Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$

The cell reaction is: $Zn_{(s)} + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu_{(s)}$
Number of electrons transferred,$n = 2$.
Using the Nernst equation at $298 \ K$:
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Substitute the given values:
$E_{cell} = 1.1 - \frac{0.0591}{2} \log \frac{0.6}{0.3}$
$E_{cell} = 1.1 - 0.02955 \log 2$
Since $\log 2 \approx 0.3010$:
$E_{cell} = 1.1 - (0.02955 \times 0.3010)$
$E_{cell} = 1.1 - 0.00889$
$E_{cell} \approx 1.0911 \ V$