Important compounds — KMnO₄, K₂Cr₂O₇ — preparation, oxidizing reactions: Question bank

True or false — justify

K M n O_{4}

always reduces to

M n^{2 +}

when it acts as an oxidiser.

False — only in acidic medium (5 e⁻ →

M n^{2 +}

). Neutral/faintly alkaline gives

M n O_{2}

(3 e⁻, brown), strong alkali gives green

M n O_{4}^{2 -}

(1 e⁻). The medium sets how many H⁺ are available and therefore how far Mn can fall.

The conversion

2 C r O_{4}^{2 -} + 2 H^{+} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O

is a redox reaction because the colour changes yellow↔orange.

False — chromium is +6 on both sides, so no electrons transfer. It is an acid–base condensation (degree of polymerisation changes), and colour change alone never proves redox.

K_{2} C r_{2} O_{7}

can oxidise substances in neutral or alkaline medium just like

K M n O_{4}

False — dichromate only exists in acidic solution; add base and it flips back to chromate (

C r O_{4}^{2 -}

), which is a far weaker oxidiser. So dichromate as an oxidiser is an acid-medium-only reagent.

E^{\circ} (M n O_{4}^{-} / M n^{2 +}) = + 1.51 V

being larger than

E^{\circ} (C r_{2} O_{7}^{2 -} / C r^{3 +}) = + 1.33 V

means permanganate is the stronger oxidiser.

True — a higher standard reduction potential means a greater pull for electrons (see Standard Electrode Potential & E° values), so

K M n O_{4}

is thermodynamically the stronger oxidiser of the two.

In the acidic half-reaction

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

, the 8 H⁺ are oxidised.

False — the H⁺ are not oxidised or reduced (H stays +1). They are consumed only to mop up the oxide oxygens as water; the electrons are what oxidise the other reactant.

Green

M n O_{4}^{2 -}

(manganate) is a higher oxidation state of Mn than purple

M n O_{4}^{-}

False — manganate is Mn(+6) and permanganate is Mn(+7). The purple ion is the more oxidised one; going green→purple in prep is an oxidation (loss of one more electron).

K_{2} C r_{2} O_{7}

is preferred over

K M n O_{4}

as a primary/secondary standard in titrations.

True — dichromate is obtainable pure, is non-hygroscopic, and does not react with cold dilute HCl, so its concentration stays known.

K M n O_{4}

can't be got 100% pure and slowly oxidises water, so it must be standardised each time.

Spot the error

A student titrates

F e^{2 +}

with

K M n O_{4}

using dilute HCl as the acid. What's wrong?

HCl's chloride is oxidised by

K M n O_{4}

C l_{2}

, so extra permanganate is consumed by a side reaction and the iron result comes out too high. Use dilute $H_{2} S O_{4}$ , whose

S O_{4}^{2 -}

is not oxidised.

A student writes the acidic dichromate half-reaction as

C r_{2} O_{7}^{2 -} + 14 H^{+} + 3 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

. Find the mistake.

Each Cr goes +6→+3 (3 e⁻) and there are two Cr atoms, so 6 electrons are gained, not 3. The correct form is

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

(see Balancing Redox Reactions (ion-electron method)).

Someone claims the KMnO₄ + oxalic acid titration needs 5 permanganate ions per 2 oxalate ions. Correct it.

It is the reverse:

2 M n O_{4}^{-}

per

5 C_{2} O_{4}^{2 -}

. Each

M n O_{4}^{-}

takes 5 e⁻ and each

C_{2} O_{4}^{2 -}

gives 2 e⁻; equal-electron balancing (LCM 10) needs

2 \times 5 = 10

gained and

5 \times 2 = 10

lost.

A prep scheme says

O_{2}

alone converts

M n O_{2}

straight to purple

K M n O_{4}

in one fusion step. Where does it fail?

Air's

O_{2}

only pushes Mn up to +6 (green manganate). The last electron (Mn⁶⁺→Mn⁷⁺) needs a stronger oxidiser than

O_{2}

— electrolytic oxidation at the anode (or

C l_{2}

O_{3}

/disproportionation). So it is a two-stage climb, not one.

A student cools a

K M n O_{4}

–oxalic acid titration to 5 °C to "slow side reactions". Why is this wrong?

The reaction is already sluggish at start; cooling makes it slower still and the endpoint drifts. It is warmed to ~60 °C, and once

M n^{2 +}

forms it autocatalyses the reaction.

Given "

4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}

", a student labels the chromium product dichromate. Fix it.

In alkaline roasting the product is yellow chromate

C r O_{4}^{2 -}

, not orange dichromate. Dichromate only appears in the next step when the melt is acidified.

Manganese is already at its maximum accessible state +7, so it has no electrons left to give — its only chemical move is to grab electrons back, i.e. to be reduced (get oxidised done to others). Being electron-hungry from a top oxidation state is what "oxidiser" means.

Why does adding acid to a yellow chromate solution turn it orange?

Added H⁺ shifts

2 C r O_{4}^{2 -} + 2 H^{+} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O

to the right by Le Chatelier's Principle, forming orange dichromate — a position shift, not a redox change.

Why does the electrolytic step in

K M n O_{4}

prep use the anode specifically?

At the anode electrons are forcibly stripped away, making it the strongest available oxidiser — strong enough to remove the last electron (

M n^{6 +} \to M n^{7 +}

) that air's

O_{2}

could not. It also avoids adding any foreign chemical reagent.

Why do

M n O_{4}^{-}

and

C r_{2} O_{7}^{2 -}

have such deep colours while

M n^{2 +}

is nearly colourless?

The intense colours come from charge-transfer / d-related transitions in the high-oxidation-state oxo-ions; when Mn is reduced to

M n^{2 +}

d^{5}

ion) the strong transitions vanish, so the solution goes pale — see Colour & d-d Transitions.

Why is dichromate reduced by

I^{-}

but permanganate titrations avoid

C l^{-}

Both halide oxidations are possible in principle, but with dichromate + iodide it is the intended reaction (

C r_{2} O_{7}^{2 -} + 6 I^{-} + 14 H^{+} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O

, iodometry), whereas in a KMnO₄–

F e^{2 +}

titration

C l_{2}

formation is an unwanted side reaction that ruins the count.

Why must the electron count be balanced (lost = gained) before writing a final redox equation?

Electrons are neither created nor destroyed; every electron one species loses another must gain. Taking the LCM of the two half-reaction electron numbers guarantees this, otherwise the equation violates charge conservation — see Volumetric Analysis / Titrations for why this fixes the mole ratio.

Why is

K_{2} C r_{2} O_{7}

chosen over

N a_{2} C r_{2} O_{7}

as the final crystalline salt?

The potassium salt is less soluble and non-hygroscopic, so it crystallises out cleanly and can be weighed accurately — ideal for a standard reagent.

Edge cases

What if you slowly add base to an acidic

K M n O_{4}

titration mid-way — does Mn still end at

M n^{2 +}

No — as H⁺ runs low the reduction product changes: through neutral you get brown

M n O_{2}

(3 e⁻), and in strong alkali green

M n O_{4}^{2 -}

(1 e⁻). The product tracks the medium, so mid-titration drift wrecks the stoichiometry.

What happens to

C r_{2} O_{7}^{2 -}

if the acid is completely neutralised during an oxidation?

It reverts to yellow

C r O_{4}^{2 -}

, which is a much weaker oxidiser, so the oxidation effectively stalls — dichromate has no "neutral" or "alkaline" oxidising mode.

In strong alkali,

M n O_{4}^{-}

gains only 1 electron. Is it still an oxidiser?

Yes, but a much milder one — it only drops to

M n O_{4}^{2 -}

(green), a one-electron change. It still removes an electron from the reductant, just far less than the 5-electron acidic mode.

Is chromate ever an oxidiser as strong as dichromate?

In practice no — the strong oxidising behaviour is tied to the acidic dichromate form; the alkaline chromate solution is a weak oxidiser, which is why oxidations are run in acid.

At exactly neutral pH, what is the permanganate product, and what do you see?

A brown precipitate of

M n O_{2}

(Mn +4, 3 e⁻ gained) — the purple colour is replaced by an insoluble brown solid rather than a clear solution.

If a "

K M n O_{4}

" bottle has stood in water for months, why might its titre be off even before use?

K M n O_{4}

slowly oxidises water (and traces of organics/dust), lowering its true concentration over time. That is exactly why it is standardised fresh and never treated as a primary standard.

Connections

Standard Electrode Potential & E° values — the +1.51 V vs +1.33 V comparison behind "which is the stronger oxidiser".
Oxidation States of Transition Metals — why +7 (Mn) and +6 (Cr) are reachable at all.
Balancing Redox Reactions (ion-electron method) — the electron-count discipline behind every "spot the error".
Volumetric Analysis / Titrations — where these traps bite in real titre calculations.
Le Chatelier's Principle — the acid/base shifts of the chromate–dichromate equilibrium.
Colour & d-d Transitions — why the ions are coloured and $M n^{2 +}$ is pale.

Question bank — Important compounds — KMnO₄, K₂Cr₂O₇ — preparation, oxidizing reactions

True or false — justify

Spot the error

Why questions

Edge cases

Connections