Question

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

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

Answer 1

Answer: (B)

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