Q70 / 180 · AIIMS 2005 · Chemistry
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
- Correct answerA$5,6,3,3$
- B$5,3,6,3$
- C$3,5,3,6$
- D$5,6,5,5$
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, $a=5, b=6, c=3, d=3$.
Hence, the correct answer is option: 'A'.