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Describe with chemical equations, what happens when K2Cr2O7 reacts with an acidified solution of Kl?

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Question

Describe with chemical equations, what happens when K2Cr2O7 reacts with an acidified solution of Kl?

Chemical Equations/Structures
Short Answer
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Solution

When acidified potassium dichromate (K2Cr2O7) reacts with an acidified solution of potassium iodide (KI), iodine (I2) is liberated. In this redox reaction:

Dichromate ion \[\ce{(Cr2O^{2-}_7)}\] is reduced to Cr3+

Iodide ion (I) is oxidised to Iodine (I2)

\[\ce{K2Cr2O7 + 14HCl + 6KI -> 3I2 + 2CrCl3 + 6KCl + 7H2O}\]

\[\ce{Cr2O^{2-}_7 + 14H+ + 6I- -> 2Cr^{3+} + 3I2 + 7H2O}\]

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Chapter 4: d-and ƒ-Block Elements - REVIEW EXERCISES [Page 484]

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Nootan Chemistry [English] Class 12 ISC
Chapter 4 d-and ƒ-Block Elements
REVIEW EXERCISES | Q 8.39 (ii) | Page 484
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