Question

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

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. $\frac{\left[ Cu ^{2+}\right]}{\left[ Zn ^{2+}\right]}=0.01$
• B. $\frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}=1.00$
• C. $\frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}=0.10$
• D. $\frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}=0.01$

Answer 1

Answer: (D)

$Zn ( s )+ Cu ^{2+}( aq ) \rightarrow Cu ( s )+ Zn ^{2+}( aq )$

$Q=\frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}, n=2$

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

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

This equation represents a straight line

$y=c-m x$

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

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

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

$\therefore 1.1591=1.10-\frac{0.0591}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$

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

$\therefore -2=\log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$

$\frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}=0.01$