Hypochlorite is an oxidizing agent. In the system the sulfate ion is oxidized to

Question

Hypochlorite is an oxidizing agent. In the system the sulfate ion is oxidized to sulfate ion. The chlorine atom in the OCl^- ion will either be reduced to chloride ion or to molecular chlorine. Write the balanced net ionic equation for the reaction between sulfite and hypochlorite to produce sulfate ion and chloride ion. Remember this is an alkaline system so hydroxide ions and water may also be involved. Now write the balanced net ionic equation for the same system assuming the products to be molecular chlorine and the sulfate ion, again with a possibility of water and/or hydroxide ion Consider the equations you wrote for Prelaboratory Question 1. Are there any trials that you can eliminate as being unlikely to produce the greatest quantity of energy per mole of sulfite ion consumed? Defend your answer. Does this represent a possibility for simplifying the Skills procedure? How will you determine experimentally which of the two equations correctly represents the react between sulfite and hypochlorite? Determine the mole ratio of hypochlorite to sulfite for each of the combinations that you will (Recall that they are the same as the volume ratios.) Enter these ratios in the appropriate of your data table. Which trial will give the greatest value of Delta H if chloride ion is one of the products? Which will identify molecular chlorine as the product? Step 3.f of the Procedure directs you to not only clean the inside of your calorimeter between runs but to dry it, as well. Suggest a reason why drying the inside might be necessary and advisable.

Explanation / Answer

1)

a)

SO32- ----------> SO42-

S is in +4 in left and +6 in right so this is oxidation ....and you must be knowing oxidation is loss of electron ...so writing 2e- in right to balance oxidation no.

SO32- ----------> SO42- + 2e-

now balance charge ....-2 in left and -4 in right .....so add 2H+ in right ...

SO32- ----------> SO42- + 2e- + 2H+

now balance H and O by adding H2O

SO32-  + H2O----------> SO42- + 2e- + 2H+ oxidation reaction

In other side HClO ----------> Cl-   chloride ion

after balancing

HClO + H+ +1e- ----------> Cl- + H2O reduction reaction

reduction and oxidation should be same so multiply with two

2HClO + 2H+ +2e- ----------> 2Cl- + 2H2O

combined equation SO32-  + H2O + 2HClO + 2H+ +2e-----------> SO42- + 2e- + 2H+ + 2Cl- + 2H2O

cancel same molecules  SO32- + 2HClO -----------> SO42- +2Cl- +H2O

b) oxidation reaction is same

SO32-  + H2O----------> SO42- + 2e- + 2H+ oxidation reaction

In other side HClO  ----------> Cl2

2HClO + 2e- ----------> Cl2 + 2OH-

SO32-  + H2O + 2HClO + 2e-  ----------> Cl2 + 2OH- + SO42- + 2e- + 2H+

SO32-  + H2O + 2HClO ----------> Cl2 + 2OH- + SO42- + 2H+

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.