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Question

Iodate ions $$(IO_{3}^{-}) $$ can be reduced to iodine by iodide ions. The half equation which represent the redox reaction are: 
           $$IO_{3}^{-} (aq.) + 6H^{+} (aq.) + 5e^{-} \rightarrow \dfrac{1}{2} I_{2} (s) + 3H_{2}O (l) $$
            $$I^{-} (aq.) \rightarrow \dfrac{1}{2} I_{2} (s) + e^{-} $$
How many moles of iodine are produced for every mole of iodate ions consumed in the reaction ?


A
0.5
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B
0.1
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C
2.5
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D
3
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Solution

The correct option is C $$3$$
$$I^{-} (aq.) \rightarrow \dfrac{1}{2} I_{2} (s) + e^{-} ] \times 5 $$
$$IO_{3} (aq.) + 6H^{+} (aq.) + 5e \rightarrow \dfrac{1}{2} I_{2} (s) + 3H_{2}O (l) $$ 
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$$5I^{-} (aq.) + IO_{3}^{-} (aq.) + 6H^{+} (aq.) \rightarrow 3I_{2} + 3H_{2}O (l) $$ 

Chemistry

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