PLEASE DEMONSTRATE WORK equation 3 C6H8O6+I2---> C6H6O6 + (2H^+) + 2I- question

Question

PLEASE DEMONSTRATE WORK

equation 3

C6H8O6+I2---> C6H6O6 + (2H^+) + 2I-

question 1. Identify the oxidizing agent and the reducing agent

question 2. In (equation 3) two moles of H+ are produced. Explain why H+ ions are formed.

question 3. A stock solution of I2 is standardized using a 0.1430 g sample of pure ascorbic acid. The titration to the starch end point required 25.45 mL of the I2 solution. What is the molarity of the I2 solution?

question 4. A vitamin C tablet was divided into three equal mass portions. One portion was crushed into a fine powder, dissolved in water and titrated with the standardized I2 solution from question 3, above. The titration to end point required 27.6 mL of the I2 solution. How many mg of Vitamin C are in the original tablet?

Explanation / Answer

Ans 1 - C6H8O6 + I2 ----> C6H6O6 + (2H+) + 2 I-

To find oxidising agent and reducing agent, we find oxidaton states of the elements.

Here, I2 is in 0 oxidation state on reactant side and it is -1 on product side (I-). So, I2 is reduced and hence

it acts as oxidising agent. Whereas other compound changes from C6H8O6 to C6H6O6 that is 2 H+ are

removed from the molecule so the substance is oxidised.

Because the substance which releases H+ or adds oxygen or which releases electrons to itself is oxidised.

And the substance which releases O ,adds H or accepts electrons to itself is said to be reduced.

   The substance whcih is reduced in the reaction acts as oxidising agent and the one whcih is oxidised is

called reducing agent.

Hence, in this reaction, I2 is oxidising agent and C6H8O6 is reducing agent.

Ans 2 - In above reaction, 2 H+ ions are released because C6H8O6 reacst with I2 which is a strong oxidising agent so it oxidises its neighbouring molecule that is C6H8O6 and hence it releases H+.

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