Answer:
well it might be 1.)Cadmium Fluoride CdF2 150.4078
2.)Cadmium(II) Perfluorate Cd(FO4) 2278.403
3.)Cadmium Ferrocyanide Cd2Fe(CN) 6436.7714
Explanation:
Answer: Ecell = -0.110volt
Explanation:
Zn--->Zn^+2 + 2e^-.........(1) oxidation
Cu^2+ 2e^- --->Cu........(2)reduction
Zn + Cu^2+ ----> Cu + Zn^+2 (overall
For an electrochemical cell, the reduction potential set up is given by
E(cell) = E(cathode) - E(anode)
E(cell) = E(oxidation) - E(reduction)
E(cathode) = E(oxidation)
E(anode) = E(reduction)
Given that
E(oxidation) = -0.763v
E(reduction) = +0.337v
E(cell) = -0.763 - (+0.337)
E(cell) = -0.763- 0.337
E(cell) = -0.110volt
The picture won’t show up
9.04354 * 10^2
The “*” identifies as a multiplication sign. Hope this helps
1. True 2. I think A and D. I’m not sure though