The reaction: 2IO3^- + xI^- + 12H^+ 6I2 + 6H2O What is the value of x?

The reaction: 2IO3^- + xI^- + 12H^+ 6I2 + 6H2O What is the value… — Solution | JEE Chemistry

MCQEasyJEE 2023Oxidation Number & Redox Reactions

The reaction: $2IO_3^- + xI^- + 12H^+ \to 6I_2 + 6H_2O$ What is the value of $x$ ?

Correct answer:B

Standard Method

Given: $2IO_3^- + xI^- + 12H^+ \to 6I_2 + 6H_2O$

Find: The value of $x$ .

The oxidation number of iodine in $IO_3^-$ is $+5$ , while in $I_2$ it is $0$ . Hence, $IO_3^-$ is reduced.

Iodide ion $I^-$ is oxidized to $I_2$ , so it acts as the reducing species.

Using the electron balance, the balanced ionic form is:

IO_3^- + 6H^+ + 5I^- \to 3I_2 + 3H_2O

To obtain $6I_2$ , multiply the entire equation by $2$ :

2IO_3^- + 12H^+ + 10I^- \to 6I_2 + 6H_2O

Comparing with $2IO_3^- + xI^- + 12H^+ \to 6I_2 + 6H_2O$ , we get $x = 10$ .

Therefore, the correct option is B.

Treating $IO_3^-$ and $I^-$ as if they undergo the same change in oxidation number is incorrect because one is reduced and the other is oxidized. First assign oxidation numbers, then balance electron gain and loss.
Balancing only atoms and ignoring charge balance is wrong in a redox reaction. Use the electron balance or half-reaction method so that both mass and charge are balanced.
Forgetting to multiply the intermediate equation by $2$ leads to the wrong value of $x$ because the product side must contain $6I_2$ , not $3I_2$ . Match the required stoichiometric coefficients before reading off $x$ .

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