Galvanic cell problem

is it true when you're given E null for two half cells that if one of them is higher than the other reduction is happening in that cell

for example

given MnO4– + 8H+ + 5e– → Mn2+ + 4H2O, E° = 1.51 V
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O, E° = 1.33 V

please explain i appreciate it

Well, in this case yes.. What you just provided are the standard reduction potentials of two reactions. Reduction occurs when an ion receives an electron, which both the MnO4- and the Cr2O7- did. Now, the standard cell potential is the expected amount of volts that would be expected by the reaction, but this is calculated by several ways. You can take the two reduction potentials and subtract the larger value by the smaller value. 1.51-1.33=0.18V. (You can also add the largest reduction potential, 1.51, to the oxidation potential of the other, which would be -1.33). If the cell potential is positive, then the reaction occurs.

Now, since the reaction occurs we can establish which reaction is being reduced. You can do this by looking at the two reactions and figure out which reaction has a higher reduction potential, which would be the MnO4– + 8H+ + 5e– → Mn2+ + 4H2O reaction. Logically, the other reaction, Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O must be oxidized since the electrons for the reduction of the MnO4- have to come from somewhere.

Basically, the reaction that has the higher reduction potential (Reduction Potential is a measurement of how well an ion reduces) should be the one being reduced. If you want to figure out whether you have a reduction potential as your E null, then just look at the reaction that is occurring. MnO4- + 5e-. That's obviously a reduction, so the E null value that you have is the reduction potential.

Reduced:
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O, E° = 1.51 V

Oxidized:
2Cr3+ + 7H2O → Cr2O72– + 14H+ + 6e–, E° = -1.33 V