Question

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

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

Answer 1

Answer: (B)

$1.0705V$

For Daniell cell $Zn+{Cu}^{2+}⟶{Zn}^{2+}+Cu$

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{\left[{Zn}^{2+}\right]}{\left[{Cu}^{2+}\right]}$

Intercept, $E_{cell}^o=1.10V$

$\therefore$ $E_{cell}=1.10-\frac{0.0591}{2}\log \frac{0.1}{0.01}$

$E_{cell}=1.10-0.0295=1.0705V$