Write the Nernst equation for the $E_{cell}$ reaction in the Daniell cell. How will the $E_{cell}$ be affected when the concentration of $Zn^{2+}$ ions is increased?
(N/A) The Nernst equation for the Daniell cell reaction $(Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s))$ is given by:
$E_{cell} = E^{o}_{cell} - \frac{0.0591}{2} \log_{10} \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
When the concentration of $Zn^{2+}$ ions increases,the ratio $\frac{[Zn^{2+}]}{[Cu^{2+}]}$ increases.
Consequently,the value of $\log_{10} \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ increases.
Since this term is subtracted from $E^{o}_{cell}$,the overall value of $E_{cell}$ decreases.
$E_{cell} = E^{o}_{cell} - \frac{0.0591}{2} \log_{10} \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$
When the concentration of $Zn^{2+}$ ions increases,the ratio $\frac{[Zn^{2+}]}{[Cu^{2+}]}$ increases.
Consequently,the value of $\log_{10} \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ increases.
Since this term is subtracted from $E^{o}_{cell}$,the overall value of $E_{cell}$ decreases.
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