Derive the Nernst equation for calculating $E_{cell}$ of a Daniell cell and explain the effect on $E_{cell}$ when there is a change in the concentration of $Zn^{2+}$ and $Cu^{2+}$ ions.

(N/A) For the Daniell cell reaction: $Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
The Nernst equation for the cell potential is given by:
$E_{cell} = E_{cell}^{\emptyset} - \frac{RT}{nF} \ln Q$
Where $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}$ and $n = 2$ (number of electrons transferred).
Substituting the values:
$E_{cell} = E_{cell}^{\emptyset} - \frac{RT}{2F} \ln \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Converting to $\log_{10}$ at $298 \ K$:
$E_{cell} = E_{cell}^{\emptyset} - \frac{0.0591}{2} \log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Effect of concentration changes:
$1$. If the concentration of $Cu^{2+}$ increases,the ratio $\frac{[Zn^{2+}]}{[Cu^{2+}]}$ decreases,making the logarithmic term smaller,which increases $E_{cell}$.
$2$. If the concentration of $Zn^{2+}$ increases,the ratio $\frac{[Zn^{2+}]}{[Cu^{2+}]}$ increases,making the logarithmic term larger,which decreases $E_{cell}$.