Calculate the number of electrons lost or gained the changes:
$3 I_{2} + 6 N a O H \to N a I O_{3} + 5 N a I + 3 H_{2} O$

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Solution

$3 I_{2} + 6 N a O H ⟶ N a I O_{3} + 5 N a I + 3 H_{2} O$
Here iodine is going from $0 \to + 5$ oxidation state,
so total lost electrons are $5$ .

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