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What advantage do the fuel cells have over primary and secondary batteries?
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Verified Answer
Reaction occuring in Daniel cell is as following:
$$
\mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}
$$
Nernst equation for Daniel cell is given as following:
$$
\begin{aligned}
&\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}\right] \\
&\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}+\frac{0.0591}{2} \log \left[\frac{\mathrm{Cu}^{2+}}{\mathrm{Zn}^{2+}}\right]
\end{aligned}
$$
According to this equation $E_{\text {cell }}$ is directly dependent on concentration of $\mathrm{Cu}^{2+}$ and inversely dependent upon concentration of $\mathrm{Zn}^{2+}$ ions.
Therefore, $E_{\text {cell }}$ decreases when concentration of $\mathrm{Zn}^{2+}$ ions is increased.
$$
\mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}
$$
Nernst equation for Daniel cell is given as following:
$$
\begin{aligned}
&\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}\right] \\
&\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}+\frac{0.0591}{2} \log \left[\frac{\mathrm{Cu}^{2+}}{\mathrm{Zn}^{2+}}\right]
\end{aligned}
$$
According to this equation $E_{\text {cell }}$ is directly dependent on concentration of $\mathrm{Cu}^{2+}$ and inversely dependent upon concentration of $\mathrm{Zn}^{2+}$ ions.
Therefore, $E_{\text {cell }}$ decreases when concentration of $\mathrm{Zn}^{2+}$ ions is increased.
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