JEE Main · 2023 · Shift-ImediumRDX-014

2IO3- + xI- + 12H+ 6I2 + 6H2O. What is the value of x?

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

$\mathrm{2IO_3^{-} + xI^{-} + 12H^{+} \rightarrow 6I_2 + 6H_2O}$ . What is the value of $x$ ?

Options

a
2
b
12
c
10
✓
d
6

Correct Answerc

Detailed Solution

Step 1 — Balance by atom count and charge:

Iodine balance:

Left: $2$ (from $\mathrm{IO_3^-}$ ) $+ x$ (from $\mathrm{I^-}$ )
Right: $6 \times 2 = 12$ I atoms (from $\mathrm{6I_2}$ ) $2 + x = 12 \Rightarrow x = 10$

Step 2 — Verify with charge balance:

Left: $2(-1) + 10(-1) + 12(+1) = -2 - 10 + 12 = 0$
Right: $0$ (neutral molecules) ✓ Charge balanced

Step 3 — Verify with oxidation states:

I in $\mathrm{IO_3^-}$ : $+5$ ; in $\mathrm{I^-}$ : $-1$ ; in $\mathrm{I_2}$ : $0$
Reduction: $\mathrm{IO_3^-}$ : $+5 \rightarrow 0$ (gain of 5e⁻ per I, total $2 \times 5 = 10$ e⁻ gained)
Oxidation: $\mathrm{I^-}$ : $-1 \rightarrow 0$ (loss of 1e⁻ per I, total $x \times 1 = x$ e⁻ lost)
For balance: $x = 10$ ✓

Answer: Option (3) — $x = 10$

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