Intuition The big picture (WHY these matter)Both Mn and Cr sit in the d-block where the metal can give up several electrons. In KMnO₄ the Mn is in its highest oxidation state +7 ; in K₂Cr₂O₇ the Cr is in +6 . A metal already stripped of electrons to its maximum is desperate to grab electrons back — that hunger for electrons is exactly what an oxidising agent is. So both compounds are powerful oxidisers, and the whole topic is just bookkeeping of "how many electrons does it grab, and in what medium."
Intuition WHY a two-step prep?The natural ore has Mn or Cr in a low/intermediate oxidation state. To reach +7 (Mn) or +6 (Cr) you must oxidise the metal. Nature won't hand you +7 for free, so industrial prep = (1) fuse/roast with an oxidiser in alkali to climb oxidation states, then (2) acidify / electrolyse to convert to the final salt.
Step 1 — Oxidise Mn(+4) → Mn(+6) (manganate) by alkaline fusion:
2 M n O 2 + 4 K O H + O 2 → fuse 2 K 2 M n O 4 + 2 H 2 O 2MnO_2 + 4KOH + O_2 \xrightarrow{\text{fuse}} 2K_2MnO_4 + 2H_2O 2 M n O 2 + 4 K O H + O 2 fuse 2 K 2 M n O 4 + 2 H 2 O
Air's O₂ (or KNO₃) is the oxidiser. Product = green manganate M n O 4 2 − MnO_4^{2-} M n O 4 2 −
Step 2 — Oxidise Mn(+6) → Mn(+7) (permanganate) by electrolytic oxidation (or by disproportionation/Cl₂/O₃):
2 M n O 4 2 − → electrolytic oxidation 2 M n O 4 − + 2 e − 2MnO_4^{2-} \xrightarrow{\text{electrolytic oxidation}} 2MnO_4^- + 2e^- 2 M n O 4 2 − electrolytic oxidation 2 M n O 4 − + 2 e −
Product = purple permanganate M n O 4 − MnO_4^- M n O 4 −
Worked example Why electrolytic and not just "more air"?O₂ stops at +6. To push the last electron off (Mn⁶⁺→Mn⁷⁺) you need a stronger oxidiser. At the anode electrons are forcibly removed — the anode is the strongest possible oxidiser, so it does what air cannot. Why this step? It cleanly converts green→purple without adding a foreign reagent.
Step 1 — Roast chromite with alkali + air, Cr(+3) → Cr(+6) chromate:
4 F e C r 2 O 4 + 8 N a 2 C O 3 + 7 O 2 → 8 N a 2 C r O 4 + 2 F e 2 O 3 + 8 C O 2 4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2 4 F e C r 2 O 4 + 8 N a 2 C O 3 + 7 O 2 → 8 N a 2 C r O 4 + 2 F e 2 O 3 + 8 C O 2
Yellow chromate C r O 4 2 − CrO_4^{2-} C r O 4 2 −
Step 2 — Acidify (chromate → dichromate):
2 N a 2 C r O 4 + H 2 S O 4 → N a 2 C r 2 O 7 + N a 2 S O 4 + H 2 O 2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O 2 N a 2 C r O 4 + H 2 S O 4 → N a 2 C r 2 O 7 + N a 2 S O 4 + H 2 O
Step 3 — Convert Na salt to K salt (K₂Cr₂O₇ is less soluble, crystallises out):
N a 2 C r 2 O 7 + 2 K C l → K 2 C r 2 O 7 + 2 N a C l Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl N a 2 C r 2 O 7 + 2 K C l → K 2 C r 2 O 7 + 2 N a C l
Intuition WHY the medium mattersM n O 4 − MnO_4^- M n O 4 − how far it falls depends on how many H⁺ are available. Acid (lots of H⁺) lets it grab 5 electrons (→Mn²⁺). Neutral/faintly alkaline (few H⁺) it grabs only 3 (→MnO₂). Strong alkali it grabs only 1 (→MnO₄²⁻).
Medium
Mn ends as
Electrons
Colour change
Acidic
M n 2 + Mn^{2+} M n 2 + 5
purple → colourless
Neutral/weak alkali
M n O 2 MnO_2 M n O 2 3
purple → brown ppt
Strong alkali
M n O 4 2 − MnO_4^{2-} M n O 4 2 − 1
purple → green
Worked example Acidic KMnO₄ + oxalic acid (a classic titration)Oxidation: C 2 O 4 2 − → 2 C O 2 + 2 e − C_2O_4^{2-} \to 2CO_2 + 2e^- C 2 O 4 2 − → 2 C O 2 + 2 e − M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O 2 M n O 4 − + 5 C 2 O 4 2 − + 16 H + → 2 M n 2 + + 10 C O 2 + 8 H 2 O 2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O 2 M n O 4 − + 5 C 2 O 4 2 − + 16 H + → 2 M n 2 + + 10 C O 2 + 8 H 2 O Why ×5 and ×2? Electrons lost (10) must equal electrons gained (10) → LCM of 5 and 2. Warm to ~60 °C because the reaction is slow at start (Mn²⁺ produced autocatalyses it).
Worked example Acidic KMnO₄ + Fe²⁺F e 2 + → F e 3 + + e − Fe^{2+} \to Fe^{3+} + e^- F e 2 + → F e 3 + + e − M n O 4 − + 5 F e 2 + + 8 H + → M n 2 + + 5 F e 3 + + 4 H 2 O MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O M n O 4 − + 5 F e 2 + + 8 H + → M n 2 + + 5 F e 3 + + 4 H 2 O
Worked example Dichromate + Fe²⁺ (volumetric estimation of iron)F e 2 + → F e 3 + + e − Fe^{2+} \to Fe^{3+} + e^- F e 2 + → F e 3 + + e − C r 2 O 7 2 − + 6 F e 2 + + 14 H + → 2 C r 3 + + 6 F e 3 + + 7 H 2 O Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O C r 2 O 7 2 − + 6 F e 2 + + 14 H + → 2 C r 3 + + 6 F e 3 + + 7 H 2 O Why ×6? One dichromate takes 6 e⁻, each Fe²⁺ gives 1 e⁻ → need 6 Fe²⁺.
Worked example Dichromate + I⁻ (iodometry)2 I − → I 2 + 2 e − 2I^- \to I_2 + 2e^- 2 I − → I 2 + 2 e − C r 2 O 7 2 − + 6 I − + 14 H + → 2 C r 3 + + 3 I 2 + 7 H 2 O Cr_2O_7^{2-} + 6I^- + 14H^+ \to 2Cr^{3+} + 3I_2 + 7H_2O C r 2 O 7 2 − + 6 I − + 14 H + → 2 C r 3 + + 3 I 2 + 7 H 2 O
Intuition WHY dichromate is asecondary standard but KMnO₄ is not used as a primary standard
K₂Cr₂O₇ can be obtained pure, is non-hygroscopic, and doesn't react with cold dilute HCl → its solution stays at known concentration. KMnO₄ is hard to get 100% pure and slowly oxidises water, so it must be standardised before use.
Common mistake Steel-manning the common trapsTrap 1: "KMnO₄ always gives Mn²⁺." Why it feels right: acidic titrations are taught first. Fix: only in acid . Neutral → MnO₂ (3e⁻), strong base → green MnO₄²⁻ (1e⁻).
Trap 2: "Chromate→dichromate is a redox/oxidation." Why it feels right: the colour changes dramatically, like a redox. Fix: Cr is +6 on both sides — it's an acid–base condensation, not redox.
Trap 3: Using HCl with KMnO₄ titrations. Why it feels right: HCl is a common acid. Fix: KMnO₄ oxidises Cl⁻ to Cl₂, consuming oxidiser → wrong result. Use dilute H₂SO₄ .
Trap 4: Balancing electrons wrong — forgetting that electrons lost = electrons gained . Fix: always take LCM (e.g. 5 & 2 → 10).
Recall Feynman: explain to a 12-year-oldImagine two greedy kids. The purple kid (KMnO₄) is so greedy he can grab 5 candies (electrons) if the room is full of candy (acid), but only 3 or 1 if the room is emptier. The orange kid (K₂Cr₂O₇) always grabs 6 candies but only plays in the acid room. When they grab candies they change colour: purple→clear, orange→green. To make them in the first place, we "charge them up" by cooking the dull ore with washing-soda and air (and a bit of electricity for the purple one) until the metal is super-greedy.
Mnemonic Remember the numbers"7→2 give Five (acid), 7→4 give Three (neutral), 7→6 give One (base)" for Mn.
"Dichromate Drinks 6, Demands acid" — 6 electrons, acidic only.
Colours: P ermanganate P urple, M anganate M ossy-green; C hromate C anary-yellow, D ichromate D eep-orange.
KMnO₄ — oxidation state of Mn +7
K₂Cr₂O₇ — oxidation state of Cr +6
Ore of manganese for KMnO₄ prep Pyrolusite, MnO₂
Ore of chromium for K₂Cr₂O₇ prep Chromite, FeCr₂O₄
Step 1 of KMnO₄ prep (equation) 2 M n O 2 + 4 K O H + O 2 → 2 K 2 M n O 4 + 2 H 2 O 2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O 2 M n O 2 + 4 K O H + O 2 → 2 K 2 M n O 4 + 2 H 2 O (green manganate)
Step 2 of KMnO₄ prep Electrolytic oxidation of MnO₄²⁻ → MnO₄⁻ (green→purple)
Colour of manganate vs permanganate Manganate green (+6), permanganate purple (+7)
Acidic KMnO₄ half-reaction M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O (5 e⁻)
Neutral KMnO₄ product & electrons MnO₂ (Mn +4), 3 electrons, brown ppt
Strong-alkali KMnO₄ product & electrons MnO₄²⁻ (green), 1 electron
Dichromate half-reaction C r 2 O 7 2 − + 14 H + + 6 e − → 2 C r 3 + + 7 H 2 O Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O C r 2 O 7 2 − + 14 H + + 6 e − → 2 C r 3 + + 7 H 2 O (6 e⁻)
Chromate ⇌ dichromate equilibrium 2 C r O 4 2 − + 2 H + ⇌ C r 2 O 7 2 − + H 2 O 2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O 2 C r O 4 2 − + 2 H + ⇌ C r 2 O 7 2 − + H 2 O (acid→orange, base→yellow)
Is chromate→dichromate a redox? No — Cr stays +6, only condensation/acid-base
Why dilute H₂SO₄ not HCl with KMnO₄? HCl's Cl⁻ is oxidised to Cl₂, consuming KMnO₄
KMnO₄ + oxalic acid balanced eqn 2 M n O 4 − + 5 C 2 O 4 2 − + 16 H + → 2 M n 2 + + 10 C O 2 + 8 H 2 O 2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O 2 M n O 4 − + 5 C 2 O 4 2 − + 16 H + → 2 M n 2 + + 10 C O 2 + 8 H 2 O K₂Cr₂O₇ + Fe²⁺ balanced eqn C r 2 O 7 2 − + 6 F e 2 + + 14 H + → 2 C r 3 + + 6 F e 3 + + 7 H 2 O Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O C r 2 O 7 2 − + 6 F e 2 + + 14 H + → 2 C r 3 + + 6 F e 3 + + 7 H 2 O Why K₂Cr₂O₇ is a primary/secondary standard Pure, stable, non-hygroscopic; KMnO₄ isn't
Why oxalic acid–KMnO₄ titration is warmed Slow initially; Mn²⁺ autocatalyses it
Standard Electrode Potential & E° values — explains why E°(MnO₄⁻)=+1.51 > E°(Cr₂O₇²⁻)=+1.33Oxidation States of Transition Metals — why +7 (Mn) and +6 (Cr) are accessibleBalancing Redox Reactions (ion-electron method) Volumetric Analysis / Titrations — permanganometry & dichrometryLe Chatelier's Principle — the chromate⇌dichromate shiftColour & d-d Transitions — why MnO₄⁻ is intensely purple (charge transfer)
roast with alkali plus air
Na to K exchange with KCl
High oxidation state Mn +7, Cr +6
Green manganate MnO4 2-, Mn +6
Purple permanganate MnO4-, Mn +7
Yellow chromate CrO4 2-, Cr +6
Orange dichromate Cr2O7 2-
Intuition Hinglish mein samjhoDekho, KMnO₄ aur K₂Cr₂O₇ dono strong oxidising agents hain. Iska reason simple hai: KMnO₄ mein Mn apni highest oxidation state +7 par hai aur K₂Cr₂O₇ mein Cr +6 par. Jab metal apne saare electrons de chuka hota hai, to wo wapas electrons grab karne ke liye bahut hungry hota hai — yahi to oxidiser ka kaam hai, doosre ko oxidise karna aur khud reduce ho jaana.
Preparation ka logic yaad rakho: ore mein metal low state mein hota hai, isliye pehle usse alkali + hawa (O₂) ke saath fuse/roast karke oxidation state badhate hain. KMnO₄ ke liye MnO₂ → green manganate (+6), phir electrolytic oxidation se purple permanganate (+7). K₂Cr₂O₇ ke liye chromite → yellow chromate (+6), phir acid daalo to orange dichromate ban jaata hai. Yaad rahe — chromate se dichromate koi redox nahi hai, Cr dono taraf +6 hi rehta hai, sirf acid-base equilibrium hai.
Ab oxidising reactions mein sabse important baat: medium kya hai . Acidic KMnO₄ → Mn²⁺ banta hai, 5 electron leta hai (purple se colourless). Neutral → MnO₂, 3 electron, brown. Strong base → green manganate, 1 electron. Dichromate hamesha acidic medium mein kaam karta hai, 6 electron leta hai, orange se green. Exam mein equations balance karte waqt yaad rakho — electrons lost = electrons gained, LCM lo. Aur ek important tip: KMnO₄ titration mein dilute H₂SO₄ use karo, HCl nahi, warna Cl⁻ oxidise hokar Cl₂ ban jaata hai aur result galat aa jaata hai.
