To determine the electromotive force (EMF) of the given cell reaction at 298 K:

Reaction: Ni(s) + 2Ag⁺ (0.001 M) → Ni²⁺ (0.001 M) + 2Ag(s)

Given data:

Standard cell potential, E°_cell = 10.5 V

Value at 298 K, $\frac{2.303\,RT}{F}$ = 0.059

Using the Nernst equation to find the cell potential, E_cell:

$ E_{cell} = E_{cell}^o - \frac{0.059}{n} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} $

Where:

• n = number of moles of electrons transferred = 2
• $[\text{Ni}^{2+}] = 0.001\, \text{M}$
• $[\text{Ag}^+] = 0.001\, \text{M}$

Substituting the values:

$ E_{cell} = 10.5 - \frac{0.059}{2} \log \frac{(10^{-3})}{(10^{-3})^2} $

$ = 10.5 - \frac{0.0295}{2} \log (10^3) $

$ = 10.5 - 0.0295 \times 3 $

$ = 10.5 - 0.0885 $

$ = 10.4115\, \text{V} $

Therefore, the EMF of the cell is 10.4115 V.