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