Iodate ions (IO3^-) can be reduced to iodine by iodide ions. The half equation which represent the redox reaction are : IO3^-(aq.) + 6H+(aq.) + 5e^- - 1/2 I2(s) + 3H2O(l) (i) ; I^-(aq.) - 1/2 I2(s)

Question

Iodate ions (IO3^-) can be reduced to iodine by iodide ions. The half equation which represent the redox reaction are : IO3^-(aq.) + 6H+(aq.) + 5e^- -> 1/2 I2(s) + 3H2O(l) (i) ; I^-(aq.) -> 1/2 I2(s) + e^- (ii) . How many moles of iodine are produced for every mole of iodate ions consumed in the reaction?

Prashant Kotian · Accepted Answer

Correct answer: D. Step 1: To find the overall stoichiometry, we must balance the electrons transferred. Multiplying the oxidation half-reaction (ii) by 5 yields: 5 I^-(aq.) ---> 2.5 I2(s) + 5e^-. Step 2: Adding this to the reduction half-reaction (i) yields the balanced overall ionic equation: IO3^-(aq.) + 5 I^-(aq.) + 6 H^+(aq.) ---> 3 I2(s) + 3 H2O(l). Step 3: From the balanced equation, 1 mole of iodate ions (IO3^-) reacts to produce 3 moles of iodine (I2). Thus, the correct option is (d).

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