Electrochemistry 2 Question 37
36. Calculate $\Delta G_{r}^{\circ}$ of the following reaction:
(a) $\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$
Given
$$ \begin{array}{cc} \hline \Delta G_{f}^{\circ}(\mathrm{AgCl}) & -109 \mathrm{~kJ} / \mathrm{mol} \ \hline \Delta G_{f}^{\circ}(\mathrm{Cl})^{-} & -129 \mathrm{~kJ} / \mathrm{mol} \ \hline \Delta G_{f}^{\circ}\left(\mathrm{Ag}^{+}\right) & 77 \mathrm{~kJ} / \mathrm{mol} \ \hline \end{array} $$
Represent the above reaction in form of a cell. Calculate $E^{\circ}$ of the cell. Find $\log {10} K{\text {sp }}$ of $\mathrm{AgCl}$.
$(2005,6 \mathrm{M})$
(b) $6.539 \times 10^{-2} \mathrm{~g}$ of metallic $\mathrm{Zn}(\mathrm{u}=65.39)$ was added to $100 \mathrm{~mL}$ of saturated solution of $\mathrm{AgCl}$. Calculate $\log _{10} \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$. Given that
$$ \begin{array}{rlrl} \mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; & & E^{\circ}=0.80 \mathrm{~V} \ \mathrm{Zn}^{2+}+2 e^{-} \longrightarrow \mathrm{Zn} ; & E^{\circ}=-76 \mathrm{~V} \end{array} $$
Also find how many moles of $\mathrm{Ag}$ will be formed?
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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 $\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
$$ \begin{array}{rlrl} & & \Delta G^{\circ} & =+57 \mathrm{~kJ} \ \Rightarrow \quad \Delta G^{\circ} & =-R T \ln K_{\text {sp }} \ \Rightarrow \quad & \log K_{\text {sp }} & =-\frac{\Delta G^{\circ}}{2.3 R T}=-\frac{57 \times 1000}{2.3 \times 8.314 \times 298}=-10 \ & & & \ \text { Now, } \quad E^{\circ} & \text { of } \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \rightleftharpoons \mathrm{AgCl} \ & & E^{\circ} & =-\frac{\Delta G^{\circ}}{n F}=\frac{57000}{96500}=0.59 \mathrm{~V} \end{array} $$
(b) The cell reaction is :
$$ \begin{aligned} & \mathrm{Zn}+2 \mathrm{Ag}^{+} \rightleftharpoons \mathrm{Zn}^{2+}+2 \mathrm{Ag} ; E^{\circ}=1.56 \mathrm{~V} \ & \Rightarrow \quad 0=E^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \ & \Rightarrow \quad \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}=\frac{2 E^{\circ}}{0.059}=\frac{2 \times 1.56}{0.059}=52.88 \end{aligned} $$
Moles of $\mathrm{Zn}$ added $=\frac{6.539 \times 10^{-2}}{65.39}=10^{-3}$
$\Rightarrow$ Moles of $\mathrm{Ag}$ formed $=2 \times 10^{-3}$.
