The answer is C, exam choice paper, has weird questions in it
Negative ions flow to the anode, positive ions flow to the cathode so for picture 1 , y = anode and z = cathode. Y has reduction potential of -0.24 however it is being oxidised so to work out x you do 0.43-0.24 = 0.19
z = 0.19 = reduction potential.
In second diagram, z is being oxidised and x is being reduced. so -0.19 + x = 1.15V
x = 1.34