Question

The emf of the cell
Ni | Ni²⁺ (1.0 M)| | Au³⁺ (1.0M)| Au is
[Given E°Ni²⁺ / Ni = – 0.25 V and E°Au³⁺/ Au = + 1.5 V]
(a) 2.00 V
(b) 1.25 V
(c) – 1.25 V
(d) 1.75 V

Answer 1

D IS CORRECT ANSWER.......

Answer 2

${\huge {\mathfrak {Answer :-}}}$

➡️ Option (D)

That's it..