How to Identify Fe²⁺ and Fe³⁺ from Precipitation Data

A fresh iron solution gives a green precipitate when sodium hydroxide is added. A few minutes later, that same precipitate starts turning brown. Does that mean the solution originally contained iron(III), or has something changed?

This is the main complication when identifying iron(II) and iron(III) ions from precipitation data.

The standard test uses aqueous sodium hydroxide, NaOH:

• Iron(II) ions, \(\mathrm{Fe}^{2+}\), produce a green precipitate of iron(II) hydroxide, \(\mathrm{Fe(OH)_2}\).
• Iron(III) ions, \(\mathrm{Fe}^{3+}\), produce a red-brown precipitate of iron(III) hydroxide, \(\mathrm{Fe(OH)_3}\).

However, iron(II) hydroxide can oxidise in contact with oxygen from the air, gradually turning brown. You therefore need to consider the initial observation, the time taken to record it, and whether another test supports your conclusion.

01The primary test: Add sodium hydroxide

Imagine you have two test tubes containing separate, colourless or pale-coloured iron solutions. You don’t know which contains iron(II) and which contains iron(III).

The simplest investigation is to add a few drops of dilute sodium hydroxide solution to each sample.

Place a small amount of each unknown solution into a clean test tube. Add NaOH dropwise, mixing gently, and observe whether a precipitate forms. Record its initial colour before adding excess reagent.

What should you observe?

Both precipitates are insoluble in excess sodium hydroxide.

This is important. Adding more NaOH does not dissolve either iron hydroxide, so excess reagent does not distinguish these two ions. The initial precipitate colour is the useful observation.

Why do the precipitates have different colours?

Sodium hydroxide supplies hydroxide ions, \(\mathrm{OH^-}\), which combine with dissolved iron ions to form insoluble hydroxides.

For iron(II):

\[
\ce{Fe^2+(aq) + 2OH-(aq) -> Fe(OH)2(s)}
\]

For iron(III):

\[
\ce{Fe^3+(aq) + 3OH-(aq) -> Fe(OH)3(s)}
\]

Notice the different mole ratios.

One iron(II) ion requires two hydroxide ions, while one iron(III) ion requires three. This follows directly from balancing the ionic charges.

The state symbols also matter:

• \((aq)\) means dissolved in water.
• \((s)\) means a solid precipitate has formed.

These are net ionic equations. Sodium ions and other spectator ions are not included because they do not participate in forming the precipitate.

02The major complication: Iron(II) oxidises in air

Now return to the original puzzle.

You add sodium hydroxide to an iron(II) solution and observe a green precipitate. You leave the test tube on the bench while recording your results.

When you look again, the precipitate has become brown.

Should you change your identification to iron(III)?

No. The original green precipitate is evidence that iron(II) was present when the test was performed.

The change occurs because iron(II) hydroxide reacts with oxygen in the air. The iron is oxidised from the \(+2\) oxidation state to the \(+3\) oxidation state.

The simplified school-level equation is:

\[
\ce{4Fe(OH)2(s) + O2(g) + 2H2O(l) ->
4Fe(OH)3(s)}
\]

Oxygen is the oxidising agent. It causes iron(II) to lose electrons and become iron(III).

Why does this matter when interpreting results?

Consider these two observations:

Observation A: A green precipitate forms immediately after adding NaOH, then gradually turns brown.

Observation B: A red-brown precipitate forms immediately after adding NaOH.

These observations do not mean the same thing.

• Observation A supports the presence of iron(II) initially, followed by oxidation.
• Observation B supports the presence of iron(III) when the reagent was added.

The timing of the observation is part of the experimental evidence.

A useful exam habit is to distinguish between the initial colour and the colour after standing. If a question says that a green precipitate turns brown after several minutes, do not identify the original ion as iron(III) simply because the final colour is brown.

The simplified equation represents the oxidation-state change. In actual aged samples, the brown solid can contain hydrated iron(III) oxides and oxyhydroxides rather than being pure iron(III) hydroxide.

03Confirmatory tests: Don’t rely on colour alone

The sodium hydroxide test is useful, but it has limitations.

A sample containing both iron oxidation states may produce a mixture of precipitates. A sample containing iron(II) may also become contaminated with iron(III) through oxidation.

How can you check your identification independently?

Two useful reactions are the iron(III) thiocyanate test and the acidified potassium permanganate test.

Test A: Potassium thiocyanate for iron(III)

Add a few drops of potassium thiocyanate solution, \(\mathrm{KSCN(aq)}\), to a fresh aliquot of the unknown solution.

If iron(III) ions are present, a deep blood-red solution forms.

This happens because iron(III) ions react with thiocyanate ions to form a coloured complex ion:

\[
\ce{Fe^3+(aq) + SCN-(aq) <=>
[FeSCN]^2+(aq)}
\]

The red colour comes from the complex ion, not from a precipitate.

The expected observations are:

This test is particularly useful because it provides evidence for iron(III) without relying on the colour of an iron hydroxide precipitate.

However, there is an important limitation.

A solution containing iron(II) may produce a faint red colour if it already contains a small amount of iron(III). This can happen through oxidation or contamination. The Royal Society of Chemistry also highlights this issue in its iron oxidation-state experiment.

Therefore, a faint red result should not automatically be interpreted as proof that the original solution contained substantial iron(III).

Test B: Acidified potassium permanganate for iron(II)

You can also test for iron(II) using potassium permanganate, \(\mathrm{KMnO_4(aq)}\).

This is a redox reaction rather than a precipitation reaction.

Take a fresh aliquot of the unknown solution and acidify it with dilute sulfuric acid, \(\mathrm{H_2SO_4}\). Then add purple potassium permanganate solution dropwise.

If iron(II) ions are present, the purple colour disappears as the permanganate ions are reduced to manganese(II) ions.

At the same time, iron(II) is oxidised to iron(III).

The balanced net ionic equation is:

\[
\begin{aligned}
&\ce{MnO4-(aq) + 5Fe^2+(aq) + 8H+(aq)}\\
&\ce{-> Mn^2+(aq) + 5Fe^3+(aq) + 4H2O(l)}
\end{aligned}
\]

The expected observations are:

Why does iron(II) cause the colour to disappear?

Iron(II) acts as a reducing agent. It loses electrons and becomes iron(III). Permanganate gains those electrons and is reduced to manganese(II), which is very pale in solution.

One important precaution: use dilute sulfuric acid, not hydrochloric acid or nitric acid, to acidify this test. Chloride can interfere with strongly acidic permanganate reactions, while nitric acid can oxidise iron(II) before the test is completed.

Also, other reducing agents can decolourise permanganate. A positive result is therefore evidence for iron(II) only when other possible reducing substances have been excluded.

Comparing the three tests

The strongest identification uses more than one type of evidence.

For example, a green hydroxide precipitate combined with a negative thiocyanate result and positive acidified permanganate result strongly supports iron(II), provided the sample is sufficiently fresh and other interfering substances are absent.

Likewise, a red-brown hydroxide precipitate combined with a deep-red thiocyanate result and no permanganate decolourisation supports iron(III).

04Sources of interference and common mistakes

Before interpreting precipitation data, check whether anything could have changed the result.

Three mistakes worth avoiding

Mistake 1: Assuming a brown precipitate always proves iron(III).

A green iron(II) hydroxide precipitate can turn brown through oxidation. The initial observation and the time taken to record it matter.

Mistake 2: Assuming excess NaOH distinguishes iron(II) from iron(III).

Both iron hydroxides remain insoluble in excess sodium hydroxide. The colour, not dissolution in excess reagent, is the main distinguishing observation.

Mistake 3: Treating sodium hydroxide and ammonia as independent confirmation.

Aqueous ammonia can also produce green iron(II) hydroxide and red-brown iron(III) hydroxide. Both remain insoluble in excess ammonia under ordinary qualitative-analysis conditions.

Repeating the same type of hydroxide precipitation using a different base does not resolve the oxidation problem. A redox or complex-ion test provides more independent evidence.

For a broader method of selecting reagents, using separate aliquots, and interpreting multiple precipitation results, see How to Identify Two Unknown Ions Using Precipitation Tests.

05Worked examples

06Questions and solutions

Question 1

A fresh aqueous iron solution produces a red-brown precipitate when dilute sodium hydroxide is added dropwise.

The precipitate remains insoluble when excess NaOH is added.

Identify the most likely iron ion and write the balanced net ionic equation for the reaction.

Solution 1

The solution most likely contains iron(III) ions, \(\mathrm{Fe}^{3+}\).

Iron(III) ions react with hydroxide ions to form insoluble red-brown iron(III) hydroxide.

The balanced net ionic equation is:

\[
\ce{Fe^3+(aq) + 3OH-(aq) -> Fe(OH)3(s)}
\]

Three hydroxide ions are needed to balance the \(+3\) charge on the iron ion.

The precipitate remaining insoluble in excess NaOH is consistent with iron(III) hydroxide. However, the identification relies primarily on the initial red-brown colour and should be checked against other possible ions if the candidate list is not restricted to iron species.

Question 2

An iron-containing solution is divided into two equal portions.

Both portions are initially near neutral and contain no other metal ions.

• Portion A is tested immediately with sodium hydroxide. A green precipitate forms, which becomes brown at the surface after three minutes.
• Portion B is left exposed to air for 30 minutes before sodium hydroxide is added. A red-brown precipitate forms immediately.

Explain the difference between the two observations.

Can you conclude that all the iron(II) ions in portion B have been converted into iron(III) ions?

Solution 2

Iron(II) ions were present in the freshly tested solution, and exposure to air caused at least some iron(II) to be oxidised to iron(III).

Portion A was tested before significant oxidation had occurred. Its initial green precipitate is iron(II) hydroxide.

The gradual browning occurs because oxygen oxidises the precipitate after it forms.

In portion B, the solution was exposed to oxygen before sodium hydroxide was added. Some dissolved iron(II) could therefore have oxidised to iron(III) before the precipitation test.

The resulting red-brown precipitate indicates that iron(III) was present when NaOH was added.

The oxidation reaction can be represented by:

\[
\ce{4Fe^2+(aq) + O2(g) + 4H+(aq) ->
4Fe^3+(aq) + 2H2O(l)}
\]

This is a simplified representation of the oxidation in acidic solution; the actual oxygen-oxidation pathway depends on the solution conditions.

You cannot conclude that all iron(II) has been converted. The red-brown precipitate may indicate that iron(III) is predominant, but some iron(II) may remain.

The important distinction is that portion A demonstrates what was present before extended air exposure, whereas portion B shows the result after the solution’s composition has had time to change.

Question 3

A student receives an acidic iron solution containing:

• \(10.0\ \mathrm{mL}\) of solution.
• An iron(III) concentration of \(0.0100\ \mathrm{mol\,L^{-1}}\).
• A hydrochloric acid concentration of \(0.0200\ \mathrm{mol\,L^{-1}}\).

The student adds \(0.500\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide but observes no precipitate.

a) Calculate the amount of hydrogen ions initially present.

b) Calculate the amount of hydroxide ions added.

c) Explain why no precipitate forms.

d) Calculate the theoretical total volume of \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH needed to neutralise the acid and precipitate all the iron(III). Assume complete precipitation and ignore any additional excess reagent.

Solution 3

The absence of a precipitate does not mean iron(III) is absent. The added hydroxide ions are insufficient even to neutralise the acid.

Part a: Calculate the hydrogen ions present.

Hydrochloric acid is a strong acid, so assume it dissociates completely.

Use:

\[
n=cV
\]

\[
n(\mathrm{H^+})=0.0200\times0.0100
\]

\[
n(\mathrm{H^+})=2.00\times10^{-4}\ \mathrm{mol}
\]

Part b: Calculate the hydroxide ions added.

Convert the volume into litres:

\[
0.500\ \mathrm{mL}=0.000500\ \mathrm{L}
\]

Then:

\[
n(\mathrm{OH^-})=0.100\times0.000500
\]

\[
n(\mathrm{OH^-})=5.00\times10^{-5}\ \mathrm{mol}
\]

Part c: Explain the absence of a precipitate.

The amount of hydroxide added is smaller than the amount of hydrogen ions initially present.

The hydrogen ions react with hydroxide ions:

\[
\ce{H+(aq) + OH-(aq) -> H2O(l)}
\]

Therefore, all the added hydroxide ions are consumed neutralising the acid.

There are no hydroxide ions available in sufficient quantity to precipitate the iron(III) ions.

This explains why a negative precipitation result does not automatically rule out the presence of iron.

Part d: Calculate the total volume of NaOH required.

First calculate the amount of iron(III):

\[
n(\mathrm{Fe^{3+}})=0.0100\times0.0100
\]

\[
n(\mathrm{Fe^{3+}})=1.00\times10^{-4}\ \mathrm{mol}
\]

The precipitation equation is:

\[
\ce{Fe^3+ + 3OH- -> Fe(OH)3}
\]

Therefore, the hydroxide required for precipitation is:

\[
n(\mathrm{OH^-})=3(1.00\times10^{-4})
\]

\[
=3.00\times10^{-4}\ \mathrm{mol}
\]

The total hydroxide required must account for both the acid neutralisation and precipitation:

\[
\begin{aligned}
n_{\text{total}}(\mathrm{OH^-})
&=2.00\times10^{-4}\\
&\quad+3.00\times10^{-4}\\
&=5.00\times10^{-4}\ \mathrm{mol}
\end{aligned}
\]

Calculate the required volume:

\[
V=\frac{n}{c}
\]

\[
V=\frac{5.00\times10^{-4}}{0.100}
\]

\[
\boxed{V=5.00\ \mathrm{mL}}
\]

The theoretical volume is 5.00 mL. In a practical qualitative test, a small additional amount may be needed to ensure sufficient hydroxide is present.

The key lesson is that sample acidity must be considered before interpreting the absence of a precipitate.

Question 4

An unknown iron solution may contain iron(II), iron(III), or both.

Three fresh aliquots are tested promptly, with negligible oxidation during sample preparation.

The following results are obtained:

Assume there are no other redox-active ions or complexing substances present.

a) Determine which iron ions are present.

b) Write the net ionic equation for the thiocyanate test.

c) Write the balanced redox equation for the permanganate test.

d) Explain why the sodium hydroxide observation alone would be insufficient to establish the composition.

Solution 4

Both iron(II) and iron(III) ions are present in the solution when the tests are performed.

The two confirmatory tests provide evidence for different oxidation states.

Part a: Identify the ions.

The deep blood-red colour with KSCN indicates iron(III).

The disappearance of purple permanganate indicates iron(II), which reduces permanganate ions.

Since both results are positive, both oxidation states are present under the stated assumptions.

Part b: Write the thiocyanate equation.

Iron(III) ions combine with thiocyanate ions to form a red complex:

\[
\ce{Fe^3+(aq) + SCN-(aq) <=>
[FeSCN]^2+(aq)}
\]

The colour change is due to complex-ion formation.

Part c: Write the permanganate equation.

In acidic solution, iron(II) reduces permanganate to manganese(II):

\[
\begin{aligned}
&\ce{MnO4-(aq) + 5Fe^2+(aq) + 8H+(aq)}\\
&\ce{-> Mn^2+(aq) + 5Fe^3+(aq) + 4H2O(l)}
\end{aligned}
\]

Iron(II) loses electrons, while manganese in permanganate gains electrons.

Part d: Explain the limitation of the precipitation result.

The initial green precipitate supports the presence of iron(II), but its subsequent browning can occur through oxidation by atmospheric oxygen.

If both iron ions are present, their hydroxides can also produce a mixture of colours.

Consequently, the final brown appearance cannot establish whether the original solution contained only iron(III), only iron(II), or both.

The thiocyanate and permanganate tests resolve this ambiguity because they investigate different chemical properties of the iron ions.

Question 5

An iron solution is collected at 9:00 am and left uncovered on a laboratory bench for 25 minutes.

At 9:25 am, the solution is divided into three fresh aliquots.

The following results are obtained:

• Adding NaOH produces a red-brown precipitate immediately.
• Adding KSCN produces a deep blood-red solution.
• Acidifying a third aliquot with dilute sulfuric acid and adding KMnO₄ causes the purple colour to disappear.

An analyst concludes:

“The original solution must have contained both iron(II) and iron(III) ions at 9:00 am because both oxidation states were detected at 9:25 am.”

Evaluate this conclusion.

Then design an experiment that would distinguish between:

• A solution that originally contained only iron(II), with some iron(II) oxidised during storage.
• A solution that already contained both iron(II) and iron(III) when collected.

State the expected observations for each possibility. Assume there are no other redox-active ions or complexing agents, and that the iron concentrations are sufficient for the tests to be detected.

Solution 5

The analyst’s conclusion is not justified.

The results establish that both iron(II) and iron(III) are detectable at 9:25 am. They do not establish which oxidation states were present at 9:00 am.

This is a problem of experimental timing.

Step 1: Interpret the results at 9:25 am.

The deep red thiocyanate result demonstrates the presence of iron(III).

The disappearance of purple permanganate demonstrates the presence of iron(II), assuming no other reducing agents are present.

Therefore, both oxidation states are present when the tests are performed.

The immediate red-brown hydroxide precipitate is consistent with iron(III), but it does not establish the original composition independently.

Step 2: Consider the competing explanation.

Suppose the solution contained only iron(II) at 9:00 am.

While the solution is exposed to air, oxygen can oxidise some iron(II) to iron(III).

The reaction produces a solution containing both oxidation states, provided some iron(II) remains unoxidised.

At 9:25 am, this solution could produce exactly the three observations recorded by the analyst:

• Red-brown iron hydroxide.
• A deep-red thiocyanate complex.
• Decolourisation of acidified permanganate.

The same results could also arise from a solution that originally contained both ions.

The two explanations are therefore consistent with the available evidence.

Step 3: Design a controlled experiment.

The best approach is to prevent or minimise oxidation during sample storage.

At the moment of collection, divide the sample into two portions.

• Portion A is transferred immediately into an airtight vial with minimal air space, using oxygen-excluded handling conditions.
• Portion B is deliberately left exposed to air as a comparison.

Test portion A immediately, using three separate fresh aliquots:

1. Add NaOH to the first aliquot.
2. Add KSCN to the second aliquot.
3. Acidify the third aliquot with dilute H₂SO₄ and add KMnO₄.

After 25 minutes, perform the same three tests on portion B.

Step 4: Predict the results for each possibility.

If the original solution contained only iron(II):

The immediately tested, oxygen-excluded portion should show:

• A green precipitate with NaOH.
• No strong blood-red colour with KSCN.
• Disappearance of purple permanganate.

The air-exposed portion may develop detectable iron(III) during storage. Its KSCN test may then produce a red colour, while its permanganate test can remain positive if some iron(II) is still present.

If the original solution already contained both iron(II) and iron(III):

The immediately tested, oxygen-excluded portion should show:

• Evidence of iron(II) and potentially a mixed-colour hydroxide precipitate with NaOH.
• A blood-red colour with KSCN.
• Disappearance of purple permanganate.

The air-exposed portion may become more iron(III)-rich as additional oxidation occurs.

Step 5: Explain what the comparison establishes.

The crucial evidence is whether iron(III) is already detectable in the oxygen-protected sample immediately after collection.

A positive thiocyanate test in that sample, alongside a positive iron(II) permanganate test, supports the conclusion that both oxidation states were present at collection.

If the oxygen-protected sample instead gives a green hydroxide precipitate, no strong thiocyanate response, and a positive permanganate test, the results support an initially iron(II)-only sample, with iron(III) developing during air exposure.

A negative thiocyanate result cannot prove that absolutely no iron(III) was present. It only indicates that iron(III) was not detected under the test conditions.

This is the important distinction: a chemical test identifies what is present when the test is performed, not necessarily what was present when the sample was originally collected.

07Where to go next

Once you can distinguish iron(II) from iron(III), the natural next step is redox stoichiometry.

The acidified permanganate reaction is particularly useful because it can be used not just to detect iron(II), but also to calculate its concentration through a redox titration.

The same electron-transfer reaction that explains the disappearance of the purple colour becomes the basis for a quantitative chemistry calculation.