Important compounds — KMnO₄, K₂Cr₂O₇ — preparation, oxidizing reactions: Exercises

A quick reminder of the two workhorse half-reactions we lean on all page:

Figure — Important compounds — KMnO₄, K₂Cr₂O₇ — preparation, oxidizing reactions

The left ladder (blue) shows Mn falling from $+ 7$ down 5 rungs to $+ 2$ in acid; the right ladder (orange) shows each Cr falling 3 rungs ( $+ 6 \to + 3$ ), two Cr making 6 rungs total. "Electron appetite" = how many rungs the species drops = how many electrons it swallows. See Balancing Redox Reactions (ion-electron method) and Standard Electrode Potential & E° values.

Level 1 — Recognition

Q1. State the oxidation state of the metal in each: (a) $K M n O_{4}$ , (b) $K_{2} M n O_{4}$ , (c) $K_{2} C r_{2} O_{7}$ , (d) $K_{2} C r O_{4}$ .

Recall Solution

The trick: the whole compound is neutral, so all charges must cancel. Oxygen is almost always $- 2$ ; potassium (K) is always $+ 1$ . (a) $K M n O_{4}$ : $(+ 1) + M n + 4 (- 2) = 0 \Rightarrow M n = + 7$ . (b) $K_{2} M n O_{4}$ : $2 (+ 1) + M n + 4 (- 2) = 0 \Rightarrow M n = + 6$ . (c) $K_{2} C r_{2} O_{7}$ : $2 (+ 1) + 2 C r + 7 (- 2) = 0 \Rightarrow 2 C r = + 12 \Rightarrow C r = + 6$ . (d) $K_{2} C r O_{4}$ : $2 (+ 1) + C r + 4 (- 2) = 0 \Rightarrow C r = + 6$ . Note (c) and (d) are both $+ 6$ — chromate and dichromate differ only in how many CrO units are stuck together, not in oxidation state. See Oxidation States of Transition Metals.

Q2. Name and give the formula of the ore used to prepare (a) KMnO₄, (b) K₂Cr₂O₇. What colour is the final product in each case?

Recall Solution

(a) Pyrolusite, $M n O_{2}$ → final product KMnO₄ is purple. (b) Chromite, $F e C r_{2} O_{4}$ → final product K₂Cr₂O₇ is orange.

Q3. Fill the colour: manganate $M n O_{4}^{2 -}$ is green; permanganate $M n O_{4}^{-}$ is purple; chromate $C r O_{4}^{2 -}$ is yellow; dichromate $C r_{2} O_{7}^{2 -}$ is orange. The colour map below shows all four side by side.

Recall Solution

Manganate = green (+6), permanganate = purple (+7), chromate = yellow (+6), dichromate = orange (+6). The intense colours come from charge-transfer/Colour & d-d Transitions absorption.

Level 2 — Application

Q4. Balance the reduction of permanganate to Mn²⁺ in acidic medium from scratch (state each electron/O/H step).

Recall Solution

Skeleton: $M n O_{4}^{-} \to M n^{2 +}$ .

Electrons: Mn goes $+ 7 \to + 2$ , a drop of 5, so add $5 e^{-}$ to the left (electrons are gained on the side being reduced): $M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}$ .
Oxygen: 4 O on the left, 0 on right → add $4 H_{2} O$ on the right: $M n O_{4}^{-} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ .
Hydrogen: now 8 H on right, 0 on left → add $8 H^{+}$ on the left: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ . Charge check: left $= (- 1) + 8 (+ 1) + 5 (- 1) = + 2$ ; right $= + 2$ . ✓

Q5. Write the balanced equation for acidic $K M n O_{4}$ oxidising $F e^{2 +}$ to $F e^{3 +}$ .

Recall Solution

Oxidation half: $F e^{2 +} \to F e^{3 +} + e^{-}$ (gives up 1 electron). Reduction half: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ (takes 5). To balance electrons, multiply the Fe half by 5 (LCM of 1 and 5 is 5): $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$

Q6. Write the balanced equation for acidic $K_{2} C r_{2} O_{7}$ oxidising $I^{-}$ to $I_{2}$ (iodometry).

Recall Solution

Oxidation half: $2 I^{-} \to I_{2} + 2 e^{-}$ (2 electrons per $I_{2}$ ). Reduction half: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ (6 electrons). LCM of 2 and 6 is 6 → multiply the iodide half by 3: $C r_{2} O_{7}^{2 -} + 6 I^{-} + 14 H^{+} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$

Level 3 — Analysis

Q7. In neutral medium, KMnO₄ is reduced to $M n O_{2}$ (Mn $+ 4$ ) instead of $M n^{2 +}$ . (a) How many electrons per Mn now? (b) Balance the neutral half-reaction — first in acid, then convert to neutral/basic. (c) Explain why it stops at $+ 4$ instead of falling to $+ 2$ .

Recall Solution

(a) Mn goes $+ 7 \to + 4$ , a drop of 3, so 3 electrons. (b) We use a two-stage trick, because balancing H is easiest with free $H^{+}$ first, then we neutralise those $H^{+}$ to fit a neutral medium where free $H^{+}$ is scarce. Stage 1 — balance in acid (pretend $H^{+}$ is available):

Skeleton: $M n O_{4}^{-} \to M n O_{2}$ . Add $3 e^{-}$ to the left (gain of 3): $M n O_{4}^{-} + 3 e^{-} \to M n O_{2}$ .
Oxygen: 4 O left, 2 O right → add $2 H_{2} O$ on the right: $M n O_{4}^{-} + 3 e^{-} \to M n O_{2} + 2 H_{2} O$ .
Hydrogen: 4 H now on the right, 0 on left → add $4 H^{+}$ on the left: $M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O (acid form)$ Stage 2 — convert to neutral/basic (WHY: there is no pool of free $H^{+}$ in neutral water). We cancel each $H^{+}$ by adding the same number of $O H^{-}$ to both sides — that is legal because $O H^{-}$ is added equally, it changes nothing chemically, but $H^{+} + O H^{-} \to H_{2} O$ ties up the free protons into water:
Add $4 O H^{-}$ to both sides: left becomes $M n O_{4}^{-} + 4 H^{+} + 4 O H^{-} + 3 e^{-}$ ; the $4 H^{+} + 4 O H^{-}$ merge into $4 H_{2} O$ .
Left: $M n O_{4}^{-} + 4 H_{2} O + 3 e^{-}$ ; Right: $M n O_{2} + 2 H_{2} O + 4 O H^{-}$ .
Cancel $2 H_{2} O$ common to both sides: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$ Charge check: left $= (- 1) + 3 (- 1) = - 4$ ; right $= 4 (- 1) = - 4$ . ✓ (c) Falling all the way to $M n^{2 +}$ requires 8 $H^{+}$ per Mn (from the acidic half-reaction). In neutral medium there simply aren't enough $H^{+}$ around to pay that cost, so the reduction halts at the $+ 4$ solid $M n O_{2}$ , which needs no free $H^{+}$ . Fewer protons ⇒ shallower fall.

Q8. Explain why dilute $H_{2} S O_{4}$ — not HCl — is used to acidify KMnO₄ in a titration. Support your answer with $E^{\circ}$ values: $E^{\circ} (M n O_{4}^{-} / M n^{2 +}) = + 1.51 V$ , $E^{\circ} (C l_{2} / C l^{-}) = + 1.36 V$ .

Recall Solution

A species will oxidise another only if its own reduction potential is higher (it "wants" electrons more). Here $E^{\circ} (M n O_{4}^{-} / M n^{2 +}) = + 1.51 V > E^{\circ} (C l_{2} / C l^{-}) = + 1.36 V$ , so permanganate is strong enough to rip electrons off chloride: $2 M n O_{4}^{-} + 16 H^{+} + 10 C l^{-} \to 2 M n^{2 +} + 5 C l_{2} + 8 H_{2} O$ This wastes KMnO₄ on the chloride instead of on your intended reductant, so the titre reads too high. $H_{2} S O_{4}$ 's $S O_{4}^{2 -}$ is not oxidised, so it is a "spectator" acid — it supplies $H^{+}$ without stealing oxidiser. See Standard Electrode Potential & E° values.

Level 4 — Synthesis

Q9. Trace the full journey of manganese from pyrolusite to purple KMnO₄. For each step give (i) the equation, (ii) the change in Mn oxidation state, (iii) why that particular oxidiser is used.

The flow diagram below summarises the journey: two boxes are joined by two arrows, the first driven by air's $O_{2}$ (orange), the second by the anode (red).

Recall Solution

Step 1 — alkaline oxidative fusion ( $+ 4 \to + 6$ ): $2 M n O_{2} + 4 K O H + O_{2} fuse 2 K_{2} M n O_{4} + 2 H_{2} O$ Mn climbs $+ 4 \to + 6$ (green manganate). Air's $O_{2}$ is a cheap, adequate oxidiser to lift Mn two steps in molten alkali. Step 2 — electrolytic oxidation ( $+ 6 \to + 7$ ). In a half-cell we still balance mass and charge. Because $M n O_{4}^{2 -}$ and $M n O_{4}^{-}$ have the same 4 oxygens and no hydrogens, no water or protons are needed — only one electron per Mn leaves at the anode: $M n O_{4}^{2 -} \to M n O_{4}^{-} + e^{-}$ Charge check: left $= - 2$ ; right $= (- 1) + (- 1) = - 2$ . ✓ (Doubling gives $2 M n O_{4}^{2 -} \to 2 M n O_{4}^{-} + 2 e^{-}$ .) Mn climbs $+ 6 \to + 7$ (purple permanganate). $O_{2}$ cannot do this last step — it isn't a strong enough oxidiser to strip the final electron. The anode forcibly removes electrons and is effectively an unlimited-strength oxidiser, so it completes the job cleanly without adding foreign chemicals. Net oxidation-state path: $M n O_{2} + 4 O_{2} M n O_{4}^{2 -} + 6 anode M n O_{4}^{-} + 7$ .

Q10. For the chromate ⇌ dichromate system $2 C r O_{4}^{2 -} + 2 H^{+} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O$ , predict the colour and shift direction when we (a) add $H_{2} S O_{4}$ , (b) add NaOH, (c) add solid NaOH to an orange dichromate solution. Justify with Le Chatelier's Principle and confirm no redox occurs.

Recall Solution

(a) Add $H_{2} S O_{4}$ : raises $[H^{+}]$ (a reactant) → equilibrium shifts right → more $C r_{2} O_{7}^{2 -}$ → solution turns orange. (b) Add NaOH: $O H^{-}$ consumes $H^{+}$ ( $H^{+} + O H^{-} \to H_{2} O$ ), removing a reactant → shift left → more $C r O_{4}^{2 -}$ → yellow. (c) Same logic as (b): adding base to orange dichromate removes $H^{+}$ , shifting left, so the orange turns yellow. No redox check: Cr is $+ 6$ in both $C r O_{4}^{2 -}$ and $C r_{2} O_{7}^{2 -}$ (verify: $C r O_{4}^{2 -}$ : $C r - 8 = - 2 \Rightarrow C r = + 6$ ; $C r_{2} O_{7}^{2 -}$ : $2 C r - 14 = - 2 \Rightarrow C r = + 6$ ). Same oxidation state ⇒ this is condensation/acid–base, not redox.

Level 5 — Mastery

Q11. $25.0 mL$ of an $F e^{2 +}$ solution is titrated with $0.0200 M$ $K M n O_{4}$ in dilute $H_{2} S O_{4}$ . The purple colour just persists after $22.0 mL$ of $K M n O_{4}$ . Find the molarity of the $F e^{2 +}$ solution.

Recall Solution

Reaction: $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ — so 1 mol $M n O_{4}^{-}$ reacts with 5 mol $F e^{2 +}$ . Moles of $K M n O_{4}$ used: $0.0200 \times 0.0220 = 4.40 \times 1 0^{- 4} mol$ . Moles of $F e^{2 +}$ : $5 \times 4.40 \times 1 0^{- 4} = 2.20 \times 1 0^{- 3} mol$ . Molarity of $F e^{2 +}$ : $\frac{2.20 \times 1 0 ^{- 3}}{0.0250} = 0.0880 M$ . See Volumetric Analysis / Titrations.

Q12. The same $F e^{2 +}$ solution ( $0.0880 M$ , $25.0 mL$ ) is instead titrated with $K_{2} C r_{2} O_{7}$ . If the dichromate is $0.0200 M$ , what volume is needed?

Recall Solution

Reaction: $C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O$ — 1 mol dichromate per 6 mol $F e^{2 +}$ . Moles $F e^{2 +}$ : $0.0880 \times 0.0250 = 2.20 \times 1 0^{- 3} mol$ . Moles $C r_{2} O_{7}^{2 -}$ needed: $\frac{2.20 \times 1 0 ^{- 3}}{6} = 3.667 \times 1 0^{- 4} mol$ . Volume: $\frac{3.667 \times 1 0 ^{- 4}}{0.0200} = 0.01833 L = 18.3 mL$ .

Q13. A $1.000 g$ impure sample of oxalic acid dihydrate ( $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ , $M = 126 g/mol$ ) is dissolved and needs $30.0 mL$ of $0.0500 M$ acidic $K M n O_{4}$ to reach the endpoint. Find the mass and percentage of oxalic acid dihydrate in the sample.

Recall Solution

Reaction: $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ — ratio 2 $M n O_{4}^{-}$ : 5 $C_{2} O_{4}^{2 -}$ . Moles $K M n O_{4}$ : $0.0500 \times 0.0300 = 1.50 \times 1 0^{- 3} mol$ . Moles oxalate: $\frac{5}{2} \times 1.50 \times 1 0^{- 3} = 3.75 \times 1 0^{- 3} mol$ . Mass of $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ : $3.75 \times 1 0^{- 3} \times 126 = 0.4725 g$ . Percentage: $\frac{0.4725}{1.000} \times 100 = 47.25 %$ .

Q14. Compare the "oxidising power per mole" of $K M n O_{4}$ (acidic) vs $K_{2} C r_{2} O_{7}$ (acidic) purely by electrons accepted per formula unit, then explain why $E^{\circ}$ (not electron count) is the true measure of strength.

Recall Solution

Electron count: 1 mol $M n O_{4}^{-}$ accepts 5 $e^{-}$ ; 1 mol $C r_{2} O_{7}^{2 -}$ accepts 6 $e^{-}$ . By raw electron capacity per mole, dichromate wins ( $6 > 5$ ). But capacity ≠ strength. Strength = how badly it wants each electron = its reduction potential. Here $E^{\circ} (M n O_{4}^{-} / M n^{2 +}) = + 1.51 V > E^{\circ} (C r_{2} O_{7}^{2 -} / C r^{3 +}) = + 1.33 V$ , so permanganate is the stronger oxidiser (higher electron-hunger) even though it grabs fewer electrons per formula unit. Analogy: a bucket that holds more water (6 vs 5) isn't necessarily the one that sucks water up harder. See Standard Electrode Potential & E° values.

Connections

Balancing Redox Reactions (ion-electron method) — the machinery behind every half-reaction here.
Standard Electrode Potential & E° values — why $+ 1.51 > + 1.36 > + 1.33$ decides Q8 and Q14.
Volumetric Analysis / Titrations — the mole-ratio calculations of Q11–Q13.
Le Chatelier's Principle — the chromate ⇌ dichromate shifts of Q10.
Oxidation States of Transition Metals · Colour & d-d Transitions — the L1 recognition facts.
Parent: topic note.