Question

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Explain about permanganate titrations ​

Answer 1

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It is a redox titration that involves the use of permanganates to measure the amount of analyte present in unknown chemical samples. ... Depending on how the titration is performed, the permanganate ion is reduced to Mnx, where x is 2+, 3+, 4+ or 6+.

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Answer 2

Answer:

Permanganometry is one of the techniques used in chemical quantitative analysis. It is a redox titration that involves the use of permanganates to measure the amount of analyte present in unknown chemical samples.[1] It involves two steps, namely the titration of the analyte with potassium permanganate solution and then the standardization of potassium permanganate solution with standard sodium oxalate solution. The titration involves volumetric manipulations to prepare the analyte solutions.[2]

Depending on how the titration is performed, the permanganate ion is reduced to Mnx, where x is 2+, 3+, 4+ or 6+. Using permanganometry we can estimate the quantitative presence of Fe2+, Mn2+, Fe2+ and Mn2+ when they are both present in a mixture, C2O42−, NO2−, H2O2 etc.

In most cases permanganometry is performed in a very acidic solution in which the following reaction occurs:[3]

{\textstyle {\ce {MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}}}{\textstyle {\ce {MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}}}

The standard potential of this electrochemical reaction is:[4]

Eo=+1.51 V

which shows that KMnO4 (in an acidic medium) is a very strong oxidizing agent. With this method we can oxidize:

Fe+2 (EoFe+3/Fe+2=+0.77 V)

Sn+2 (EoSn+4/Sn+2=+0.2 V)

and even

Cl− ( EoCl2/Cl−=+1.36 V) etc.

In weak acidic medium MnO4− can not accept 5 electrons to form Mn+2, this time it accepts only 3 electrons and forms MnO2(s) by the following electrochemical reaction:

{\displaystyle {\ce {MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}}}{\displaystyle {\ce {MnO4- + 4H+ + 3e- -> MnO2 + 2H2O}}}

The standard potential is Eo=+1.69 V.

And if the solution has a concentration c(NaOH)>1 mol dm−3 the following reaction occurs:

MnO4− + e− → MnO42− Eo=+0.56 V.[5]

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