So, we know the reduction potentials of the electrodes :
Electrode A = -0.21 V
Electrode B = -0.15 V
We want our cell to be spontaneous - this means that the voltage has to be positive. In order to do so, we need to turn the more negative reduction half reaction to be an oxidation half reaction.
-0.21 is more negative than -0.15, so electrode A will be the oxidation half reaction.
Anode is where the oxidation half reaction takes place, so electrode A is the correct answer . A fun way to remember this is An(ode)Ox(idation) = AnOx and Red(uction)Cat(hode) = RedCat
The E° cell will be 0.21 -0.15 V = 0.6 V
