The standardization of KMnO 4 solution, if the magnetic stir bar was dropped int

Question

  The standardization of KMnO4 solution, if the magnetic stir bar was dropped into the flask, and some FAS solution splashed out, how would the calculation for the molarity of KMNO4 be affected?    The calculated molarity of KMnO4 would be:

Lower.   Less FAS would need less MnO4-, therefore, the reduced volume would result in a smaller M.

Unchanged.

Lower.   More FAS would need less MnO4-, therefore, the increased volume would result in a smaller M.

Higher.  Less FAS would need less MnO4-, therefore, the reduced volume would result in a larger M.

Higher.  More FAS would need less MnO4-, therefore, the increased volume would result in a smaller M.

The following redox reaction, determine what is oxidized and is also the reducing agent:

3 Cu (s) + 8 HNO3 (aq) 3 Cu2+ (aq) + 2 NO (g) + 6 NO3- (aq) + 4 H2O (l)

H2O

NO

Cu

HNO3

None of the above

2)

he following redox reaction, determine what is reduced and is also the oxidizing agent:

3 SO2 (aq) + Cr2O72- (aq) + 2 H+ (aq) 3 SO42- (aq) + 2 Cr3+ (aq) + H2O (l)

H2O

SO42-

SO2

Cr2O72-

None of the above

2 MnO4- (aq) + 5 H2O2 (aq) + 6 H+(aq) 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l)

3.4%

24%

100%

7.9%

12%

Please help, this is a new investigation and I am having trouble understanding the concepts. Please show work :)

Lower.   Less FAS would need less MnO4-, therefore, the reduced volume would result in a smaller M.

Unchanged.

Lower.   More FAS would need less MnO4-, therefore, the increased volume would result in a smaller M.

Higher.  Less FAS would need less MnO4-, therefore, the reduced volume would result in a larger M.

Higher.  More FAS would need less MnO4-, therefore, the increased volume would result in a smaller M.

Explanation / Answer

1)

Higher.  Less FAS would need less MnO4-, therefore, the reduced volume would result in a larger M.

M1V1=M2V2

volume reduced. So, molaity will be high to make MV constant

2) Cu . It is oxidised to Cu+2 . The compound which is oxidised acts as a reducing agent

3) Cr2O7-2 . Cr changed its oxidaion state from +6 to +3. The compound which is reduced acts as a oxidising agent

4) only equation given. No question

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