Write the Nernst equation for the $E_{cell}$ of a Daniell cell.
(N/A) The Daniell cell reaction is given by: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
For this cell,the Nernst equation at $298 \ K$ is expressed as:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$ (number of electrons transferred in the redox reaction).
Thus,the equation becomes: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
For this cell,the Nernst equation at $298 \ K$ is expressed as:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$ (number of electrons transferred in the redox reaction).
Thus,the equation becomes: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
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