Calculate the emf of the cell in the following reaction: Ni(s) + 2Ag⁺(0.002M) → Ni²⁺(0.160M) + 2Ag(s) (Given that E⁰cell = 1.05V)

The Nernst equation is used to calculate the emf of the cell under non-standard conditions:

Ecell = E⁰cell - (RT/nF)lnQ

where:

• Ecell is the cell potential under non-standard conditions
• E⁰cell is the standard cell potential (1.05 V)
• R is the ideal gas constant (8.314 J/mol·K)
• T is the temperature in Kelvin (assume 298 K)
• n is the number of moles of electrons transferred (2 in this case)
• F is Faraday's constant (96485 C/mol)
• Q is the reaction quotient

For the given reaction, the reaction quotient Q is:

Q = [Ni²⁺]/[Ag⁺]² = (0.160)/(0.002)² = 40000

Substituting the values into the Nernst equation:

Ecell = 1.05 V - (8.314 J/mol·K × 298 K / (2 × 96485 C/mol)) × ln(40000)

Ecell = 1.05 V - (0.01285 V) × ln(40000)

Ecell ≈ 1.05 V - (0.01285 V) × 10.5966

Ecell ≈ 1.05 V - 0.136 V

Ecell ≈ 0.914 V

Therefore, the emf of the cell is approximately 0.91 V.