Question

The emf of the cell: Ni / Ni2+ (1.0 M) // Au3+ (1.0 M) / Au (E° = -0.25 V for Ni2+/Ni; E° = 1.5 V for Au3+/Au)​

Answer 1

Answer:

Explanation:

In a electrochemical cell, anode undergo oxidation(lose electrons) and cathode undergo reduction(gain electron).Hence, in the above question, Ni act as anode and Au act as cathode.

E=E0-0.06/n log(anodic conc.)/(cathodic conc.), where n= no. of electrons involved(here 6)

E0=E0(cathode)-Eo(Anode)=1.5-(-0.25)=1.75V

E=1.75-0.06/6 log1=1.75-0.01=1.74V