Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag(s) [Given that E⁰_cell = 1.05 V]

Correct option is B. 0.91V

Given,
[Ag⁺] = 0.002M
[Ni²⁺] = 0.160M
n = 2

According to Nernst equation:
Ecell = E⁰ - (0.0591/n)log₁₀([Ni²⁺]/[Ag⁺]²)
∴ Ecell = 1.05 - (0.0591/2)log₁₀(0.160/(0.002)²)
∴ Ecell = 1.05 - 0.02955log₁₀(4 × 10⁴)
∴ Ecell = 0.91V

Hence, the correct option is B