Fe2+(aq) + IO3−(aq) → Fe3+(aq) + I2(s)Consider the above unbalanced redox reaction occurring under acidic conditions.  Which of the following correctly represents the balanced reaction?A.Fe2+(aq) + 2 IO3−(aq) + 12 H+(aq) → Fe3+(aq) + I2(s) + 6 H2O(l)B.5 Fe2+(aq) + IO3−(aq) + 6 H+(aq) → 5 Fe3+(aq) + I2(s) + 3 H2O(l)C.10 Fe2+(aq) + 2 IO3−(aq) + 12 H+(aq) → 10 Fe3+(aq) + I2(s) + 6 H2O(l)D.10 Fe2+(aq) + 2 IO3−(aq) + 6 H2O(l) → 10 Fe3+(aq) + I2(s) + 12 OH−(aq)

The correct answer is A.

Here's how you balance the reaction:

Step 1: Separate the reaction into two half-reactions:

Fe2+ → Fe3+ (oxidation half-reaction) IO3− → I2 (reduction half-reaction)

Step 2: Balance all atoms except H and O:

For the oxidation half-reaction, Fe is already balanced. For the reduction half-reaction, balance I by adding a coefficient of 2 in front of I2: IO3− → 2I2

Step 3: Balance O atoms by adding H2O:

For the oxidation half-reaction, there are no O atoms. For the reduction half-reaction, add 3 H2O to the right side: IO3− → 2I2 + 3H2O

Step 4: Balance H atoms by adding H+:

For the oxidation half-reaction, there are no H atoms. For the reduction half-reaction, add 6 H+ to the left side: IO3− + 6H+ → 2I2 + 3H2O

Step 5: Balance charges by adding electrons (e−):

For the oxidation half-reaction, add 1 e− to the right side: Fe2+ → Fe3+ + e− For the reduction half-reaction, add 6 e− to the right side: IO3− + 6H+ + 6e− → 2I2 + 3H2O

Step 6: Make the electron loss equal to the electron gain:

Multiply the oxidation half-reaction by 6 and the reduction half-reaction by 1: 6Fe2+ → 6Fe3+ + 6e− IO3− + 6H+ + 6e− → 2I2 + 3H2O

Step 7: Add the half-reactions together and simplify:

6Fe2+ + IO3− + 6H+ → 6Fe3+ + 2I2 + 3H2O

This is the balanced redox reaction under acidic conditions. The correct answer is A: Fe2+(aq) + 2 IO3−(aq) + 12 H+(aq) → Fe3+(aq) + I2(s) + 6 H2O(l).