced equation. The half-reaction method is the standard approach in Alberta Chemistry 30. You split the overall reaction into two half-reactions — one for oxidation, one for reduction — balance each separately for atoms and charge, then combine them so electrons cancel.

Consider the titration of iron(II) ions,

\text{Fe}^{2+}

, with permanganate in acidic solution. Step 1: Write the two unbalanced half-reactions. Reduction:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+}

. Oxidation:

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}

. Step 2: Balance atoms other than H and O first. Mn and Fe are already balanced one-to-one. Step 3: Balance O by adding

\text{H}_2\text{O}

to the side that needs oxygen. Reduction half-reaction:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O}

. Step 4: Balance H by adding

\text{H}^+

to the side that needs hydrogen. Reduction:

\text{MnO}_4^- + 8\,\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O}

. Step 5: Balance charge by adding electrons. Left side of reduction:

-1 + 8 = +7

; right side:

+2

. Add

5\,e^-

to the left:

\text{MnO}_4^- + 8\,\text{H}^+ + 5\,e^- \rightarrow \text{Mn}^{2+} + 4\,\text{H}_2\text{O}

. For oxidation: left

= +2

, right

= +3

; add

1\,e^-

to the right:

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

.

Step 6: Multiply each half-reaction so electrons cancel. The reduction half-reaction involves 5 electrons; the oxidation involves 1. Multiply the oxidation half-reaction by 5. Step 7: Add the two half-reactions and cancel electrons and any species that appear on both sides. The result is the balanced net ionic equation used for all calculations in this titration.

\text{MnO}_4^-(aq) + 8\,\text{H}^+(aq) + 5\,\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\,\text{Fe}^{3+}(aq) + 4\,\text{H}_2\text{O}(l)

Split into oxidation and reduction half-reactions.
Balance non-H, non-O atoms first; then O with H₂O; then H with H⁺; then charge with electrons.
Multiply half-reactions so the number of electrons lost equals the number gained.
Add half-reactions and cancel electrons to get the balanced overall equation.
Always verify: atom counts and total charge must be equal on both sides.

The Four-Step Calculation Strategy

Every redox titration calculation follows the same logical path. Once you have the balanced equation and the lab data, apply these four steps in order. Step 1 — Find moles of the known substance (titrant): use

n_{\text{titrant}} = c_{\text{titrant}} \times V_{\text{titrant}}

. Always convert volume from mL to L by dividing by 1000. Step 2 — Apply the mole ratio: read the coefficients of the titrant and the analyte from the balanced equation, then calculate moles of analyte as moles of titrant multiplied by the ratio of the analyte coefficient to the titrant coefficient. Step 3 — Solve for the unknown: if the question asks for concentration, use

c_{\text{analyte}} = \frac{n_{\text{analyte}}}{V_{\text{analyte}}}

. If the question asks for mass, use

m = n \times M

where

M

is molar mass. Step 4 — Check significant digits and units: your answer should match the least number of significant digits in the given data, and units must be correct.

This four-step strategy works for any redox titration problem, regardless of which substances are involved. The only parts that change from problem to problem are the balanced equation (and therefore the mole ratio) and which variable is unknown. Always identify what is known and what is unknown before you begin calculating.

A note on significant digits: volumes read from a burette are typically reported to two decimal places (e.g., 18.62 mL), giving four significant digits. Concentrations are often given to three or four significant digits. Round only at the final step to avoid rounding errors that accumulate through the calculation.

c_{\text{analyte}} = \frac{n_{\text{titrant}} \times \text{mole ratio}}{V_{\text{analyte}}}

Step 1: $n_{\text{titrant}} = c_{\text{titrant}} \times V_{\text{titrant}}$ (V in litres).
Step 2: $n_{\text{analyte}} = n_{\text{titrant}} \times \text{mole ratio}$ .
Step 3: Solve for the requested unknown — concentration, volume, or mass.
Step 4: Apply correct significant digits and include units at every step.
Round only at the final answer to prevent accumulated rounding error.

Reading a Redox Titration Problem: Diploma-Style Reasoning

Alberta diploma-style questions often give you extra information (such as the volume of acid added to make the solution acidic) and expect you to identify which data are relevant. In a

\text{KMnO}_4

titration, the acid is added to create the acidic conditions required for the reaction, but the acid itself does not appear in your stoichiometry calculation — only the volume and concentration of

\text{KMnO}_4

and the analyte matter.

Questions may also ask you to calculate the percentage purity of a sample. In that case, you first find the mass of pure analyte from stoichiometry, then divide by the total mass of the impure sample and multiply by 100%. Always label which quantity is the stoichiometric amount and which is the actual sample mass.

Some problems provide a burette reading as both an initial reading and a final reading. The volume of titrant used is always the difference:

V_{\text{used}} = V_{\text{final}} - V_{\text{initial}}

. Double-check that you subtract in the correct order; the final reading is always larger than or equal to the initial reading on a standard burette.

\%\,\text{purity} = \frac{m_{\text{pure}}}{m_{\text{sample}}} \times 100\%

Identify which data belong to the titrant and which to the analyte before calculating.
Acid added to create acidic conditions is not part of the stoichiometry calculation.
Volume of titrant used = final burette reading − initial burette reading.
For percentage purity: $\%\,\text{purity} = \frac{m_{\text{pure}}}{m_{\text{sample}}} \times 100\%$ .
Re-read the question after finishing to confirm you answered what was asked.

Comparing Key Features of Acid-Base and Redox Titrations

Feature	Acid-Base Titration	Redox Titration
What transfers	Protons (H⁺)	Electrons (e⁻)
Equation must balance	Atoms only	Atoms and charge
Common titrant example	NaOH(aq) or HCl(aq)	KMnO₄(aq) or K₂Cr₂O₇(aq)
Indicator role	Separate indicator (e.g., phenolphthalein)	KMnO₄ is self-indicating; others need a separate indicator
Mole ratio source	Balanced molecular equation	Balanced net ionic equation from half-reactions

Worked example

Finding the Concentration of an Iron(II) Solution by Permanganate Titration

A student dissolves an iron(II) sulfate sample in dilute sulfuric acid to create an acidic analyte solution. A 25.00 mL portion of this

\text{Fe}^{2+}

solution is titrated with 0.02000 mol/L

\text{KMnO}_4

. The burette reads 1.25 mL at the start and 19.87 mL at the end. Calculate the molar concentration of

\text{Fe}^{2+}

in the analyte solution.

Step 1 – Write the balanced net ionic equation
The half-reactions give the overall equation. Permanganate (5 electrons gained) reacts with iron(II) (1 electron lost each), so 5 Fe²⁺ are needed per MnO₄⁻. This equation was derived in the lesson above and is verified by checking atoms and charge on both sides.
$\text{MnO}_4^-(aq) + 8\,\text{H}^+(aq) + 5\,\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\,\text{Fe}^{3+}(aq) + 4\,\text{H}_2\text{O}(l)$
Step 2 – Find the volume of KMnO₄ used
Subtract the initial burette reading from the final burette reading. Both readings are in mL; convert to L for the next step.
$V_{\text{KMnO}_4} = 19.87\,\text{mL} - 1.25\,\text{mL} = 18.62\,\text{mL} = 0.01862\,\text{L}$
Step 3 – Calculate moles of KMnO₄ (titrant)
Use $n = cV$ . The concentration of KMnO₄ is given as 0.02000 mol/L. Multiply by the volume in litres.
$n_{\text{KMnO}_4} = 0.02000\,\text{mol/L} \times 0.01862\,\text{L} = 3.724 \times 10^{-4}\,\text{mol}$
Step 4 – Apply the mole ratio to find moles of Fe²⁺
From the balanced equation, the mole ratio of Fe²⁺ to MnO₄⁻ is 5 : 1. Multiply moles of KMnO₄ by this ratio. Every one mole of MnO₄⁻ reacts with exactly five moles of Fe²⁺.
$n_{\text{Fe}^{2+}} = 3.724 \times 10^{-4}\,\text{mol} \times \frac{5}{1} = 1.862 \times 10^{-3}\,\text{mol}$
Step 5 – Calculate the concentration of Fe²⁺
The analyte volume is 25.00 mL = 0.02500 L. Divide moles of Fe²⁺ by this volume. Round to four significant digits to match the precision of the given data.
$c_{\text{Fe}^{2+}} = \frac{1.862 \times 10^{-3}\,\text{mol}}{0.02500\,\text{L}} = 0.07448\,\text{mol/L}$

Answer: The molar concentration of

\text{Fe}^{2+}

in the analyte solution is

0.07448\,\text{mol/L}

.

Check: Verify by working backwards:

0.07448\,\text{mol/L} \times 0.02500\,\text{L} = 1.862 \times 10^{-3}\,\text{mol Fe}^{2+}

. Dividing by the mole ratio:

1.862 \times 10^{-3} \div 5 = 3.724 \times 10^{-4}\,\text{mol KMnO}_4

. Dividing by 0.01862 L gives

0.02000\,\text{mol/L}

, which matches the given concentration of KMnO₄. The answer is consistent.

Common mistakes and how to avoid them

Using an unbalanced or incorrectly balanced redox equation, leading to a wrong mole ratio (e.g., assuming 1 : 1 for MnO₄⁻ to Fe²⁺ instead of 1 : 5).

Correction: Always derive the mole ratio from the fully balanced net ionic equation. For the permanganate–iron(II) reaction, the correct ratio is 1 mol MnO₄⁻ : 5 mol Fe²⁺. Verify your equation by checking that atoms and total charge balance on both sides.

Forgetting to convert volume from mL to L before using

n = cV

, which gives an answer 1000 times too large.

Correction: Divide mL by 1000 to get L before multiplying by concentration. For example, 18.62 mL becomes 0.01862 L.

Using the total volume in the flask (analyte + any acid added) instead of just the analyte volume when calculating analyte concentration.

Correction: Only the volume of the analyte solution matters for finding its concentration. Acid is added to create reaction conditions, not as a stoichiometric reactant in the titration calculation.

Applying the mole ratio upside down — for example, multiplying moles of KMnO₄ by (1/5) instead of (5/1) to get moles of Fe²⁺.

Correction: Write the ratio so the unknown substance is in the numerator:

n_{\text{Fe}^{2+}} = n_{\text{MnO}_4^-} \times \frac{5\,\text{mol Fe}^{2+}}{1\,\text{mol MnO}_4^-}

.

Rounding intermediate values to three significant digits and then reporting the final answer to four, introducing accumulated rounding error.

Correction: Keep at least one extra digit in all intermediate steps. Round only the final answer to the correct number of significant digits.

Lesson summary

A redox titration measures the concentration or amount of an unknown substance by reacting it completely with a standard solution; electrons transfer from the reducing agent to the oxidizing agent.
The balanced net ionic equation, derived from balanced half-reactions, is essential because it gives the correct mole ratio of titrant to analyte.
The four-step calculation strategy is: (1) find moles of titrant using $n = cV$ ; (2) apply the mole ratio; (3) solve for the unknown (concentration, volume, or mass); (4) report with correct units and significant digits.
Permanganate ion, $\text{MnO}_4^-$ , acts as its own indicator: the endpoint is signalled by a permanent faint pink colour after the equivalence point.
Volume of titrant used equals final burette reading minus initial burette reading; always convert mL to L before using $n = cV$ .
For percentage purity problems, calculate the mass of pure substance from stoichiometry, then divide by the total sample mass and multiply by 100%.

Check your understanding

Question 1

In the reaction

\text{MnO}_4^-(aq) + 8\,\text{H}^+(aq) + 5\,\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\,\text{Fe}^{3+}(aq) + 4\,\text{H}_2\text{O}(l)

, which species is the reducing agent?

$\text{MnO}_4^-$ because it gains electrons
$\text{H}^+$ because it is added to make the solution acidic
$\text{Fe}^{2+}$ because it loses electrons and is oxidized
$\text{Mn}^{2+}$ because it is the product of reduction

Show answer and explanation

\text{Fe}^{2+}

because it loses electrons and is oxidized

\text{Fe}^{2+}

loses one electron per ion to become

\text{Fe}^{3+}

, so it is oxidized. The substance that is oxidized is the reducing agent.

\text{MnO}_4^-

gains electrons (is reduced), making it the oxidizing agent.

Question 2

A 20.00 mL sample of

\text{Fe}^{2+}

solution requires 12.50 mL of 0.01000 mol/L

\text{KMnO}_4

to reach the endpoint. Using the 1 : 5 mole ratio (

\text{MnO}_4^-

:

\text{Fe}^{2+}

), what is the concentration of

\text{Fe}^{2+}

?

$0.001250\,\text{mol/L}$
$0.006250\,\text{mol/L}$
$0.03125\,\text{mol/L}$
$0.06250\,\text{mol/L}$

Show answer and explanation

0.03125\,\text{mol/L}

n_{\text{KMnO}_4} = 0.01000\,\text{mol/L} \times 0.01250\,\text{L} = 1.250 \times 10^{-4}\,\text{mol}

. Applying the 5 : 1 ratio:

n_{\text{Fe}^{2+}} = 1.250 \times 10^{-4}\,\text{mol} \times 5 = 6.250 \times 10^{-4}\,\text{mol}

. Concentration:

c = 6.250 \times 10^{-4}\,\text{mol} \div 0.02000\,\text{L} = 0.03125\,\text{mol/L}

.

Question 3

In a redox titration using

\text{KMnO}_4

, 3.00 mL of sulfuric acid is added to the analyte flask to make the solution acidic. A student uses a total flask volume of 28.00 mL (analyte + acid) to calculate the analyte concentration. What error does this introduce?

No error — total volume must always be used when acid is present.
The calculated concentration will be too high because the denominator is too small.
The calculated concentration will be too low because the denominator is too large.
The mole ratio will be incorrect because acid changes the stoichiometry.

Show answer and explanation

The calculated concentration will be too low because the denominator is too large.

The acid is not the analyte; it is added only to create acidic conditions for the reaction. The concentration of the analyte is calculated using only the analyte volume (25.00 mL). Using 28.00 mL makes the denominator too large, so the calculated concentration will be too low.

Question 4

An impure iron wire sample with a mass of 0.3760 g is dissolved and titrated. Stoichiometry shows the sample contains

3.360 \times 10^{-3}

mol

\text{Fe}^{2+}

. The molar mass of Fe is 55.85 g/mol. What is the percentage purity of the iron wire?

$\approx 39.9\%$
$\approx 49.9\%$
$\approx 79.9\%$
$\approx 99.9\%$

Show answer and explanation

\approx 49.9\%

Mass of pure Fe:

m = 3.360 \times 10^{-3}\,\text{mol} \times 55.85\,\text{g/mol} = 0.1877\,\text{g}

. Percentage purity:

(0.1877\,\text{g} \div 0.3760\,\text{g}) \times 100\% = 49.9\%

. Option A would result from an arithmetic error; options C and D are too high to be consistent with these values.

Key terms

Titrant: The solution of known concentration added from the burette during a titration.
Analyte: The solution of unknown concentration placed in the flask; the substance being measured.
Equivalence point: The point in a titration at which the moles of titrant have reacted exactly and completely with the moles of analyte according to the balanced equation.
Endpoint: The point at which the indicator changes colour, signalling that the titration is complete; ideally very close to the equivalence point.
Oxidizing agent: The substance that gains electrons in a redox reaction; it is reduced.
Reducing agent: The substance that loses electrons in a redox reaction; it is oxidized.
Half-reaction: An equation that shows either only the oxidation process or only the reduction process, including electrons as a reactant or product.
Mole ratio: The ratio of coefficients of two substances in a balanced chemical equation, used to convert moles of one substance into moles of another.

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Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Chemistry (CHEM30), expectation 30-B1.8k. It is a study resource, not an official curriculum publication.

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30-B1.8k · Calculate quantities of substances in redox titrations

Alberta Grade 12 Chemistry

Alberta Chemistry 30 – Unit B1, Outcome 8k

What you will learn

Prerequisite Review: Titration and Stoichiometry Basics

What Makes a Redox Titration Different

Balancing the Redox Equation Using Half-Reactions

The Four-Step Calculation Strategy

Reading a Redox Titration Problem: Diploma-Style Reasoning

Comparing Key Features of Acid-Base and Redox Titrations

Finding the Concentration of an Iron(II) Solution by Permanganate Titration

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review