Question

The reaction below was carried out in an acidic solution. Upper I minus, plus upper I upper O minus subscript 3 right arrow upper I subscript 2. Which statement is true about this equation?

Answer 1

Answer:

It has been balanced by using the half-reaction method.

Explanation:

I- and IO3- gives I2

We divide the reaction into two half-reactions

(2 I- >> I2 + 2e-) x5 ( oxidation  : I goes from -1 to 0 )

2 IO3- + 12H+ + 10e- >> I2 + 6H2O ( reduction : I goes from +5 to 0 )

10 I- >> 5I2 + 10e-

2IO3- + 12H+ + 10e- >> I2 + 6H2O

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10 I- + 2IO3- + 12H+ >> 6I2 + 6H2O

To get the smallest numbers we divide by 2 :

5 I- + IO3- + 6H+ >> 3I2 + 3H2O

Answer 2

It has been balanced by using the half-reaction method. Hence, option B is correct.

What is a redox reaction?

A chemical reaction that takes place between an oxidizing substance and a reducing substance.

We divide the reaction into two half-reactions

$(2 I^-$ → $I_2 + 2e^-)$ x 5 ( oxidation  : I goes from -1 to 0 )

$2 IO^{3-} + 12H^+ + 10e^-$ → $I_2 + 6H_2O$ ( reduction : I goes from +5 to 0 )

$10 I^-$ →$5I_2 + 10e^-$

$2IO^{3-} + 12H^+ + 10e^-$ → $I_2 + 6H_2O$

$10 I^- + 2IO^{3-} + 12H^{+}$→ $6I_2 + 6H_2O$

To get the smallest numbers we divide by 2 :

$5 I^- + IO^{3-} + 6H^+$ → $3I_2 + 3H_2O$

Hence, option B is correct.

Learn more about the redox reaction here:

https://brainly.com/question/14881952

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