The following diagram on the left represents the copper nitrate solution in the half‑cell before the galvanic cell starts to operate. In the following diagram on the right, draw a representation of the solution in the same half‑cell after the galvanic cell has been operating for some time and is still producing a voltage. Before reaction : 4 Cu2+ molecules and 8NO3-

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

the ratio of the conjugate base to the acid required to attain the desired ph.