Skip to content
Question 68 of 72

Q.How many moles of I2I_2 are liberated when 1 mole of potassium dichromate react with potassium iodide ?

(a) 3
(b) 1
(c) 4
(d) 2
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2025MCQ· 1mImportance★★★★★
94% · 68/72 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Potassium dichromate is a 6-electron oxidant per formula unit (each of its two Cr atoms is reduced from +6 to +3); since forming one I2I_2 from 2I−2I^- needs only 2 electrons, one mole of dichromate liberates 3 moles of I2I_2.

Reduction half-reaction: Cr2O72−+14H++6e−→2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O Each dichromate ion accepts 6 electrons (2 Cr atoms ×\times 3 electrons each, as Cr goes from +6+6 to +3+3).

Oxidation half-reaction: 2I−→I2+2e−2I^- \rightarrow I_2 + 2e^- Each I2I_2 molecule releases 2 electrons.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.