Question

\(2 I O^ − _3 + x I ^− + 12 H ^+ ⟶ 6 I_ 2 + 6 H _2 O\) What is the value of x?

• A. 2
• B. 10
• C. 6
• D. 12

Answer 1

The Correct Option is B

To solve the given chemical equation and find the value of \(x\), we must ensure the equation is balanced. The given equation is:

\(2 IO_3^- + x I^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2O\) 

To balance this equation, we follow these steps:

1. Identify the oxidation states: The iodine atoms in \(\text{IO}_3^-\) are in the +5 oxidation state, while in \(I^-\) they are in the -1 oxidation state. In \(I_2\), they are in the 0 oxidation state.
2. Determine the change in oxidation states:
   • Iodine in \(\text{IO}_3^-\) goes from +5 to 0 in \(I_2\).
   • Iodine in \(I^-\) goes from -1 to 0 in \(I_2\).
3. Balance the elements (excluding \(H\) and \(O\)):

The oxygen and hydrogen will be balanced later. For now, let's focus on iodine:

• In \(2 \, \text{IO}_3^-\), there are \(2 \times 1 = 2\) iodine atoms.
• In \(x \, I^-\), there are \(x\) iodine atoms.
• In \(6 \, I_2\), there are \(6 \times 2 = 12\) iodine atoms.

1. Calculate \(x\):

The total number of iodine atoms on the reactant side is \(2 + x\). So, \(2 + x = 12\). Solving for \(x\), we get:

\(x = 12 - 2 = 10\)

1. Balance the charge, hydrogen, and oxygen:

Now that \(x\) has been determined, verify the charge, hydrogen, and oxygen balance. The equation:

\(2 IO_3^- + 10 I^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2O\)

Charge Balance: Reactants side has a charge of \((-2 -10 +12 = 0\)) and the products side is neutral (0 = 0).

Hydrogen and Oxygen Balance: Already balanced in the solution.

 

Thus, the value of \(x\) is correctly given as 10.