Calculate the emf of the following cell:Ni(s)+2Ag+(0.01 M)⟶Ni2+(0.1 M)+2Ag(s)Given: Ecell∘=1.05 V, log 10 = 1.Answer: E_cell ≈ 0.96 VSolutionNernst equation (n=2):Ecell=1.05−20.059log[Ag+]2[Ni2+]=1.05−0.0295log(0.01)20.1=1.05−0.0295log103=1.05−0.0885Ecell=0.96 VCBSE Class 12 Board · 2024 · Subjective