Find the emf of the cell in which the following reaction takes place at 298 K Ni(s) + 2 Ag+ (0.001 M) → Ni2+ (0.001 M) + 2 Ag (s)

To find the emf of the cell, we need to use the Nernst equation. The Nernst equation is used to calculate the potential of a half-cell in an electrochemical cell, or to calculate the total voltage (electromotive force) for a full electrochemical cell.

The Nernst equation is: E = E° - (RT/nF) * lnQ

Where:

• E is the cell potential (emf)
• E° is the standard cell potential
• R is the gas constant (8.314 J/(mol*K))
• T is the temperature in Kelvin
• n is the number of moles of electrons transferred in the balanced redox reaction
• F is the Faraday constant (96485 C/mol)
• Q is the reaction quotient

First, we need to find the standard cell potential (E°). The standard cell potential is the difference between the standard reduction potentials of the two half-cells. For the given reaction, the standard reduction potential of Ni2+/Ni is -0.25 V and that of Ag+/Ag is 0.80 V. So, E° = E°(Ag+/Ag) - E°(Ni2+/Ni) = 0.80 V - (-0.25 V) = 1.05 V.

Next, we need to find the number of moles of electrons transferred in the balanced redox reaction (n). In this case, n = 2.

Then, we need to find the reaction quotient (Q). The reaction quotient is the ratio of the concentrations of the products to the concentrations of the reactants. For the given reaction, Q = [Ni2+]/[Ag+]^2 = (0.001 M)/(0.001 M)^2 = 1.

Now, we can substitute these values into the Nernst equation to find the cell potential (E):

E = 1.05 V - (8.314 J/(mol*K) * 298 K / (2 * 96485 C/mol)) * ln(1) E = 1.05 V - 0 E = 1.05 V

So, the emf of the cell is 1.05 V.