\(2 I O^ − _3 + x I ^− + 12 H ^+ ⟶ 6 I_ 2 + 6 H _2 O\) What is the value of x?
- 2
- 10
- 6
- 12
The Correct Option is B
Solution and Explanation
To balance the given redox reaction: The oxidation number of I in IO$_3^-$ is +5, while in I$_2$, it is 0. The n-factor for IO$_3^-$ is 5 because each IO$_3^-$ gains 5 electrons to become I$_2$. \item I$^-$ is oxidized to I$_2$, so its n-factor is 1. \end{itemize} To determine the value of $x$, we use the molar ratio of IO$_3^-$ to I$^-$, which is 1:5: \[ \text{IO}_3^- + 6\text{H}^+ + 5\text{I}^- \to 3\text{I}_2 + 3\text{H}_2\text{O} \] To obtain 6 moles of I$_2$, the equation is multiplied by 2: \[ 2\text{IO}_3^- + 12\text{H}^+ + 10\text{I}^- \to 6\text{I}_2 + 6\text{H}_2\text{O} \] Thus, $x = 10$.
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