why should MnO4- and Cr2O72- be acidified to be an Oxidizing

why should MnO4- and Cr2O72- be acidified to be an Oxidizing Agent ?
The main reason for MnO4- and Cr2O72- to be acidified before use is to increase their oxidizing power. Since the half equation for MnO4- is written as follows MnO4- + 8H+ + 5e- 出 Mn2++ 4H2O.
By applying the Nernst equation E = Eo + 0.059/n*log[Ox]a/[Red]b, we have
E =+1.52 + 0.059/5*log[MnO4- ][H+]8/[ Mn2+]where Eo = +1.52V.
For simplicity, we now take both [MnO4- ]and[ Mn2+] as 1M.
Hence the equation becomes E =1.52 + 0.059/5*log[1 ][H+]8/[1]
and E =1.52 + 0.059/5*log[H+]8

You can now put any values to [H+] for example 0.0001 and 10.
The equation becomes E =1.52 + 0.059/5*log[0.0001]8 (即是0.0001的八次方之後再log一log這個數)and E =1.52 + 0.059/5*log[10]8 (即是10的八次方之後再log一log這個數)
What you will find is the higher the concentration of hydrogen ion, [H+] , the more +ve of the value, E, is. The more +ve E implies its oxidizing power is increased. Hence by acidifiying MnO4- and Cr2O72- , their oxidizing power can be increased.
+ve = positive ; * = 乘

I hope you will find my answer helpful.

Because MnO4- and Cr2O72- oxidize others only when there is H+ ions:

MnO4- + 8H+ + 5e- -----> Mn2+ + 4H2O

Cr2O72- + 14H+ + 6e- -------> 2Cr3+ + 7H2O

When acid is added to the solution, the acid ionizes and gives out H+ ions for the reaction.