Chemistry › Redox Reactions › Balancing of redox reactions
In the balanced chemical reaction $\mathrm{IO}_3^{-}+a \mathrm{I}^{-}+b \mathrm{H}^{+} \rightarrow c…
In the balanced chemical reaction
$\mathrm{IO}_3^{-}+a \mathrm{I}^{-}+b \mathrm{H}^{+} \rightarrow c \mathrm{H}_2 \mathrm{O}+d \mathrm{I}_2$
$a, b, c$ and $d$ respectively corresponds to
$5,6,3,3$ $5,3,6,3$ $3,5,3,6$ $5,6,5,5$ Solution
$\begin{aligned}
& \text { }: \mathrm{IO}_3^{-}+a \mathrm{I}^{-}+b \mathrm{H}^{+} \rightarrow c \mathrm{H}_2 \mathrm{O}+d \mathrm{I}_2 \\
& \text { Step } 1: \mathrm{I}^{-} \rightarrow \mathrm{I}_2 \text { (oxidation) } \\
& \quad \mathrm{IO}_3^{-} \rightarrow \mathrm{I}_2 \text { (reduction) } \\
& \text { Step } 2: 2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+} \rightarrow \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O} \\
& \text { Step } 3: 2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+}+10 e^{-} \rightarrow \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O} \\
& \quad 2 \mathrm{I}^{-} \rightarrow \mathrm{I}_2+2 e^{-} \\
& \text {Step } 4: 2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+}+10 e^{-} \rightarrow \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O} \\
& \quad\left[2 \mathrm{I}^{-} \rightarrow \mathrm{I}_2+2 e\right] \times 5
\end{aligned}$
$\begin{aligned}
& \text { Step } 5: 2 \mathrm{IO}_3^{-}+10 \mathrm{I}^{-}+12 \mathrm{H}^{+} \rightarrow 6 \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O} \\
& \mathrm{IO}_3^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}
\end{aligned}$
On comparing, $a=5, b=6, c=3, d=3$.
Asked in: NEET 2009 (Screening)
Practice more Redox Reactions questions on Aicharya