Answer
a)
$E^{0}_{X} = -0.25 V$
$E^{0}_{Y} = +0.34 V$
b)
$E^{0}_{cell} = 0.59 V$
Work Step by Step
a)
In a galvanic cell, electrons flow from anode to cathode. Here electrons flow from X to Y, so here X is an anode and Y is the cathode. It's also given that X when connected to SHE, electrons flow from X to SHE, so here also X is the anode and has a negative sign.
If Y acts as an anode with SHE, it also has a negative sign(- 0.34 V), but it never happens since it is said that electron flows from X to Y. Then Y has the possibility of a cathode with a positive sign.
b)
$E^{0}_{cell} = E^{0}_{Y^{2+}/Y} - E^{0}_{X^{2+}/X}$
$E^{0}_{cell} = 0.34-(-0.25) = 0.59 V$
