Question

If the potential of a Daniell cell, initially containing 1 litre solution of each 1.0 M copper (II) ion and 1.0 M zinc (II) ion after passage of $1\times {10}^5$ coulomb of charge is $x$ V, then $100x$ is :
Given $E_{Zn/Z{n}^{2+}}^o$ and $E_{Cu/C{u}^{2+}}^o$ are 0.76 and -0.34 V respectively

Answer 1

The expression for the standard cell potential is $E^o=E_{Cu}^o-E_{Zn}^o=0.337V-(-0.763V)=1.1V$
The number of moles of faraday passed is $\frac{1\times {10}^5}{96500}=1.036$
The number of moles of Zn oxidized is $\frac{1.036}{2}=0.518$
$\left[Z{n}^{2+}\right]=1.0+0.518=1.518M$
$\left[C{u}^{2+}\right]=1.0-0.518=0.482M$
The expression for the cell potential is
$E=E^o-\frac{0.0592}{n}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
Substitute values in the above expression.
$E=1.1-\frac{0.0592}{2}log\frac{1.518}{0.482}=1.1-0.01475=1.09=x$
Hence, $100x=109$.