Question

The standard oxidation potential of Zn referred to SHE is 0.76 V and that of Cu is $-0.34V$ at ${25}^{\circ }C$. When excess of Zn is added to $CuS{O}_4,Zn$ displaces $C{u}^{2+}$ till equilibrium is reached. What is the ratio of $Z{n}^{2+}$ to $C{u}^{2+}$ ions at equilibrium?
Express the ans in the form $\left[Z{n}^{2+}\right]/\left[C{u}^{2+}\right]=k\times {10}^{37}$, and find k :

• A. $1.94$
• B. $2.34$
• C. $1.45$
• D. none of these

Answer 1

The correct option is C $1.45$

$E_{cell}^0=E^{0}Cu-E_{Zn}^0=0.34-(-0.76)=1.1V$
At equilibrium, $E_{cell}=0$
$E_{cell}=E_{Cell}^0-\frac{0.0592}{n}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
Substitute values in the above expression.
$0=1.1-\frac{0.0592}{2}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
$log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=37.16$
$\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=1.45\times {10}^{37}=k\times {10}^{37}$
Hence, $k=1.45$