Question

For the cell of $298K$,
$Zn(s)+C{u}^{2+}(aq)\rightleftharpoons Cu(s)+Z{n}^{2+}(aq)$

Variation of $E_{\text{cell }}$ with $\log Q$ (where $Q$ is reaction quotient) is of the type
At what value of ratio of molar concentration of ions, $E_{\text{cell }}$ would be $1.1591$ ?

• A. $\frac{\left[C{u}^{2+}\right]}{\left[Z{n}^{2+}\right]}=0.01$
• B. $\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=1.00$
• C. $\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=0.10$
• D. $\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=0.01$

Answer 1

Answer: (D)

$Zn(s)+C{u}^{2+}(aq)\to Cu(s)+Z{n}^{2+}(aq)$

$Q=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]},n=2$

$\therefore E_{\text{cell }}=E_{cell}^0-\frac{2.303RT}{nF}\log Q$

$E_{\text{cell }}=E_{\text{cell }}^0-\frac{0.0591}{2}\log Q$

This equation represents a straight line

$y=c-mx$

At $Q=1,E_{\text{cell }}=E_{\text{cell }}^{\circ }$

Thus, $y=c=$ intercept $=OA=E_{\text{cell }}^{\circ }$

Thus, $E_{\text{cell }}^{\circ }=1.10V$

$\therefore 1.1591=1.10-\frac{0.0591}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$\therefore +0.0591=-\frac{0.0591}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$\therefore -2=\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=0.01$