Question

The emf of the cell
Ni/Ni+2 (1.0 M)|| Au+3 (0.1M)/Au
[E° for Ni+2/Ni = - 0.25, E° for Au+3/Au = 1.50V] is
given as:

Answer 1

Answer:

E_cell= 1.7303V

reduction at RHE: Au+3 + 3e--> Au --(1)

reduction at LHE: Ni+2+2e--> Ni --(2)

for net cell reaction do as following

(1) *2 -(2) *3

from this onwards there is an attached pic

Hope this would help.