Question

In the following reaction, which element is oxidized?
$8NaI+5H_{2}S{O}_4\to 4I_2+H_{2}S+4N{a}_{2}S{O}_4+4H_{2}O$

• A. Sodium
• B. Iodine
• C. Sulfur
• D. Hydrogen

Answer 1

The correct option is B Iodine

$NaI$

Oxidation state of $Na$ is $+1$.

$+1+x=0x=-1.$

$\therefore$ Oxidation state of $I$ before reaction is $-1$. After reaction it is $0$. Because in nature, it is available as $I_2$.

$\therefore$ Iodine is oxidised from $-1$ to $0$.