Question

# Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox Titration. Some half – cell reactions and their standard potential are given below:  MnO−4(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(I);                   E∘=1.51 V Cr2 O2−7 (aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(I);              E∘=1.38V Fe3+(aq)+e−→Fe2+(aq);                                      E∘=0.77V Cl2(g)+2e−→2Cl−(aq);                               E∘=1.40V Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2.

A
MnO4 can be used in aqueous HCl
B
Cr2O27 can be used in aqueous HCl
C
MnO4 can be used in aqueous H2SO4
D
Cr2O27 can be used in aqueous H2SO4

Solution

## The correct option is A MnO−4 can be used in aqueous HClMnO−4 will oxidize Cl− into Cl2 according to the following equation:               2 MnO−4+16 H+10 Cl−→2Mn2++8 H2O+5 Cl2 The corresponding cell is                          PtCl2(1 atm) |Cl−| |MnO−4,Mn2+,H+ |Pt                          E∘cell=+1.51−1.40=+0.11 V A positvie value confirms that the above process is spontaneous. Thus, MnO−4  will oxidize Cl− as well as Fe(NO3)2 simultaneously.

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