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Electrochemistry

Question
CBSEENCH12005944

E0Ni2+/Ni and E0Cu2+/Cu are -0.25 V and + 0.34 respectively at 298 K. Formulate the self operating galvanic cell for this electrode pair. What reaction takes place in its operation? How is the ΔG° for this reaction related to the cell e.m.f.?

Solution
The cell may be represented as:
Ni(s) +Ni2+(aq)  Cu2+ (aq)+ Cu(s)At anode : Ni(s)  Ni2+(aq) +2e-At cathode : Cu2+(aq) +2e- Cu(s)Net cell reaction :Ni(s) +Cu2+(Aq)  Ni2+ (Aq)+Cu(s)

we have given that 
E0Ni2+/Ni   =-0.25V E0Cu2+/Cu = +0.34VEcell0 =0.34-(-0.25)       = 0.59
Relationship between G  and cell e.m.f.
G = -nE°F

here n=2
F=96500 
E0cell= 0.59
plug in above equation we get 

G  =-2 x 96500 x 0.59 =-113870 Kj/mol