Question

EMF of Daniell cell was found using different concentrations of $Zn ^{2+}$ ion and $Cu ^{2+}$ ion. A graph was then plotted between $E_{\text {cell }}$ and $\log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$. The plot was found to be linear with intercept on $E_{\text {cell }}$ axis equal to $1.10 V$. $E_{\text {cell }}$ for $Zn / Zn ^{2+}$ $(0.1 M ) \| Cu ^{2+}(0.01 M ) \mid Cu$ will be

• A. 1.10 V
• B. 1.0705 V
• C. 0.93 V
• D. 0.078V

Answer 1

Answer: (B)

For Daniell cell, $Zn + Cu ^{2+} \to Zn^{2-} + Cu$
$E_{\text{cell}} = E^{\circ}_{\text{cell}} = \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Intercept $ = E^{\circ}_{\text{cell}} = 1.10\,V$
$\therefore E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log \frac{0.1}{0.01}$
$E_{\text{cell}} = 1.10 - 0.0295 = 1.0705\,V$