Solution:
The image shows a chemical equation that appears to be unbalanced:\[ IO_{2}^{-} + 10I^{-} + 12H^{+} \rightarrow 6I_{2} + 6H_{2}O \]To balance this equation, you will need to ensure that the number of atoms of each element is the same on both sides of the equation. Additionally, the charge should also be balanced.1. Balance the iodine atoms: - There are 10 iodine (I) atoms in the form of I- on the reactant side and 12 I atoms in I2 on the product side, so you will need to adjust the coefficients. Start by changing the coefficient of I2 to 5 to match the number of I atoms: \[ IO_{2}^{-} + 10I^{-} + 12H^{+} \rightarrow 5I_{2} + 6H_{2}O \] This will give us 10 I atoms from I- plus 1 I atom from IO2-, totaling 11 I atoms on the reactants' side, which corresponds to 5.5 I2 on the product side. Since we cannot have a fraction of a molecule, we will need to multiply the entire equation by 2 to get whole numbers: \[ 2IO_{2}^{-} + 20I^{-} + 24H^{+} \rightarrow 11I_{2} + 12H_{2}O \]2. Balance the oxygen atoms: - There are 4 oxygen (O) atoms on the reactant side coming from 2 IO2- ions. To balance the oxygen, we adjust the water molecules on the product side: \[ 2IO_{2}^{-} + 20I^{-} + 24H^{+} \rightarrow 11I_{2} + 4H_{2}O \]3. Balance the hydrogen atoms: - There are 8 hydrogen (H) atoms in the form of H2O on the product side and 24 H+ on the reactant side. Since 4 H2O contribute 8 H atoms, we will need to adjust the water molecules to have 12 H2O: \[ 2IO_{2}^{-} + 20I^{-} + 24H^{+} \rightarrow 11I_{2} + 12H_{2}O \]4. Balance the charges: - The total charge on the reactant side is 2(-1) from IO2- + 20(-1) from I- + 24(+1) from H+, which equals 2 negative charges. - The total charge on the product side is neutral. - To balance the charges, reduce the H+ ions to balance the negative charges from the iodine species. This makes the equation: \[ 2IO_{2}^{-} + 20I^{-} + 22H^{+} \rightarrow 11I_{2} + 12H_{2}O \]Now each element is balanced in terms of the number of atoms on both sides, and the charges are balanced as well.