Transition metals exhibit variable oxidation states in their compounds. Changes in oxidation state are accompanied by changes in the d-electron configuration of the metal ion, which directly alters the wavelengths of light absorbed and results in distinctive colour changes. You must learn the redox chemistry, balanced half-equations, and colour transitions for key processes involving manganese, chromium, and iron.
A half-equation is a balanced chemical equation showing either the oxidation or the reduction process of a redox reaction separately, with the gained or lost electrons written explicitly.
An oxidation state (or oxidation number) is a value assigned to an atom in a molecule or ion that represents the hypothetical charge the atom would carry if all bonds to different elements were completely ionic.
An oxidising agent is a chemical reactant that oxidises another substance by accepting electrons, causing its own oxidation state to decrease (it undergoes reduction).
1. Redox Chemistry of Iron
Iron commonly exists in the \(+2\) and \(+3\) oxidation states in aqueous solution. Yellow-brown aqueous iron(III) ions, \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\), can be reduced to pale green iron(II) ions, \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\), by adding a suitable reducing agent such as solid zinc under acidic conditions:
The colour change observed is from yellow/brown to pale green, accompanied by the effervescence of hydrogen gas (as zinc also reacts with the acid) and the dissolution of the grey zinc solid.
Conversely, pale green iron(II) can be oxidised back to yellow-brown iron(III) by reaction with an oxidising agent, such as acidified potassium manganate(VII) solution or hydrogen peroxide (\(\text{H}_2\text{O}_2\)):
2. Dichromate(VI) Reduction
Chromium is stable in the \(+6\) oxidation state as the orange dichromate(VI) ion, \(\text{Cr}_2\text{O}_7^{2-}\). In strongly acidic solutions, dichromate(VI) acts as a powerful oxidising agent and is reduced to green chromium(III) ions, \(\text{Cr}^{3+}\). The half-equation for this reduction is:
The characteristic colour change is orange to green. This reaction is commonly used in organic chemistry to oxidise primary and secondary alcohols.
3. Manganate(VII) Reduction
Manganese(VII) in the form of the deep purple manganate(VII) ion, \(\text{MnO}_4^-\), is a widely used oxidising agent in redox titrations. In acidic solution (typically acidified with dilute sulfuric acid), it accepts five electrons to form manganese(II) ions, \(\text{Mn}^{2+}\). The half-equation is:
The colour change observed is from purple to colourless (or extremely pale pink, which appears colourless in dilute titration solutions). This dramatic change makes potassium manganate(VII) self-indicating in titrations.
🔑 Key Principle: Summary of Redox Colour Changes
Make sure you memorise the following standard species and their colours in acidic solution:
- \(\text{MnO}_4^-\) (purple, \(+7\)) \(\rightarrow\) \(\text{Mn}^{2+}\) (colourless/pale pink, \(+2\))
- \(\text{Cr}_2\text{O}_7^{2-}\) (orange, \(+6\)) \(\rightarrow\) \(\text{Cr}^{3+}\) (green, \(+3\))
- \(\text{Fe}^{3+}\) (yellow/brown, \(+3\)) \(\rightarrow\) \(\text{Fe}^{2+}\) (pale green, \(+2\))
- \(\text{I}_2\) (brown, \(0\)) \(\rightleftharpoons\) \(\text{I}^-\) (colourless, \(-1\))
Oxidation State Ladders
The diagram below displays the key oxidation states for manganese, chromium, and iron, showing the corresponding chemical formulas and colours at each level.
Transition Metal Colour Palette
Use the card below as a quick reference for the characteristic colours of common aqueous complexes and substituted species in the syllabus.
A common error is producing an equation that balances for elements but not for charge. For example, in the combined manganate(VII) and iron(II) equation: \[ \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) + 5\text{Fe}^{3+}(\text{aq}) \] Let's check the charge balance:
- Left side: \((-1) + 8(+1) + 5(+2) = -1 + 8 + 10 = +17\)
- Right side: \((+2) + 0 + 5(+3) = 2 + 15 = +17\)
Step 1: Write down both half-equations
Reduction: \[ \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \]
Oxidation: \[ \text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^- \]
Step 2: Equalise the electrons
The reduction equation requires 5 electrons. Multiply the oxidation equation by 5:
\[ 5\text{Fe}^{2+}(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + 5\text{e}^- \]
Step 3: Combine and cancel electrons
\[ \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) + 5\text{Fe}^{3+}(\text{aq}) \]
Step 4: Observations
The purple manganate(VII) solution is decolourised (turns colourless / very pale pink). The pale green iron(II) solution turns yellow/brown due to the formation of iron(III) ions.
Step 1: Write down both half-equations
Reduction: \[ \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{e}^- \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l}) \]
Oxidation: \[ \text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^- \]
Step 2: Equalise the electrons
The reduction equation requires 6 electrons. Multiply the oxidation equation by 6:
\[ 6\text{Fe}^{2+}(\text{aq}) \rightarrow 6\text{Fe}^{3+}(\text{aq}) + 6\text{e}^- \]
Step 3: Combine and cancel electrons
\[ \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 6\text{Fe}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l}) \]
Get flashcards and quizzes in ChemEasy, or plan your revision with ChemPlan IB.