Chemistry · Redox reactions and electrochemistry · Grade 8

Redox reactions

Why does permanganate go colourless in acid but turn green in alkali?

The situation

The “margansovka” sold at the pharmacy is potassium permanganate KMnO₄: its crystals are dark violet, almost black; dissolved in water they give a beautiful violet-pink solution. We pour some of this solution into three test tubes, add sulfuric acid to one, water to the second and alkali to the third, and add sodium sulfite drop by drop to each. The first becomes colourless, a brown precipitate forms in the second, the third turns green. Why does one oxidising agent give three different results, and how many electrons does it take up in each case?

Topic

Oxidation state and redox reactions. Permanganate (KMnO₄): decolourised in acid, a brown precipitate in water, green in alkali — one oxidising agent, three different outcomes

Key ideas

Oxidation state
Electron balance
Effect of the medium

Formulas

\sum_{i} n_{i} x_{i} = 0 (molecule), \sum_{i} n_{i} x_{i} = q (ion)

Sum of oxidation states: zero in a molecule, the charge of the ion in an ion.

Fe^{0} - 2 e^{-} \overset{}{\to} Fe^{+ 2} Cu^{+ 2} + 2 e^{-} \overset{}{\to} Cu^{0}

Oxidation is the loss of electrons (the oxidation state increases), reduction is the gain of electrons (the oxidation state decreases).

MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \overset{}{\to} Mn^{2 +} + 4 H_{2} O

Half-equation (acidic medium): O is balanced with H₂O, H with H⁺, charge with electrons.

2 KMnO_{4} + 5 Na_{2} SO_{3} + 3 H_{2} SO_{4} \overset{}{\to} 2 MnSO_{4} + 5 Na_{2} SO_{4} + K_{2} SO_{4} + 3 H_{2} O

In an acidic medium KMnO₄ is decolourised (Mn²⁺).

E = \frac{M}{n (e^{-})}

Equivalent mass of the oxidising agent: for KMnO₄, 158/5 = 31.6; 158/3 = 52.7; 158/1 = 158 g/mol.

Theory

Oxidation state (§18)

The oxidation state is the conventional charge that would belong to an atom if all the bonds in the compound were ionic. Rules: 0 in an element; fluorine −1; oxygen usually −2 (−1 in peroxides, +2 in OF₂); hydrogen usually +1 (−1 in metal hydrides); alkali metals +1, alkaline-earth metals +2, Al +3. In a molecule the sum of all the states is zero, in an ion it is the charge of the ion. Examples: S is +6 in H₂SO₄, Mn is +7 in KMnO₄, Cr is +6 in K₂Cr₂O₇, and in NH₄NO₃ there are two kinds of nitrogen: −3 (NH₄⁺) and +5 (NO₃⁻). The oxidation state is not the same as valency: in methanol, formaldehyde and formic acid carbon is tetravalent, but its states are −2, 0 and +2 (−4 in CH₄, +4 in CO₂). On the scales in the scene the positive contributions are on the left pan and the negative ones on the right — if the sum is right, the scales balance.

Redox reactions (§19)

Reactions in which oxidation states change are redox reactions. Oxidation is the loss of electrons (the oxidation state increases), reduction is the gain of electrons (the oxidation state decreases); the two always go together. A substance that gives up electrons is a reducing agent (it is itself oxidised): metals, H₂, C, CO, H₂S, sulfites. A substance that takes up electrons is an oxidising agent (it is itself reduced): O₂, Cl₂, HNO₃, KMnO₄, K₂Cr₂O₇. An atom in its highest state (Mn⁺⁷, S⁺⁶ in H₂SO₄) can only be an oxidising agent, one in its lowest (S⁻², N⁻³) only a reducing agent, and those in between (S⁺⁴) can be either.

The electron balance method (§20)

Writing an equation: 1) find the atoms whose oxidation state has changed; 2) write the electron equations: $Cu^{0} - 2 e^{-} \overset{}{\to} Cu^{+ 2}$ , $N^{+ 5} + 3 e^{-} \overset{}{\to} N^{+ 2}$ ; 3) the electrons given and taken must be equal — find the lowest common multiple (6); the multipliers (3 and 2) are the coefficients in front of the oxidising and reducing agents; 4) balance the remaining atoms (the part that forms the salt, then H and O): $3 Cu + 8 HNO_{3} \overset{}{\to} 3 Cu (NO_{3})_{2} + 2 NO ↑ + 4 H_{2} O$ — of the 8 HNO₃ only 2 are oxidising agents. Always check the number of atoms and the electrons given and taken.

The half-equation (ion–electron) method (§28)

More convenient for reactions in solution: the real particles (MnO₄⁻, SO₃²⁻) are written and each half-equation is balanced separately. In an acidic medium excess O is bound by H⁺ (forming H₂O), and missing O is supplied by H₂O (releasing H⁺). In a neutral medium only H₂O is written on the left, and H⁺ or OH⁻ forms on the right. In an alkaline medium, if O is missing, 2OH⁻ supplies it and H₂O is released; excess O is bound by H₂O to give 2OH⁻. Then the charge is balanced with electrons, the numbers of electrons are equalised and the half-equations are added. It is not necessary to know the oxidation states.

Effect of the medium and equivalent mass (§29–30, lab work)

The reduction product of KMnO₄ depends on the medium: acidic — Mn²⁺ (almost colourless, 5 e⁻), neutral — MnO₂ (brown precipitate, 3 e⁻), alkaline — MnO₄²⁻ (green, 1 e⁻). The equivalent mass of an oxidising agent is the mass per electron: $E = M / n (e^{-})$ , for KMnO₄ 31.6; 52.7 and 158 g/mol. In the experiment: each test tube has 2 mL of 0.005 mol/L KMnO₄ (0.010 mmol) + 1 mL of the medium; 0.05 mol/L Na₂SO₃ is added dropwise (1 drop = 0.05 mL = 0.0025 mmol) until the colour change is complete. Each SO₃²⁻ gives 2 electrons: $n (e^{-}) = 2 n (S O_{3}^{2 -})$ , so $n (e^{-}) / n (M n O_{4}^{-})$ and $E$ are found from the number of drops. Safety: wear safety goggles; KMnO₄ stains skin and clothes and is a strong oxidising agent; be careful with acid and alkali, and pour acid into water; experiments that release Cl₂ and NO₂ only in a fume cupboard.

About the model

The solutions are taken to be ideal, and the reactions to go instantly and completely as each drop falls; the oxidation of Na₂SO₃ by air is not taken into account. In an alkaline medium, if there is an excess of sulfite, the manganate can slowly be reduced further to MnO₂ — the model shows the first stage of the colour change. The colours of the solutions are approximate. The scales are an illustration for showing the sum (charge is not “weight”); the ion models and the flying electrons are diagrams; MnO₂ is a solid, with no separate molecule.

Did you know?

A very dilute (pale pink) solution of potassium permanganate is used in medicine as an antiseptic: it is a strong oxidising agent and so kills microbes. A concentrated solution, however, burns the skin and leaves brown stains.
In 2020 a new plant producing up to 500 thousand tonnes of nitric acid a year was commissioned at Navoiyazot. Nitric acid production begins with the oxidation of ammonia over a catalyst (4NH₃ + 5O₂ → 4NO + 6H₂O), and most of the acid is converted into ammonium nitrate fertiliser.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Electrolysis and corrosion — Grade 9
Galvanic cells and corrosion — Year 1