Write balanced net ionic equation for the following reaction in acidic solution.
$I O_{3}^{-} (a q) + R e (s) \to R e O_{4}^{-} (a q) + I^{-} (a q)$

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Solution

Given equation
${I O}_{3}^{-} (a q) + R e (s) ⟶ {R e O}_{4}^{-} (a q) + I^{-} (a q)$
Solving by ion electron method
We get, the balanced equation is
${7 I O}_{3}^{-} + 6 R e + 3 H_{2} O ⇌ 6 {R e O}_{4}^{-} + 7 I^{-} + 6 H^{+}$
Hence balanced both sides same number of atoms and charge.

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