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. I^-(aq) -> 1/2 I2(s) + e^- } x 5. IO3^-(aq) + 6H+(aq) + 5e^- -> 1/2 I2(s) + 3H2O(l). 5I^-(aq) + IO3^-(aq) + 6H+(aq) -> 3I2 + 3H2O(l). 3 moles of iodine are produced.

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