Use these halfcell potentials Cu2 aq 2e Cu s 0034 Zn2 aq 2

Use these half-cell potentials: Cu2+ (aq) + 2e- Cu s 0:0.34 Zn2+ (aq) + 2e- Zn (s) =-0.78 Calculate Go for the voltaic cell shown above. n= moles of electrons transferred F= 96500 Coulomb (charge on one mol of electrons) Use the units kJ in your answer

Solution

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Cu+2/Cu - Eo Zn+2/Zn

Eo cell = 0.34 - (-0.78)

Eo cell = 1.12 V

now

n = 2 , as 2 moles of electrons are transferred

now

dGo - n x F x Eo cell

dGo = - 2 x 96500 x 1.12 J

dGo = -216160 J

dGo = -216.16 kJ