Assume a cell with the following reaction Cu$$_{(s)}$$ + 2Ag$$^+$$(1 $$\times 10^{-3}$$M) $$\to$$ Cu$$^{2+}$$(0.250M) + 2Ag$$_{(s)}$$
E$$^\circ_{cell}$$ = 2.97 V
E$$_{cell}$$ for the above reaction is ___ V.
(Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]

For the cell reaction $$\text{Cu(s)} + 2\text{Ag}^+(1 \times 10^{-3}\text{ M}) \to \text{Cu}^{2+}(0.250\text{ M}) + 2\text{Ag(s)}$$, we use the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$$, where $$n = 2$$ (two electrons transferred).

The reaction quotient is $$Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.250}{(1 \times 10^{-3})^2} = \frac{0.250}{10^{-6}} = 2.50 \times 10^5$$.

Now $$\log Q = \log(2.50 \times 10^5) = \log 2.5 + 5 = 0.3979 + 5 = 5.3979$$.

Substituting: $$E_{\text{cell}} = 2.97 - \frac{0.0591}{2} \times 5.3979 = 2.97 - 0.02955 \times 5.3979 = 2.97 - 0.1595 = 2.81 \text{ V}$$.

Rounding to the nearest integer, the answer is $$3$$.