Question

calculate the cell potential,E, at 25°c for the cell if the initial concentration of Ni(NO3)2 is 0.1 molar and the initial concentration of AgNO3 is 1.00 molar.[E°Ni 2+/Ni=0.25v;E° ag3+/ag=0.80v]​

Answer 1

Answer:

At anode : Ni→Ni2++2e− 

At cathode :  [Ag++e−→Ag]×2 

Cell reaction : Ni+2Ag+→Ni2++2Ag 

 Ecell∘=Ecathode∘−Eanode∘ 

       =EAg+/Ag∘−ENi2+/Ni∘=0.80V−(−0.25V) 

       Ecell∘=1.05V 

ii. Ecell=Ecell∘−n0.059log[Ag+]2[Ni2+] 

   Here , n=2,Ecell∘=1.05V,[Ni2+]=0.1M,[Ag+]=