Electrochemistry Result Question 36
36. Calculate $\Delta G_r^{\circ}$ of the following reaction:
(a) $\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$
Given
| $\Delta G_f^{\circ}(\mathrm{AgCl})$ | $-109 \mathrm{~kJ} / \mathrm{mol}$ |
|---|---|
| $\Delta G_f^{\circ}(\mathrm{Cl})^{-}$ | $-129 \mathrm{~kJ} / \mathrm{mol}$ |
| $\Delta G_f^{\circ}\left(\mathrm{Ag}^{+}\right)$ | $77 \mathrm{~kJ} / \mathrm{mol}$ |
Represent the above reaction in form of a cell. Calculate $E^{\circ}$ of the cell. Find $\log {10} K{\mathrm{sp}}$ of AgCl .
$(2005,6 M$ )
(b) $6.539 \times 10^{-2} \mathrm{~g}$ of metallic $\mathrm{Zn}(\mathrm{u}=65.39)$ was added to 100 mL of saturated solution of AgCl . Calculate $\log _{10} \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}$. Given that
$ \begin{array}{rlr} \mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; & E^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Zn}^{2+}+2 e^{-} \longrightarrow \mathrm{Zn} ; & E^{\circ}=-76 \mathrm{~V} \end{array} $
Also find how many moles of Ag will be formed?
$(2005,6 M)$
Show Answer
Solution:
- (a) $\Delta G^{\circ}=\Sigma \Delta G_f^{\circ}$ (products) $-\Sigma \Delta G_f^{\circ}$ (reactants)
$ =-109-(-129+77) \mathrm{kJ}=-57 \mathrm{~kJ} $
Cell : $\quad \mathrm{Ag}\left|\mathrm{AgCl}, \mathrm{Cl}^{-} | \mathrm{Ag}^{+}\right| \mathrm{Ag}$
For $K_{\text {sp }}$; reaction is $\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
$ \begin{aligned} & \Delta G^{\circ}=+57 \mathrm{~kJ} \\ & \Rightarrow \quad \Delta G^{\circ}=-R T \ln K_{\text {sp }} \\ & \Rightarrow \quad \log K_{\text {sp }}=-\frac{\Delta G^{\circ}}{2.3 R T}=-\frac{57 \times 1000}{2.3 \times 8.314 \times 298}=-10 \\ \end{aligned} $
Now, $\quad E^{\circ}$ of $\mathrm{Ag}^{+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{AgCl}$
$ E^{\circ}=-\frac{\Delta G^{\circ}}{n F}=\frac{57000}{96500}=0.59 \mathrm{~V} $
(b) The cell reaction is :
$ \begin{aligned} & \mathrm{Zn}+2 \mathrm{Ag}^{+} \rightleftharpoons \mathrm{Zn}^{2+}+2 \mathrm{Ag} ; E^{\circ}=1.56 \mathrm{~V} \\ & \Rightarrow \quad 0=E^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \Rightarrow \quad \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}=\frac{2 E^{\circ}}{0.059}=\frac{2 \times 1.56}{0.059}=52.88 \\ & \text { Moles of } \mathrm{Zn} \text { added }=\frac{6.539 \times 10^{-2}}{65.39}=10^{-3} \\ & \Rightarrow \quad \text { Moles of } \mathrm{Ag} \text { formed }=2 \times 10^{-3} . \end{aligned} $
BETA
