Chapter 4 · The d- and f-Block Elements

4.2 Important Compounds: Potassium Dichromate and Potassium Permanganate

Transition metals in high oxidation states hide inside oxoanions. Orange dichromate and purple permanganate are the two most important: how they are made, why their colours and structures are what they are, and how many electrons each one takes in acidic, neutral and alkaline solution.

An old roadside breath-test tube held orange crystals that turned green when a driver who had been drinking blew through them: dichromate, $\mathrm{Cr_{2}O_{7}^{2-}}$, was being reduced to $\mathrm{Cr^{3+}}$ by ethanol. A pinch of purple potassium permanganate turns a bucket of well water pink and pulls dissolved iron out as a brown sludge. In the school laboratory, a burette of permanganate titrates iron(II) or oxalate solutions with no indicator at all: the first drop in excess stays pink. These are everyday uses of two compounds in which chromium ($+6$) and manganese ($+7$) are in their highest oxidation states, the states we met at the top of the "staircase" in Section 4.1.

Such high oxidation states are not found as simple cations. A $\mathrm{Mn^{7+}}$ ion would be impossibly small and highly charged. Instead the metal is bonded to oxygen atoms in an oxide or an oxoanion, where oxygen's ability to form multiple bonds stabilises it. This section first surveys the oxides of the 3d metals and how their acid–base character changes with oxidation state. It then follows potassium dichromate and potassium permanganate from their ores to the titration flask, and shows how to balance every one of their redox reactions by counting electrons.

Oxides and Oxoanions of the 3d Metals

The 3d metals react with oxygen at high temperatures to form oxides. Every metal of the series except scandium forms an ionic oxide $\mathrm{MO}$. Higher oxides are known for most of them, and the highest oxidation state in an oxide equals the group number from $\mathrm{Sc_{2}O_{3}}$ (group 3) to $\mathrm{Mn_{2}O_{7}}$ (group 7). Beyond group 7, no oxide of iron higher than $\mathrm{Fe_{2}O_{3}}$ is known, although iron(VI) occurs in ferrate ions, $\mathrm{FeO_{4}^{2-}}$, formed in alkaline solution and easily decomposed.

📐 Rule 4.6 — Oxidation state and the character of oxides
  1. As the oxidation state of the metal rises, its oxide becomes less ionic and more covalent. $\mathrm{Mn_{2}O_{7}}$ is a covalent green oil; $\mathrm{CrO_{3}}$ and $\mathrm{V_{2}O_{5}}$ have low melting points.
  2. At the same time the oxide becomes more acidic: the lowest oxides are basic, the intermediate ones amphoteric and the highest acidic.
    Metalbasicamphotericacidic
    Cr$\mathrm{CrO}$ (+2)$\mathrm{Cr_{2}O_{3}}$ (+3)$\mathrm{CrO_{3}}$ (+6)
    Mn$\mathrm{MnO}$ (+2)$\mathrm{MnO_{2}}$ (+4)$\mathrm{Mn_{2}O_{7}}$ (+7)
    V$\mathrm{VO}$, $\mathrm{V_{2}O_{3}}$$\mathrm{V_{2}O_{4}}$, $\mathrm{V_{2}O_{5}}$ (mainly acidic)—
  3. Acidic oxides give oxoacids and oxoanions: $\mathrm{Mn_{2}O_{7}}$ gives $\mathrm{HMnO_{4}}$; $\mathrm{CrO_{3}}$ gives $\mathrm{H_{2}CrO_{4}}$ and $\mathrm{H_{2}Cr_{2}O_{7}}$; $\mathrm{V_{2}O_{5}}$ gives $\mathrm{VO_{4}^{3-}}$ salts with alkalis and $\mathrm{VO_{2}^{+}}$ salts with acids. $\mathrm{V_{2}O_{4}}$ dissolves in acids to give $\mathrm{VO^{2+}}$ salts.

The reason is the same as for the non-metals of Class 11: a small, highly charged metal centre pulls electron density from oxygen, so its oxide behaves like a non-metal oxide. Besides the oxoanions, high states also appear in oxocations such as $\mathrm{VO_{2}^{+}}$ (V(V)), $\mathrm{VO^{2+}}$ (V(IV)) and $\mathrm{TiO^{2+}}$ (Ti(IV)).

📘 Definition 4.6 — Oxoanion
An oxoanion is a negative ion in which a central atom is bonded only to oxygen atoms, for example $\mathrm{CrO_{4}^{2-}}$, $\mathrm{Cr_{2}O_{7}^{2-}}$, $\mathrm{MnO_{4}^{2-}}$, $\mathrm{MnO_{4}^{-}}$ and $\mathrm{VO_{4}^{3-}}$. For the 3d metals the central atom is usually in a high oxidation state ($+5$ to $+7$), and many of these oxoanions are intensely coloured, strong oxidising agents.
✏️ Example 1 — Acid, base or both?
(a) Classify $\mathrm{MnO}$, $\mathrm{Cr_{2}O_{3}}$, $\mathrm{CrO_{3}}$ and $\mathrm{Mn_{2}O_{7}}$ as acidic, basic or amphoteric, giving the oxidation state of the metal in each. (b) Write balanced equations for the reactions of $\mathrm{Cr_{2}O_{3}}$ with hydrochloric acid and with fused sodium hydroxide, and of $\mathrm{V_{2}O_{5}}$ with aqueous sodium hydroxide (giving sodium vanadate, $\mathrm{Na_{3}VO_{4}}$).
(a) $\mathrm{MnO}$: Mn $+2$, basic. $\mathrm{Cr_{2}O_{3}}$: Cr $+3$, amphoteric. $\mathrm{CrO_{3}}$: Cr $+6$, acidic. $\mathrm{Mn_{2}O_{7}}$: Mn $+7$ ($2x-14=0$), strongly acidic. The higher the oxidation state, the more acidic the oxide.
(b) With acid $\mathrm{Cr_{2}O_{3}(s)+6HCl(aq)\rightarrow2CrCl_{3}(aq)+3H_{2}O(l)}$: the oxide acts as a base.
(b) With alkali $\mathrm{Cr_{2}O_{3}(s)+2NaOH(l)\rightarrow2NaCrO_{2}(s)+H_{2}O(g)}$ on fusion (sodium chromite): the oxide acts as an acid. Together the two reactions show that $\mathrm{Cr_{2}O_{3}}$ is amphoteric.
(b) V₂O₅ $\mathrm{V_{2}O_{5}(s)+6NaOH(aq)\rightarrow2Na_{3}VO_{4}(aq)+3H_{2}O(l)}$. Check: V 2, Na 6, H 6, O $5+6=11=8+3$ ✓.

Potassium Dichromate, $\mathrm{K_{2}Cr_{2}O_{7}}$

Preparation. Dichromates are made from chromite ore, $\mathrm{FeCr_{2}O_{4}}$ (that is, $\mathrm{FeO\cdot Cr_{2}O_{3}}$). The process has three stages.

  1. Oxidative fusion. The powdered ore is fused with sodium carbonate (or sodium hydroxide) in free access of air. Chromium(III) is oxidised to chromium(VI), giving yellow sodium chromate:
    $$\begin{aligned}&\mathrm{4FeCr_{2}O_{4}(s)+8Na_{2}CO_{3}(s)+7O_{2}(g)}\\&\quad\mathrm{\rightarrow8Na_{2}CrO_{4}(s)+2Fe_{2}O_{3}(s)+8CO_{2}(g)}\end{aligned}$$
  2. Acidification. The chromate is extracted with water, filtered, and acidified with sulphuric acid; orange sodium dichromate crystallises:
    $$\mathrm{2Na_{2}CrO_{4}(aq)+2H^{+}(aq)\rightarrow Na_{2}Cr_{2}O_{7}(aq)+2Na^{+}(aq)+H_{2}O(l)}$$
  3. Conversion to the potassium salt. Sodium dichromate is more soluble than potassium dichromate, so adding potassium chloride to a hot concentrated solution gives orange crystals of $\mathrm{K_{2}Cr_{2}O_{7}}$ on cooling:
    $$\mathrm{Na_{2}Cr_{2}O_{7}(aq)+2KCl(aq)\rightarrow K_{2}Cr_{2}O_{7}(s)+2NaCl(aq)}$$
📐 Rule 4.7 — The chromate–dichromate equilibrium

In water, yellow chromate and orange dichromate are interconverted by changing the pH:

$$\mathrm{2CrO_{4}^{2-}(aq)+2H^{+}(aq)\rightleftharpoons Cr_{2}O_{7}^{2-}(aq)+H_{2}O(l)}$$
$$\mathrm{Cr_{2}O_{7}^{2-}(aq)+2OH^{-}(aq)\rightleftharpoons2CrO_{4}^{2-}(aq)+H_{2}O(l)}$$

Adding acid shifts the equilibrium to the right (orange dichromate), adding alkali to the left (yellow chromate), exactly as Le Chatelier's principle predicts. Chromium stays in the $+6$ state throughout: this is an acid–base reaction, not a redox reaction.

Structures. The chromate ion is tetrahedral, with chromium at the centre. The dichromate ion consists of two $\mathrm{CrO_{4}}$ tetrahedra sharing one corner oxygen; the $\mathrm{Cr{-}O{-}Cr}$ bond angle is about $126^{\circ}$. The bridging $\mathrm{Cr{-}O}$ bonds are longer than the six terminal ones, which have partial double-bond character.

Interactive 1 — Chromate or Dichromate? The pH Switch

Set the pH and the total chromium(VI) concentration. The curve shows the fraction of chromium present as orange $\mathrm{Cr_{2}O_{7}^{2-}}$ (the rest is yellow $\mathrm{CrO_{4}^{2-}}$), from the equilibrium $\mathrm{2CrO_{4}^{2-}+2H^{+}\rightleftharpoons Cr_{2}O_{7}^{2-}+H_{2}O}$ with $K\approx3\times10^{14}$ at $298\ \mathrm{K}$. The flask mixes the two colours. Press Play to add acid and alkali in turn. This is a simplified two-species model (it ignores $\mathrm{HCrO_{4}^{-}}$), but it shows the key point: a few pH units switch the colour completely, and diluting the solution favours chromate.

4.00
$[\mathrm{CrO_{4}^{2-}}]$ (mol/L)—
$[\mathrm{Cr_{2}O_{7}^{2-}}]$ (mol/L)—
Cr as dichromate—
half-way pH—
📐 Rule 4.8 — Dichromate as an oxidising agent

In acid solution dichromate is a strong oxidising agent. Each chromium is reduced from $+6$ to $+3$, so one dichromate ion takes six electrons, and the orange solution turns green ($\mathrm{Cr^{3+}}$):

$$\mathrm{Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow2Cr^{3+}(aq)+7H_{2}O(l)}$$

with $E^{\circ}=+1.33\ \mathrm{V}$. The electrons come from the reducing agent; typical half-equations are

$$\mathrm{6I^{-}(aq)\rightarrow3I_{2}(aq)+6e^{-}}$$
$$\mathrm{6Fe^{2+}(aq)\rightarrow6Fe^{3+}(aq)+6e^{-}}$$
$$\mathrm{3Sn^{2+}(aq)\rightarrow3Sn^{4+}(aq)+6e^{-}}$$
$$\mathrm{3H_{2}S(g)\rightarrow6H^{+}(aq)+3S(s)+6e^{-}}$$

Adding a reduction half-equation and an oxidation half-equation that involve the same number of electrons gives the full ionic equation. Potassium dichromate is not hygroscopic and can be weighed out pure, so it is used as a primary standard in volumetric analysis (for example to estimate iron(II)). It is also used in leather tanning and, in organic chemistry, to oxidise alcohols (Chapter 7).

✏️ Example 2 — Following chromium through its manufacture
(a) Find the oxidation states of iron and chromium in chromite, $\mathrm{FeCr_{2}O_{4}}$, and of chromium in $\mathrm{Na_{2}CrO_{4}}$ and $\mathrm{K_{2}Cr_{2}O_{7}}$. (b) In the fusion step, which elements are oxidised and which is reduced? Show that the electrons balance for the equation as written. (c) Why is the acidification step not a redox reaction?
(a) $\mathrm{FeCr_{2}O_{4}}=\mathrm{FeO\cdot Cr_{2}O_{3}}$: Fe $+2$, Cr $+3$ (check $2+2(3)-8=0$). $\mathrm{Na_{2}CrO_{4}}$: $2+x-8=0$, Cr $+6$. $\mathrm{K_{2}Cr_{2}O_{7}}$: $2+2x-14=0$, Cr $+6$.
(b) Who changes Cr: $+3\rightarrow+6$ (oxidised); Fe: $+2\rightarrow+3$ in $\mathrm{Fe_{2}O_{3}}$ (oxidised); O in $\mathrm{O_{2}}$: $0\rightarrow-2$ (reduced). Na, C stay $+1$, $+4$.
(b) Electron count Electrons lost: 8 Cr × 3 = 24, plus 4 Fe × 1 = 4; total 28. Electrons gained: 7 $\mathrm{O_{2}}$ × 4 = 28 ✓. That is why the equation needs exactly $\mathrm{7O_{2}}$ for $\mathrm{4FeCr_{2}O_{4}}$.
(c) Chromium is $+6$ both in $\mathrm{CrO_{4}^{2-}}$ and in $\mathrm{Cr_{2}O_{7}^{2-}}$, and no other element changes its oxidation state. The acid only removes oxide ions as water (two chromates condense, sharing one oxygen): an acid–base condensation.
✏️ Example 3 — Dichromate in action
Write balanced ionic equations for the oxidation, by acidified dichromate, of (a) iron(II) ions, (b) hydrogen sulphide, (c) tin(II) ions. (d) What mass of $\mathrm{K_{2}Cr_{2}O_{7}}$ ($294\ \mathrm{g\,mol^{-1}}$) is needed to oxidise $0.0300\ \mathrm{mol}$ of $\mathrm{Fe^{2+}}$?
(a) Six $\mathrm{Fe^{2+}}$ supply the six electrons: $$\begin{aligned}&\mathrm{Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6Fe^{2+}(aq)}\\&\quad\mathrm{\rightarrow2Cr^{3+}(aq)+6Fe^{3+}(aq)+7H_{2}O(l)}\end{aligned}$$ Charge: $-2+14+12=+24$ and $+6+18=+24$ ✓.
(b) Three $\mathrm{H_{2}S}$ give $6e^{-}$ and $6\mathrm{H^{+}}$, which cancel six of the fourteen $\mathrm{H^{+}}$: $$\begin{aligned}&\mathrm{Cr_{2}O_{7}^{2-}(aq)+8H^{+}(aq)+3H_{2}S(g)}\\&\quad\mathrm{\rightarrow2Cr^{3+}(aq)+3S(s)+7H_{2}O(l)}\end{aligned}$$ A pale yellow precipitate of sulphur forms.
(c) $$\begin{aligned}&\mathrm{Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+3Sn^{2+}(aq)}\\&\quad\mathrm{\rightarrow2Cr^{3+}(aq)+3Sn^{4+}(aq)+7H_{2}O(l)}\end{aligned}$$
(d) From (a), $1\ \mathrm{mol}$ $\mathrm{Cr_{2}O_{7}^{2-}}$ oxidises $6\ \mathrm{mol}$ $\mathrm{Fe^{2+}}$: $n=\dfrac{0.0300}{6}=5.00\times10^{-3}\ \mathrm{mol}$; mass $=5.00\times10^{-3}\times294=1.47\ \mathrm{g}$.

Potassium Permanganate, $\mathrm{KMnO_{4}}$

Preparation. Manganese(IV) oxide, from the ore pyrolusite, is first oxidised to green potassium manganate, $\mathrm{K_{2}MnO_{4}}$ (Mn $+6$), by fusing it with potassium hydroxide and an oxidising agent such as air or potassium nitrate:

$$\mathrm{2MnO_{2}(s)+4KOH(l)+O_{2}(g)\rightarrow2K_{2}MnO_{4}(s)+2H_{2}O(g)}$$

The dark green melt is extracted with water. In neutral or acidic solution the manganate ion disproportionates: some of it is oxidised to purple permanganate and some reduced to manganese(IV) oxide.

$$\mathrm{3MnO_{4}^{2-}(aq)+4H^{+}(aq)\rightarrow2MnO_{4}^{-}(aq)+MnO_{2}(s)+2H_{2}O(l)}$$

Commercially, the alkaline manganate solution is instead oxidised electrolytically at the anode, so that none of the manganese is lost as $\mathrm{MnO_{2}}$:

$$\mathrm{MnO_{4}^{2-}(aq)\xrightarrow{\text{electrolytic oxidation}}MnO_{4}^{-}(aq)+e^{-}}$$

In the laboratory, a manganese(II) salt can be oxidised directly to permanganate by peroxodisulphate:

$$\begin{aligned}&\mathrm{2Mn^{2+}(aq)+5S_{2}O_{8}^{2-}(aq)+8H_{2}O(l)}\\&\quad\mathrm{\rightarrow2MnO_{4}^{-}(aq)+10SO_{4}^{2-}(aq)+16H^{+}(aq)}\end{aligned}$$
📘 Definition 4.7 — Disproportionation
A disproportionation reaction is a redox reaction in which one species is simultaneously oxidised and reduced, so that an element in one oxidation state ends up in a higher and a lower state. Examples: $\mathrm{MnO_{4}^{2-}}$ (Mn $+6$) $\rightarrow\mathrm{MnO_{4}^{-}}$ ($+7$) $+\ \mathrm{MnO_{2}}$ ($+4$); $\mathrm{Cu^{+}}$ ($+1$) $\rightarrow\mathrm{Cu^{2+}}$ ($+2$) $+\ \mathrm{Cu}$ ($0$) (Section 4.1).

Properties and structure. Potassium permanganate forms dark purple, almost black, crystals, isostructural with $\mathrm{KClO_{4}}$. It is only moderately soluble in water ($6.4\ \mathrm{g}$ per $100\ \mathrm{g}$ of water at $293\ \mathrm{K}$). Heated to $513\ \mathrm{K}$ it decomposes and releases oxygen:

$$\mathrm{2KMnO_{4}(s)\xrightarrow{513\ \mathrm{K}}K_{2}MnO_{4}(s)+MnO_{2}(s)+O_{2}(g)}$$

Both the manganate and permanganate ions are tetrahedral, the $\pi$ bonding taking place by overlap of oxygen $p$ orbitals with manganese $d$ orbitals. The green manganate ion ($\mathrm{Mn^{6+}}$, $d^{1}$) is paramagnetic with one unpaired electron; the purple permanganate ion ($\mathrm{Mn^{7+}}$, $d^{0}$) is diamagnetic. Permanganate has no $d$ electrons, so its intense colour cannot come from a $d$–$d$ transition. It comes from a charge-transfer transition: light moves an electron from the oxygen atoms to the manganese.

📐 Rule 4.9 — Permanganate as an oxidising agent in different media

(a) Acidic solution (dilute $\mathrm{H_{2}SO_{4}}$): Mn $+7\rightarrow+2$, five electrons; the purple solution becomes almost colourless ($\mathrm{Mn^{2+}}$ is very pale pink).

$$\mathrm{MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)}$$

$E^{\circ}=+1.52\ \mathrm{V}$. The acid used is sulphuric acid. Hydrochloric acid is avoided because permanganate oxidises $\mathrm{Cl^{-}}$ to chlorine, and nitric acid because it is itself an oxidising agent.

(b) Neutral or faintly alkaline solution: Mn $+7\rightarrow+4$, three electrons; a brown precipitate of $\mathrm{MnO_{2}}$ forms.

$$\mathrm{MnO_{4}^{-}(aq)+2H_{2}O(l)+3e^{-}\rightarrow MnO_{2}(s)+4OH^{-}(aq)}$$

$E^{\circ}=+0.59\ \mathrm{V}$.

(c) Strongly alkaline solution: Mn $+7\rightarrow+6$, one electron; the purple solution turns green.

$$\mathrm{MnO_{4}^{-}(aq)+e^{-}\rightarrow MnO_{4}^{2-}(aq)}$$

The same reagent therefore takes $5$, $3$ or $1$ electrons per formula unit, depending on the medium. This number, the electrons gained (by an oxidising agent) or lost (by a reducing agent) per formula unit, is often called the n-factor in competitive-exam problems. It fixes the mole ratio in which an oxidant and a reductant react: $n_{\text{ox}}\times(\text{mol oxidant})=n_{\text{red}}\times(\text{mol reductant})$.

✏️ Example 4 — Permanganate in acid
Write balanced ionic equations for the reaction of acidified permanganate with (a) oxalate ions (giving carbon dioxide), (b) sulphite ions, (c) nitrite ions. (d) When oxalic acid is titrated with permanganate at about $333\ \mathrm{K}$, the first few drops are decolourised slowly, but later drops instantly. Suggest why.
Method Reduction: $\mathrm{MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)}$ (5 electrons). Each reductant here loses 2 electrons per formula unit, so multiply the permanganate half by 2 and the reductant half by 5 (10 electrons each), add, and cancel $\mathrm{H^{+}}$ and $\mathrm{H_{2}O}$.
(a) $\mathrm{C_{2}O_{4}^{2-}(aq)\rightarrow2CO_{2}(g)+2e^{-}}$: $$\begin{aligned}&\mathrm{5C_{2}O_{4}^{2-}(aq)+2MnO_{4}^{-}(aq)+16H^{+}(aq)}\\&\quad\mathrm{\rightarrow2Mn^{2+}(aq)+8H_{2}O(l)+10CO_{2}(g)}\end{aligned}$$
(b) $\mathrm{SO_{3}^{2-}(aq)+H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+2H^{+}(aq)+2e^{-}}$; five waters and ten $\mathrm{H^{+}}$ cancel against the permanganate side: $$\begin{aligned}&\mathrm{5SO_{3}^{2-}(aq)+2MnO_{4}^{-}(aq)+6H^{+}(aq)}\\&\quad\mathrm{\rightarrow2Mn^{2+}(aq)+3H_{2}O(l)+5SO_{4}^{2-}(aq)}\end{aligned}$$
(c) $\mathrm{NO_{2}^{-}(aq)+H_{2}O(l)\rightarrow NO_{3}^{-}(aq)+2H^{+}(aq)+2e^{-}}$, in exactly the same way: $$\begin{aligned}&\mathrm{5NO_{2}^{-}(aq)+2MnO_{4}^{-}(aq)+6H^{+}(aq)}\\&\quad\mathrm{\rightarrow2Mn^{2+}(aq)+5NO_{3}^{-}(aq)+3H_{2}O(l)}\end{aligned}$$
(d) The reaction is catalysed by one of its own products, $\mathrm{Mn^{2+}}$ (autocatalysis). At the start there is almost no $\mathrm{Mn^{2+}}$, so the reaction is slow; as $\mathrm{Mn^{2+}}$ builds up, it speeds up. Warming also helps.
Check (a) Charge: $-10-2+16=+4$ and $2\times(+2)=+4$ ✓. O: $20+8=28$ and $8+20=28$ ✓.
✏️ Example 5 — Permanganate in neutral or faintly alkaline solution
In neutral or faintly alkaline solution permanganate is reduced to $\mathrm{MnO_{2}}$. Balance the reactions in which it oxidises (a) iodide to iodate, $\mathrm{IO_{3}^{-}}$; (b) thiosulphate, $\mathrm{S_{2}O_{3}^{2-}}$, to sulphate; (c) manganese(II) ions to $\mathrm{MnO_{2}}$ (in the presence of zinc oxide, which removes the $\mathrm{H^{+}}$ formed). In each, how many moles of $\mathrm{MnO_{4}^{-}}$ react per mole of reductant?
(a) $\mathrm{MnO_{4}^{-}}$ takes $3e^{-}$; $\mathrm{I^{-}}$ ($-1$) $\rightarrow\mathrm{IO_{3}^{-}}$ ($+5$) gives $6e^{-}$. So $2\mathrm{MnO_{4}^{-}}$ per $\mathrm{I^{-}}$: $$\begin{aligned}&\mathrm{2MnO_{4}^{-}(aq)+H_{2}O(l)+I^{-}(aq)}\\&\quad\mathrm{\rightarrow2MnO_{2}(s)+2OH^{-}(aq)+IO_{3}^{-}(aq)}\end{aligned}$$ Ratio $2:1$.
(b) Each S goes from $+2$ to $+6$: $\mathrm{S_{2}O_{3}^{2-}}$ gives $8e^{-}$. LCM of 3 and 8 is 24: $8\mathrm{MnO_{4}^{-}}$ and $3\mathrm{S_{2}O_{3}^{2-}}$. $$\begin{aligned}&\mathrm{8MnO_{4}^{-}(aq)+3S_{2}O_{3}^{2-}(aq)+H_{2}O(l)}\\&\quad\mathrm{\rightarrow8MnO_{2}(s)+6SO_{4}^{2-}(aq)+2OH^{-}(aq)}\end{aligned}$$ Ratio $8:3$.
(c) $\mathrm{Mn^{2+}}$ ($+2$) $\rightarrow\mathrm{MnO_{2}}$ ($+4$): $2e^{-}$. LCM 6: $2\mathrm{MnO_{4}^{-}}$ and $3\mathrm{Mn^{2+}}$. $$\begin{aligned}&\mathrm{2MnO_{4}^{-}(aq)+3Mn^{2+}(aq)+2H_{2}O(l)}\\&\quad\mathrm{\rightarrow5MnO_{2}(s)+4H^{+}(aq)}\end{aligned}$$ Ratio $2:3$. This is a comproportionation (the reverse of disproportionation): $+7$ and $+2$ meet at $+4$.
Check (a) charge $-2-1=-3$, $-2-1=-3$ ✓; (b) $-8-6=-14$, $-12-2=-14$ ✓; (c) $-2+6=+4$, $+4$ ✓.
✏️ Example 6 — How much iron is in an iron tablet?
(a) $0.1340\ \mathrm{g}$ of sodium oxalate, $\mathrm{Na_{2}C_{2}O_{4}}$ ($134.0\ \mathrm{g\,mol^{-1}}$), dissolved in dilute sulphuric acid and warmed, needs $20.00\ \mathrm{mL}$ of a $\mathrm{KMnO_{4}}$ solution. Find the concentration of the permanganate. (b) An iron tablet of mass $1.20\ \mathrm{g}$, containing iron(II) sulphate, is dissolved in dilute sulphuric acid and needs $22.50\ \mathrm{mL}$ of this permanganate. Find the mass of iron in the tablet and the percentage by mass of $\mathrm{FeSO_{4}\cdot7H_{2}O}$ ($278\ \mathrm{g\,mol^{-1}}$; $\mathrm{Fe}=55.8\ \mathrm{g\,mol^{-1}}$). (c) Why is no indicator needed?
(a) $n(\mathrm{C_{2}O_{4}^{2-}})=\dfrac{0.1340}{134.0}=1.000\times10^{-3}\ \mathrm{mol}$. From Example 4(a), $2\ \mathrm{mol}\ \mathrm{MnO_{4}^{-}}$ per $5\ \mathrm{mol}$ oxalate: $n(\mathrm{MnO_{4}^{-}})=0.400\times10^{-3}\ \mathrm{mol}$. $c=\dfrac{0.400\times10^{-3}}{0.02000\ \mathrm{L}}=0.0200\ \mathrm{mol\,L^{-1}}$.
(b) Moles $n(\mathrm{MnO_{4}^{-}})=0.0200\times0.02250=4.50\times10^{-4}\ \mathrm{mol}$. $\mathrm{MnO_{4}^{-}}$ takes $5e^{-}$ and $\mathrm{Fe^{2+}}$ gives $1e^{-}$, so $n(\mathrm{Fe^{2+}})=5\times4.50\times10^{-4}=2.25\times10^{-3}\ \mathrm{mol}$.
(b) Masses Iron: $2.25\times10^{-3}\times55.8=0.126\ \mathrm{g}=126\ \mathrm{mg}$. $\mathrm{FeSO_{4}\cdot7H_{2}O}$: $2.25\times10^{-3}\times278=0.6255\ \mathrm{g}$, i.e. $\dfrac{0.6255}{1.20}\times100=52.1\%$ of the tablet.
(c) Permanganate is its own indicator (a self-indicator): while reductant remains, each drop is reduced to almost colourless $\mathrm{Mn^{2+}}$; the first drop in excess turns the whole solution permanently pale pink.
Interactive 2 — Redox Balancer for Permanganate and Dichromate

Pick the oxidising agent and medium, then a reducing agent. Press Play (or Next step) to watch the half-equation method: write the two half-equations, multiply them so that the electrons match (the least common multiple of the two n-factors), add, cancel $\mathrm{H^{+}}$ and $\mathrm{H_{2}O}$, and, in neutral or alkaline solution, turn $\mathrm{H^{+}}$ into water with $\mathrm{OH^{-}}$. The readouts give the n-factors, the mole ratio and the volume of $0.0200\ \mathrm{mol\,L^{-1}}$ oxidant needed for $1.00\ \mathrm{mmol}$ of reductant. Every equation is checked for atoms and charge.

step 1 of 4
n-factor of oxidant—
n-factor of reductant—
mol oxidant : mol reductant—
mL of 0.0200 M oxidant per mmol reductant—
🚀 Beyond the board syllabus (JEE Advanced) — equivalents and normality

In JEE Advanced 2026 (normality); not in the CBSE syllabus, JEE Main or NEET.

Many competitive-exam problems count equivalents instead of balancing equations. The number of equivalents of a substance is (moles) × (n-factor), and the normality of a solution is $N=M\times n$. Because the electrons lost must equal the electrons gained, equivalents of oxidant equal equivalents of reductant at the end point: $N_{1}V_{1}=N_{2}V_{2}$. For example, $0.0200\ \mathrm{M}$ $\mathrm{KMnO_{4}}$ used in acid ($n=5$) is $0.100\ \mathrm{N}$, but the same solution used in neutral solution ($n=3$) is $0.0600\ \mathrm{N}$: the normality depends on the reaction, which is why the board course works with balanced equations and moles instead.
⚠️ Common mistakes with dichromate and permanganate

1. Treating the chromate–dichromate change as redox. Chromium is $+6$ in both ions; only the pH changes.

2. Using one n-factor for permanganate. $\mathrm{MnO_{4}^{-}}$ takes $5e^{-}$ in acid, $3e^{-}$ in neutral or faintly alkaline solution and $1e^{-}$ in strongly alkaline solution. Read the medium first.

3. Acidifying with HCl. Permanganate titrations use dilute $\mathrm{H_{2}SO_{4}}$; $\mathrm{HCl}$ would itself be oxidised to $\mathrm{Cl_{2}}$ and give a wrong titre.

4. Calling permanganate's colour a $d$–$d$ transition. $\mathrm{Mn(VII)}$ is $d^{0}$; the colour is charge transfer from oxygen to manganese. Likewise dichromate ($\mathrm{Cr(VI)}$, $d^{0}$).

Exercises

I
Oxides and oxoanions. Include state symbols in your equations.
  • 1 Give the oxidation state of the metal and the acid–base character of $\mathrm{V_{2}O_{5}}$, $\mathrm{CrO}$, $\mathrm{Mn_{2}O_{7}}$ and $\mathrm{MnO_{2}}$.
  • 2 Why is $\mathrm{Mn_{2}O_{7}}$ acidic while $\mathrm{MnO}$ is basic?
  • 3 $\mathrm{MnO}$ is an ionic solid; $\mathrm{Mn_{2}O_{7}}$ is a covalent liquid (an oil). Explain.
  • 4 Write the formula of the oxoanion of (a) chromium(VI) in alkaline solution, (b) manganese(VI), (c) manganese(VII), (d) vanadium(V) in alkaline solution.
  • 5 Write the equation for the reaction of $\mathrm{Mn_{2}O_{7}}$ with water.
  • 6 Write equations for the reactions of $\mathrm{CrO_{3}}$ with water and with aqueous sodium hydroxide.
  • 7 In which of $\mathrm{VO^{2+}}$ and $\mathrm{VO_{2}^{+}}$ is vanadium in the $+5$ state?
1 $\mathrm{V_{2}O_{5}}$: $+5$, amphoteric (mainly acidic). $\mathrm{CrO}$: $+2$, basic. $\mathrm{Mn_{2}O_{7}}$: $+7$, acidic. $\mathrm{MnO_{2}}$: $+4$, amphoteric.
2 In $\mathrm{Mn_{2}O_{7}}$ the very small, highly charged Mn(VII) centre pulls electron density strongly from oxygen; the oxide behaves like a non-metal oxide and gives an oxoacid ($\mathrm{HMnO_{4}}$) with water. In $\mathrm{MnO}$, $\mathrm{Mn^{2+}}$ is a large, low-charge cation and the oxide ion is free to act as a base.
3 A high oxidation state means high polarising power, so the Mn–O bonds in $\mathrm{Mn_{2}O_{7}}$ are covalent and it consists of discrete molecules (weak forces between them). $\mathrm{MnO}$ is a lattice of $\mathrm{Mn^{2+}}$ and $\mathrm{O^{2-}}$ ions.
4 (a) $\mathrm{CrO_{4}^{2-}}$; (b) $\mathrm{MnO_{4}^{2-}}$; (c) $\mathrm{MnO_{4}^{-}}$; (d) $\mathrm{VO_{4}^{3-}}$.
5 $\mathrm{Mn_{2}O_{7}(l)+H_{2}O(l)\rightarrow2HMnO_{4}(aq)}$.
6 $\mathrm{CrO_{3}(s)+H_{2}O(l)\rightarrow H_{2}CrO_{4}(aq)}$; $\mathrm{CrO_{3}(s)+2NaOH(aq)\rightarrow Na_{2}CrO_{4}(aq)+H_{2}O(l)}$.
7 $\mathrm{VO_{2}^{+}}$: $x-4=+1$, $x=+5$. In $\mathrm{VO^{2+}}$, $x-2=+2$, $x=+4$.
II
Potassium dichromate. $E^{\circ}_{\mathrm{Cr_{2}O_{7}^{2-}/Cr^{3+}}}=+1.33\ \mathrm{V}$.
  • 1 Describe, with equations, the three stages in the preparation of potassium dichromate from chromite ore.
  • 2 Why is potassium dichromate, rather than sodium dichromate, used as a primary standard?
  • 3 What is seen when sodium hydroxide solution is added to orange potassium dichromate solution, and then excess dilute sulphuric acid? Write the equations.
  • 4 Describe the structure of the dichromate ion: how many terminal and how many bridging Cr–O bonds are there, and what is the Cr–O–Cr angle?
  • 5 Balance the oxidation of iodide ions by acidified dichromate.
  • 6 Balance the oxidation of sulphite ions to sulphate by acidified dichromate.
  • 7 What volume of $0.0500\ \mathrm{mol\,L^{-1}}$ $\mathrm{K_{2}Cr_{2}O_{7}}$ oxidises $25.0\ \mathrm{mL}$ of $0.120\ \mathrm{mol\,L^{-1}}$ $\mathrm{FeSO_{4}}$ in acid?
  • 8 Breath test. In an early breath-alcohol tester, ethanol is oxidised to ethanal ($\mathrm{CH_{3}CHO}$) by acidified dichromate. Write the balanced equation and state the colour change.
1 (i) Fusion with $\mathrm{Na_{2}CO_{3}}$ in air: $\mathrm{4FeCr_{2}O_{4}(s)+8Na_{2}CO_{3}(s)+7O_{2}(g)\rightarrow8Na_{2}CrO_{4}(s)+2Fe_{2}O_{3}(s)+8CO_{2}(g)}$. (ii) Extraction and acidification: $\mathrm{2Na_{2}CrO_{4}(aq)+2H^{+}(aq)\rightarrow Na_{2}Cr_{2}O_{7}(aq)+2Na^{+}(aq)+H_{2}O(l)}$. (iii) With KCl: $\mathrm{Na_{2}Cr_{2}O_{7}(aq)+2KCl(aq)\rightarrow K_{2}Cr_{2}O_{7}(s)+2NaCl(aq)}$.
2 Sodium dichromate is very soluble and absorbs moisture from the air (deliquescent), so it cannot be weighed accurately. Potassium dichromate is less soluble, crystallises pure and is not hygroscopic.
3 With alkali the solution turns yellow: $\mathrm{Cr_{2}O_{7}^{2-}(aq)+2OH^{-}(aq)\rightarrow2CrO_{4}^{2-}(aq)+H_{2}O(l)}$. With excess acid it turns orange again: $\mathrm{2CrO_{4}^{2-}(aq)+2H^{+}(aq)\rightarrow Cr_{2}O_{7}^{2-}(aq)+H_{2}O(l)}$.
4 Two $\mathrm{CrO_{4}}$ tetrahedra sharing one oxygen: six terminal Cr–O bonds and two bridging Cr–O bonds (one from each Cr to the shared O); $\mathrm{Cr{-}O{-}Cr}\approx126^{\circ}$.
5 $\mathrm{Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6I^{-}(aq)\rightarrow2Cr^{3+}(aq)+3I_{2}(aq)+7H_{2}O(l)}$.
6 $\mathrm{SO_{3}^{2-}(aq)+H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+2H^{+}(aq)+2e^{-}}$ (×3): $\mathrm{Cr_{2}O_{7}^{2-}(aq)+3SO_{3}^{2-}(aq)+8H^{+}(aq)\rightarrow2Cr^{3+}(aq)+3SO_{4}^{2-}(aq)+4H_{2}O(l)}$.
7 $n(\mathrm{Fe^{2+}})=0.0250\times0.120=3.00\times10^{-3}\ \mathrm{mol}$; $n(\mathrm{Cr_{2}O_{7}^{2-}})=\dfrac{3.00\times10^{-3}}{6}=5.00\times10^{-4}\ \mathrm{mol}$; $V=\dfrac{5.00\times10^{-4}}{0.0500}=0.0100\ \mathrm{L}=10.0\ \mathrm{mL}$.
8 $\mathrm{CH_{3}CH_{2}OH\rightarrow CH_{3}CHO+2H^{+}+2e^{-}}$ (×3): $\mathrm{3CH_{3}CH_{2}OH+Cr_{2}O_{7}^{2-}+8H^{+}\rightarrow3CH_{3}CHO+2Cr^{3+}+7H_{2}O}$. Orange to green.
III
Potassium permanganate. $\mathrm{KMnO_{4}}=158\ \mathrm{g\,mol^{-1}}$; $E^{\circ}_{\mathrm{MnO_{4}^{-}/Mn^{2+}}}=+1.52\ \mathrm{V}$, $E^{\circ}_{\mathrm{Cl_{2}/Cl^{-}}}=+1.36\ \mathrm{V}$.
  • 1 Write equations for the preparation of potassium permanganate from pyrolusite, $\mathrm{MnO_{2}}$, by (a) alkaline oxidative fusion, (b) disproportionation of the manganate in acid.
  • 2 The manganate ion is green and paramagnetic, the permanganate ion purple and diamagnetic. Explain both differences.
  • 3 Write the equation for the decomposition of $\mathrm{KMnO_{4}}$ at $513\ \mathrm{K}$, and find the mass of oxygen released by $3.16\ \mathrm{g}$ of $\mathrm{KMnO_{4}}$.
  • 4 Write the reduction half-equations of permanganate in acidic, neutral and strongly alkaline solution, with the n-factor and the colour change in each.
  • 5 Balance the oxidation of hydrogen sulphide by acidified permanganate.
  • 6 Balance the oxidation of iron(II) by acidified permanganate.
  • 7 Balance the oxidation of iodide by acidified permanganate.
  • 8 Why are permanganate titrations not carried out in the presence of hydrochloric acid?
  • 9 Under standard conditions, which of acidified $\mathrm{KMnO_{4}}$ and acidified $\mathrm{K_{2}Cr_{2}O_{7}}$ can oxidise chloride ions to chlorine? Explain.
1 (a) $\mathrm{2MnO_{2}(s)+4KOH(l)+O_{2}(g)\rightarrow2K_{2}MnO_{4}(s)+2H_{2}O(g)}$. (b) $\mathrm{3MnO_{4}^{2-}(aq)+4H^{+}(aq)\rightarrow2MnO_{4}^{-}(aq)+MnO_{2}(s)+2H_{2}O(l)}$.
2 Manganate has Mn(VI), $d^{1}$: one unpaired electron (paramagnetic). Permanganate has Mn(VII), $d^{0}$: no unpaired electrons (diamagnetic). Both colours come from charge-transfer (O → Mn) absorption, which falls at different wavelengths in the two ions.
3 $\mathrm{2KMnO_{4}(s)\rightarrow K_{2}MnO_{4}(s)+MnO_{2}(s)+O_{2}(g)}$. $n(\mathrm{KMnO_{4}})=\dfrac{3.16}{158}=0.0200\ \mathrm{mol}$, giving $0.0100\ \mathrm{mol}$ $\mathrm{O_{2}}$ $=0.320\ \mathrm{g}$.
4 Acid: $\mathrm{MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)}$, $n=5$, purple → colourless. Neutral: $\mathrm{MnO_{4}^{-}(aq)+2H_{2}O(l)+3e^{-}\rightarrow MnO_{2}(s)+4OH^{-}(aq)}$, $n=3$, purple → brown precipitate. Strongly alkaline: $\mathrm{MnO_{4}^{-}(aq)+e^{-}\rightarrow MnO_{4}^{2-}(aq)}$, $n=1$, purple → green.
5 $\mathrm{5H_{2}S(g)+2MnO_{4}^{-}(aq)+6H^{+}(aq)\rightarrow2Mn^{2+}(aq)+8H_{2}O(l)+5S(s)}$.
6 $\mathrm{5Fe^{2+}(aq)+MnO_{4}^{-}(aq)+8H^{+}(aq)\rightarrow Mn^{2+}(aq)+4H_{2}O(l)+5Fe^{3+}(aq)}$.
7 $\mathrm{10I^{-}(aq)+2MnO_{4}^{-}(aq)+16H^{+}(aq)\rightarrow2Mn^{2+}(aq)+8H_{2}O(l)+5I_{2}(aq)}$.
8 Permanganate oxidises chloride ions to chlorine ($E^{\circ}$ $1.52\gt1.36\ \mathrm{V}$), so part of the permanganate would be used up by the acid itself and the titre would be too high. Dilute sulphuric acid is used instead.
9 Only permanganate: $E^{\circ}_{cell}=1.52-1.36=+0.16\ \mathrm{V}\gt0$. For dichromate $1.33-1.36=-0.03\ \mathrm{V}\lt0$, so under standard conditions it does not oxidise $\mathrm{Cl^{-}}$ (this is why dichromate titrations of iron can be done in dilute HCl).
IV
Redox titrations. Molar masses: $\mathrm{KMnO_{4}}$ $158$, $\mathrm{Na_{2}C_{2}O_{4}}$ $134.0$, Mohr's salt $\mathrm{FeSO_{4}\cdot(NH_{4})_{2}SO_{4}\cdot6H_{2}O}$ $392$, $\mathrm{Fe}$ $55.8\ \mathrm{g\,mol^{-1}}$.
  • 1 What mass of $\mathrm{KMnO_{4}}$ is needed to make $250\ \mathrm{mL}$ of a $0.0200\ \mathrm{mol\,L^{-1}}$ solution?
  • 2 What volume of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ reacts with $25.0\ \mathrm{mL}$ of $0.0500\ \mathrm{mol\,L^{-1}}$ oxalic acid in acid solution?
  • 3 $20.0\ \mathrm{mL}$ of an iron(II) solution needs $16.0\ \mathrm{mL}$ of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ in acid. Find the concentration of $\mathrm{Fe^{2+}}$.
  • 4 What volume of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$, in neutral solution, oxidises $1.00\ \mathrm{mmol}$ of iodide to iodate?
  • 5 $0.392\ \mathrm{g}$ of Mohr's salt is dissolved in dilute sulphuric acid. What volume of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ is needed?
  • 6 A $0.500\ \mathrm{g}$ sample of impure sodium oxalate needs $28.0\ \mathrm{mL}$ of $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ in acid. Find the percentage purity.
  • 7 Water treatment. Well water contains $2.0\ \mathrm{mg\,L^{-1}}$ of $\mathrm{Fe^{2+}}$. Permanganate oxidises it in near-neutral solution: $\mathrm{3Fe^{2+}(aq)+MnO_{4}^{-}(aq)+7H_{2}O(l)\rightarrow3Fe(OH)_{3}(s)+MnO_{2}(s)+5H^{+}(aq)}$. Check that the equation is balanced, and find the mass of $\mathrm{KMnO_{4}}$ needed to treat $1000\ \mathrm{L}$ of water.
Check: amount of reductant from a titre
mmol oxidant—
mmol electrons—
mmol reductant—
1 $n=0.0200\times0.250=5.00\times10^{-3}\ \mathrm{mol}$; mass $=5.00\times10^{-3}\times158=0.790\ \mathrm{g}$.
2 Oxalate $=1.25\ \mathrm{mmol}$; $\mathrm{MnO_{4}^{-}}=\tfrac25\times1.25=0.500\ \mathrm{mmol}$; $V=\dfrac{0.500}{0.0200}=25.0\ \mathrm{mL}$.
3 $\mathrm{MnO_{4}^{-}}=0.0200\times16.0=0.320\ \mathrm{mmol}$; $\mathrm{Fe^{2+}}=5\times0.320=1.60\ \mathrm{mmol}$; $c=\dfrac{1.60}{20.0}=0.0800\ \mathrm{mol\,L^{-1}}$. (The check-lab above starts with this titre.)
4 Ratio $2\ \mathrm{MnO_{4}^{-}}:1\ \mathrm{I^{-}}$ ($n=3$ against $n=6$): $2.00\ \mathrm{mmol}$, $V=\dfrac{2.00}{0.0200}=100\ \mathrm{mL}$.
5 $\dfrac{0.392}{392}=1.00\times10^{-3}\ \mathrm{mol}$ $\mathrm{Fe^{2+}}$; $\mathrm{MnO_{4}^{-}}=0.200\ \mathrm{mmol}$; $V=10.0\ \mathrm{mL}$.
6 $\mathrm{MnO_{4}^{-}}=0.0200\times28.0=0.560\ \mathrm{mmol}$; oxalate $=\tfrac52\times0.560=1.40\ \mathrm{mmol}$, mass $=1.40\times10^{-3}\times134.0=0.1876\ \mathrm{g}$; purity $=\dfrac{0.1876}{0.500}\times100=37.5\%$.
7 Check: Fe 3/3, Mn 1/1, O $4+7=11=9+2$, H $14=9+5$, charge $+6-1=+5$ ✓. $\mathrm{Fe^{2+}}$ in $1000\ \mathrm{L}$: $2000\ \mathrm{mg}$, $\dfrac{2.000}{55.8}=0.03584\ \mathrm{mol}$; $\mathrm{KMnO_{4}}=\dfrac{0.03584}{3}=0.01195\ \mathrm{mol}$, mass $=0.01195\times158=1.89\ \mathrm{g}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks (a) What happens when the pH of a potassium dichromate solution is raised? Write the equation. (b) Permanganate contains manganese with no $d$ electrons, yet it is intensely coloured. Why? (c) Write the ionic equation for the reaction of acidified permanganate with iron(II) ions.

(a) Orange dichromate changes to yellow chromate: $\mathrm{Cr_{2}O_{7}^{2-}(aq)+2OH^{-}(aq)\rightarrow2CrO_{4}^{2-}(aq)+H_{2}O(l)}$.
(b) The colour is due to charge transfer: visible light promotes an electron from the oxygen atoms to the manganese(VII) centre. No $d$–$d$ transition is involved.
(c) $\mathrm{MnO_{4}^{-}(aq)+5Fe^{2+}(aq)+8H^{+}(aq)\rightarrow Mn^{2+}(aq)+5Fe^{3+}(aq)+4H_{2}O(l)}$.

E2. JEE Main Numerical value. The volume (in mL) of $0.100\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$ needed to oxidise completely, in acid solution, $100\ \mathrm{mL}$ of $0.500\ \mathrm{mol\,L^{-1}}$ iron(II) oxalate, $\mathrm{FeC_{2}O_{4}}$, is ______.

Idea Both ions of iron(II) oxalate are oxidised: $\mathrm{Fe^{2+}\rightarrow Fe^{3+}}$ ($1e^{-}$) and $\mathrm{C_{2}O_{4}^{2-}\rightarrow2CO_{2}}$ ($2e^{-}$), so each formula unit gives $3e^{-}$.
Count $n(\mathrm{FeC_{2}O_{4}})=0.0500\ \mathrm{mol}$, electrons $=0.150\ \mathrm{mol}$; each $\mathrm{MnO_{4}^{-}}$ takes $5$, so $n(\mathrm{MnO_{4}^{-}})=0.0300\ \mathrm{mol}$.
Answer $V=\dfrac{0.0300}{0.100}=0.300\ \mathrm{L}=\mathbf{300}\ \mathrm{mL}$.

E3. NEET In neutral or faintly alkaline solution, permanganate oxidises iodide ions to (A) $\mathrm{I_{2}}$  (B) $\mathrm{IO_{3}^{-}}$  (C) $\mathrm{IO_{4}^{-}}$  (D) $\mathrm{IO^{-}}$

Idea $\mathrm{2MnO_{4}^{-}(aq)+H_{2}O(l)+I^{-}(aq)\rightarrow2MnO_{2}(s)+2OH^{-}(aq)+IO_{3}^{-}(aq)}$; in acid the product would be $\mathrm{I_{2}}$.
Answer (B) iodate.

E4. JEE Advanced One or more correct. Which statement(s) about the manganate ion, $\mathrm{MnO_{4}^{2-}}$, and the permanganate ion, $\mathrm{MnO_{4}^{-}}$, are correct? (A) Both ions are tetrahedral. (B) Manganate is paramagnetic and permanganate diamagnetic. (C) The colour of permanganate is due to a $d$–$d$ transition. (D) In acidic solution manganate disproportionates into permanganate and manganese(IV) oxide.

(A) True Both are $\mathrm{MO_{4}}$ tetrahedra with $p_{\pi}$–$d_{\pi}$ bonding.
(B) True Mn(VI) $d^{1}$, one unpaired electron; Mn(VII) $d^{0}$.
(C) False $d^{0}$: the colour is charge transfer.
(D) True · Answer $\mathrm{3MnO_{4}^{2-}(aq)+4H^{+}(aq)\rightarrow2MnO_{4}^{-}(aq)+MnO_{2}(s)+2H_{2}O(l)}$. Correct: (A), (B) and (D).

E5. NEET Board Assertion–reason. Assertion (A): An orange solution of potassium dichromate turns yellow when sodium hydroxide is added. Reason (R): In alkaline solution chromium(VI) is reduced to chromium(III). (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Check A is true (dichromate → yellow chromate). R is false: chromium stays $+6$; the change is an acid–base equilibrium, $\mathrm{Cr_{2}O_{7}^{2-}(aq)+2OH^{-}(aq)\rightleftharpoons2CrO_{4}^{2-}(aq)+H_{2}O(l)}$.
Answer (C).

E6. Board · Case study In a school laboratory a student titrates $10.0\ \mathrm{mL}$ of an iron(II) sulphate solution, acidified with dilute sulphuric acid, against $0.0200\ \mathrm{mol\,L^{-1}}$ $\mathrm{KMnO_{4}}$. The mean titre is $12.5\ \mathrm{mL}$. (a) Why is dilute sulphuric acid used and not hydrochloric acid? (1 mark) (b) Calculate the concentration of iron(II) in the solution. (2 marks) (c) How does the student recognise the end point? (1 mark)

(a) Permanganate would oxidise chloride ions to chlorine, using up extra permanganate and giving a wrong (too large) titre.
(b) $n(\mathrm{MnO_{4}^{-}})=0.0200\times12.5=0.250\ \mathrm{mmol}$; $\mathrm{MnO_{4}^{-}}:\mathrm{Fe^{2+}}=1:5$, so $n(\mathrm{Fe^{2+}})=1.25\ \mathrm{mmol}$; $c=\dfrac{1.25}{10.0}=0.125\ \mathrm{mol\,L^{-1}}$.
(c) No indicator is needed: the first drop of excess permanganate gives a permanent pale pink colour.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How dirty is a river? Environmental laboratories answer with the chemical oxygen demand (COD), measured by boiling the water with excess acidified dichromate. How does the method turn a dichromate titration into "milligrams of oxygen per litre", and why are laboratories looking for alternatives to it?

Why it matters. Organic pollution from sewage and industry uses up the dissolved oxygen that fish need. Discharge limits for factories and sewage plants (in India set by the Central Pollution Control Board) are written in terms of COD, often a few hundred $\mathrm{mg\,L^{-1}}$ at most for treated effluent.

What to do.

  1. Show that one mole of $\mathrm{Cr_{2}O_{7}^{2-}}$ accepts as many electrons as $1.5$ mol of $\mathrm{O_{2}}$ (each $\mathrm{O_{2}}$ takes $4e^{-}$ when it becomes water).
  2. A $20.0\ \mathrm{mL}$ water sample is refluxed with $10.0\ \mathrm{mL}$ of acidified $\mathrm{K_{2}Cr_{2}O_{7}}$. The unused dichromate needs $15.0\ \mathrm{mL}$ of $0.100\ \mathrm{mol\,L^{-1}}$ $\mathrm{Fe^{2+}}$ (Mohr's salt); a blank with pure water needs $25.0\ \mathrm{mL}$. Find the electrons taken from the pollutants and the COD in $\mathrm{mg\,O_{2}\,L^{-1}}$.
  3. Find out why silver sulphate and mercury(II) sulphate are added in the standard method.
  4. Chromium(VI) is toxic and carcinogenic. Look up what happens to the chromium waste, and one alternative oxidant or method being developed.

Tools: Calculator; Interactive 2 (dichromate with $\mathrm{Fe^{2+}}$ gives the $1:6$ ratio); a published standard method for COD, or a water-quality report from a city water board.

Hand in: Your COD calculation set out step by step, a one-paragraph summary of the role of each added chemical, and two sentences on the environmental cost of the method.

Going further. Some countries report a "permanganate index" for drinking water instead. Why is permanganate, which is a stronger oxidising agent on paper ($E^{\circ}=1.52\ \mathrm{V}$), less complete than boiling dichromate at oxidising organic matter? (Think about the conditions and the rate.)

Instructor note: $\mathrm{Cr_{2}O_{7}^{2-}}$ takes $6e^{-}$, $\mathrm{O_{2}}$ takes $4e^{-}$, so $6/4=1.5$. Fe²⁺ difference $=(25.0-15.0)\times0.100=1.00\ \mathrm{mmol}$ of electrons taken by the pollutants, equivalent to $0.250\ \mathrm{mmol}$ $\mathrm{O_{2}}=8.00\ \mathrm{mg}$ in $20.0\ \mathrm{mL}$, so COD $=400\ \mathrm{mg\,L^{-1}}$. $\mathrm{Ag_{2}SO_{4}}$ catalyses the oxidation of resistant organics; $\mathrm{HgSO_{4}}$ binds chloride, which would otherwise be oxidised and inflate the result. The permanganate test is run at lower temperature for a short time, so many organics are only partly oxidised (kinetics, not thermodynamics).

🧪 Self-check quiz

0/5 correct

Q1. The oxidation states of chromium in $\mathrm{K_{2}Cr_{2}O_{7}}$ and in chromite, $\mathrm{FeCr_{2}O_{4}}$, are

Q2. In neutral or faintly alkaline solution, the number of electrons gained per $\mathrm{MnO_{4}^{-}}$ ion is

Q3. When dilute acid is added to a yellow solution of sodium chromate,

Q4. Permanganate titrations are acidified with

Q5. In acidic solution the green manganate ion disproportionates to give