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JAMB Chemistry 1995 Question 27

Question 27

Redox Reactions

Cr2O72−(aq) + 14H+(aq) + 6I−(aq) → 2Cr3+(aq) + 3I2(g) + 7H2O(l). The change in the oxidation number of oxygen in the equation above is

  1. A.0
  2. B.1
  3. C.2
  4. D.7
Show the answer and explanation

Answer: A (0)

Explanation

Oxygen is −2 in Cr2O72− and still −2 in H2O, so its oxidation number does not change. Only chromium and iodine change.