Answer: A, B, C
Moles of Zn(NOX3)X2=18.9/189=0.100; in 0.500 L this is 0.200 M (A). The dilute (0.005 M) side is the anode (C). E=20.0591log0.0050.20=0.02955×log40=0.02955×1.602=0.0473 V≈47 mV (B). Diluting both ten-fold keeps the ratio 40, so the EMF is unchanged; (D) is wrong.