Question

0.598 V 1. (4.0 points, show your work) What is the mass of palladium (106.42 g/mole) in the 400.0 mL solution if the measured potential in is 0.598 Volt? Pd Electrode Pd SO Silver Silver Chloride electrode 400 ml

Answer 1

2. 1 1. The reduction reaction is - Pd2+(aq) + 24 → Pb(s) n = 2 E = E pd24/Pd -0.059 log e dety [at. 298 K ] on 0.598 = 0.951 = 0.05910g [edzt] [Ed=pg=0.951] on 0.052log [Pazn = (0.951-0.598) = 0.353 op.log [Pd?t] = -(200353 ) =-11966 cor [Pd2+] = to 108x10 CM) .400 ml solution contains 1.08x102x400 met. Pd2+ =1108x1©2x 400 x106:42. 4 Pb2+ - 1000 = 4.60X17"g cans) 1.966 -12 10 0 8 x C 1000