Question

The emf of the cell,
Ni|Ni2+(1.0 M)||Ag+(1.0 M)|Ag (E° of Ni2+|Ni = -0.25
volt, Eº for Ag | Ag = 0.80 volt)​

Answer 1

Answer:EMFcell°=E°cathode-E°anode

E°cell=0.80-(-0.25)

E°cell=0.8+0.25

E°cell=1.05v

Explanation:

Hope it mayhelps u frnd....