Q14 of 47 Page 1

Calculate the emf of the following cell at 25°C :

Al (s) |Al3+ (0.001 M) || (0.1) Ni2+ |Ni (s)


Given : E°(Ni2+ /Ni)= – 0.25 V


E° (Al3+ / Al)=– 1.66 V


[ log 2 = 0.3010, log 3 = 0.4771 ]



At anode: [Oxidation]


At cathode: [Reduction]


Total electrons transferred (n) = 6.


Also, E°cell= E°cathode+ E°anode= -0.25V –(-1.66V) = 1.41 V


From Nernst equation, we know,


Ecell = E0cell – () [Coxid = concentration at the oxidation cell (0.001 M) and Cred = concentration at the reduction cell(0.1M)]


= 1.41 – ()


= 1.41 + 0.0295 = 1.4395V


So, the value of Ecellis 1.4395V.


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