Question

Iodate ions $\left(I{O}_3^{-}\right)$ can be reduced to iodine by iodide ions. The half equation which represent the redox reaction are:
$I{O}_3^{-}(aq.)+6H^{+}(aq.)+5e^{-}\to \frac{1}{2}{I}_2\left(s\right)+3H_{2}O\left(l\right)$
$I^{-}(aq.)\to \frac{1}{2}{I}_2\left(s\right)+e^{-}$
How many moles of iodine are produced for every mole of iodate ions consumed in the reaction ?

• A. $0.5$
• B. $0.1$
• C. $2.5$
• D. $3$

Answer 1

Answer: (D)

$3$

$I^{-}(aq.)\to \frac{1}{2}{I}_2\left(s\right)+e^{-}]\times 5$
$I{O}_3(aq.)+6H^{+}(aq.)+5e\to \frac{1}{2}{I}_2\left(s\right)+3H_{2}O\left(l\right)$
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$5I^{-}(aq.)+I{O}_3^{-}(aq.)+6H^{+}(aq.)\to 3I_2+3H_{2}O\left(l\right)$