Answer:
There is too much here for me to attempt in one sitting, but I will get you started and give suggestions that may help you proceed on your own.
a) As far as I can tell, you didn't provide the reduction potentials in the table below. So, I looked them up and find them to be
Cu2+ +2e- ==> Cu(s) Eº = +0.34 V
Zn2+ + 2e- ==> Zn(s) Eº = -0.76 V
Net ionic equation: Cu2+(aq) + Zn(s) ==> Cu(s) + Zn2+(aq)
b) Eºcell = 0.34 V + 0.76 V = 1.10 V
c) ∆Gº = -nFEº where n = number of electrons transferred (2) and F = 96,485 C/mol e- and Eº = 1.1 V
d) Again, there is no diagram
e) same
f) Use the Nernst equation Ecell = Eºcell - 2.303RT/nF log Q where R = 8.314; T = Kelvin; F = Faraday const. and Q = reaction quotient, i.e. products/reactants
g) the cell voltage is an intensive property. That should help you answer this part of the question
Explanation:
its not much but i hoped this helped
