For the Daniel cell, $$ \mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \rightleftharpoo

Question

For the Daniel cell, $$ \mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} ightleftharpoons \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})} $$ The reaction quotient is given by: $$ Q=\frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]} $$ When $\mathrm{E}_{	ext {cell }}$ is plotted vs $\log \mathrm{Q}$, the following graph is obtained.\ Given $\mathrm{OA}=1.10$ volt and $\mathrm{E}_{	ext {cell }}=1.159$ volts, then:\

Professor PC Thomas Classes · Answer

$\mathrm{E}=\mathrm{E}^{0}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Q}$ Here, $\mathrm{OA}=\mathrm{E}^{0}=1.10 \mathrm{~V}$ and $\mathrm{E}=1.159 \mathrm{~V}$ and $\mathrm{n}=2$\ Therefore, $1.159=1.1-\frac{0.0591}{2} \log \mathrm{Q}$\ $\frac{0.0591}{2} \log \mathrm{Q}=1.1-1.159=-0.059$\ $\log \mathrm{Q}=-2, \quad \mathrm{Q}=10^{-2}=0.01$\