Question no. 3 Here oxalic acid is the reducing agent and potassium permanganate

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Question no. 3

Here oxalic acid is the reducing agent and potassium permanganate is the oxidizing agent. The reaction between oxalic acid and potassium permanganate occurs in acidic medium, as permanganate ion is a powerful oxidizing agent in an acidic medium.

The acidic nature is introduced by adding dilute H2SO4

The reactions are,

2KMnO4 + 3H2SO4 = K2SO4 + 2MnSO4 + 3H2O + 5[O]

5H2C2O4,2H2O + 5[O] = 10CO2 + 15H2O

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2KMnO4 + 3H2SO4 + 5H2C2O4, 2H2O = K2SO4 + 2MnSO4 + 18H2O + 10CO2

Thus for each KMnO4we need 5/2 equivalent of oxalic acid.

Question No. 4

As you have seen that KMnO4 is converted to Mn2+ ion during this redox titration. KMnO4 is violet in color while Mn2+ is nearly colorless (very diluted pink color). Thus titration is monitored by the disappearance of violet color of KMnO4. However, if we use faster addition of KMnO4, concentration of KMnO4 in the solution would be high. Thus only partial electron transfer is observed, i.e. KMnO4 is converted to Mn4+ ion (as MnO2) which is brown in color. You cannot detect the titration point where the violet color is changed to colorless rather you will find a brown solution of MnO2.

Question no. 5

The last drop of permanganate gives a light pink color of Mn2+ ion on reaching the end point.

Hope this will help. Cheers!!

Explanation / Answer

It's my third time to post this and ask you guys to type your answers, not to write on paper!

Next time I'll complain and report that to the to chegg manegment.

Please type all answers.

This is only general questions.

3. For each mole of KMnOs that reacts in this experiment, how many moles of H C,0 2H2O react? 4. Why should the KMn04 be added very slowly at the beginning of the titration? 5. What is the endpoint of this titration?

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