مسائل خارجية

احسب جهد اختزال قطب النحاس غير القياسي عند 25C في محلول تركيز أيونات النحاس فيه تساوي 0.01M مع العلم:

Eο _Cu++/Cu = + 0.34 V

$$\begin{gathered} {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}{\mathrm{E}}^{\mathrm{o}}\mathrm{Cu}++/\mathrm{Cu}=+0.34\mathrm{V} \\ {\mathbf{E}}_{\mathbf{Cu}}={\mathbf{E}}^{\mathbf{o}}{\mathrm{Cu}}_{++/\mathbf{Cu}}- \frac{0.026}{\mathrm{n}}\text{ In }\frac{1}{{\displaystyle \left[{\mathrm{Cu}}^{++}\right]}} \\ {\mathrm{E}}_{\mathrm{Cu}}=0.34-\frac{{\displaystyle 0.026}}{2}\mathrm{x}\text{ In }\frac{{\displaystyle 1}}{{\displaystyle \text{0.01}}} \\ {E}_{Cu}=0.34-0.013\times \text{In }\frac{{\displaystyle 1}}{{\displaystyle \text{0.01}}} \\ {E}_{Cu}=0.34-0.013\times 4.6=0.2802V \end{gathered}$$

للخلية Sn | Sn²⁺ || Ag⁺ | Ag جهد = 1.07V فإذا كان قطب الفضة في ظروفه القياسية، احسب [Sn⁺²].

حيث:

Eο _Sn++/Sn = - 0.14 V Eο _Ag+/Ag = + 0.80 V

$\begin{array}{cl}\mathrm{Sn}\to {\mathrm{Sn}}^{++}+2\mathrm{e}& {\mathrm{E}}^0\text{ anode }=+0.14\mathrm{V}\\ & \\ {\mathbf{Ag}}^{+}+2{\mathrm{e}}^{-}\to 2\mathrm{Ag}& {\mathrm{E}}^0\text{ Cathode }=+0.80\mathrm{V}\end{array}$

$\begin{array}{r}\mathrm{Sn}+2{\mathbf{Ag}}^{+}\to {\mathbf{Sn}}^{++}+2\mathrm{Ag}\\ \\ {\mathbf{E}}^0\text{ cell }={\mathbf{E}}^0\text{ anode }+{\mathbf{E}}^0\text{ cathode }\\ \\ {\mathbf{E}}^0\text{ cell }=(+0.14\mathrm{V})+(+0.80\mathrm{V})\\ \\ {\mathbf{E}}^0\text{ cell }=+0.94\mathrm{V}\end{array}$

$$\begin{gathered} {\mathbf{E}}_{\text{Cell }}={\mathbf{E}}^0\text{ Cell - }\frac{{\displaystyle 0.026}}{{\displaystyle 2}}\text{ln}\frac{{\displaystyle \left[{\mathbf{Sn}}^{++}\right]}}{{\displaystyle {\left[{\mathbf{Ag}}^{+}\right]}^2}} \\ 1.07=\text{0.94 - }\frac{{\displaystyle 0.026}}{{\displaystyle 2}}\text{ln}\frac{{\displaystyle \left[{\mathbf{Sn}}^{++}\right]}}{{\displaystyle {\left[\mathbf{1}\right]}^2}} \end{gathered}$$

$\begin{array}{r}0.13=-0.013\times \ln \left[{\mathrm{Sn}}^{++}\right]\\ \\ \ln \left[{\mathrm{Sn}}^{++}\right]=-10\\ \\ \ln \left[{\mathrm{Sn}}^{++}\right]=2.303\log \left[{\mathrm{Sn}}^{++}\right]\\ \\ 2.303\log \left[{\mathrm{Sn}}^{++}\right]=-10\\ \\ \log \left[{\mathrm{Sn}}^{++}\right]=-4.34\\ \\ \left[{\mathrm{Sn}}^{++}\right]=4.5\times {10}^{-5}\mathrm{M}\end{array}$

${10}^{0.66}=4.5$

خفف محلول قطب الكاثود لخلية دانيال القياسي بالماء المقطر فانخفض جهد القطب بمقدار 0.026V عن جهده القياسي، احسب تركيز أيونات القطب حينئذ.

قطب الكاثود في خلية دانيال يمثله النحاس.

$\begin{array}{r}{\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Cu}\\ \\ {\mathbf{E}}_{\mathrm{Cu}}={\mathbf{E}}^{\mathrm{o}}{\mathrm{Cu}}^{++/\mathrm{Cu}}-\mathbf{0}\mathbf{.}\mathbf{026}\\ \\ {\mathbf{E}}_{\mathrm{Cu}}={\mathbf{E}}^{\mathrm{o}}\mathrm{Cu}+/\mathrm{Cu}-\frac{{\displaystyle 0.026}}{{\displaystyle \text{ n }}}\text{ In }\frac{1}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}\\ \\ {\mathbf{E}}^0\mathrm{Cu}++/\mathrm{Cu}-0.026={\mathbf{E}}^0\mathrm{Cu}++/\mathrm{Cu} -\frac{{\displaystyle 0.026}}{{\displaystyle \text{ 2}}}\text{ In }\frac{1}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}\\ \\ 0.026=0.013\times \mathrm{In} \frac{{\displaystyle 1}}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}\\ \\ \frac{{\displaystyle 1}}{{\displaystyle \left[{\mathbf{Cu}}^{+}\right]}}=2.303\log \frac{{\displaystyle 1}}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}\\ \\ 2=2.303\log \frac{{\displaystyle 1}}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}\\ \\ \log \frac{{\displaystyle 1}}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}=0.87\\ \\ {\log }^{-1}0.87=7.4\\ \\ 7.4=\frac{{\displaystyle 1}}{{\displaystyle \left[{\mathbf{Cu}}^{++}\right]}}\\ \\ \left[{\mathrm{Cu}}^{++}\right]=0.13\mathrm{mol}/\mathrm{L}\end{array}$

للخلية Cr | Cr⁺³(0.1M) || Sn⁺²(0.01M) | Sn في 25C وΔG =- 324240 J جد جهدها القياسي؟

$\begin{array}{r}2\mathrm{Cr}\to 2{\mathrm{Cr}}^{3+}+6\mathrm{e}\\ \\ 3{\mathrm{Sn}}^{++}+6{\mathrm{e}}^{\to }\to 3\mathrm{Sn}\\ \\ 2\mathrm{Cr}+3{\mathrm{Sn}}^{++}\to 2{\mathrm{Cr}}^{3+}+3\mathrm{Sn}\end{array}$

$\mathrm{\Delta }G=-nF{E}_{\text{Cell }}\Rightarrow -324240\mathrm{J}=-6\times 96500\times E_{\text{Cell }}\Rightarrow E_{\text{Cell }}=+0.56$

$$\begin{gathered} {\mathbf{E}}_{\text{Cell }}={\mathbf{E}}^0\text{ Cell -}\frac{{\displaystyle 0.026}}{\mathrm{n}}\text{ln}\frac{{\displaystyle {\left[{\mathrm{Cr}}^{+3}\right]}^2}}{{\displaystyle {\left[{\mathbf{Sn}}^{++}\right]}^3}} \\ +0.56={\mathbf{E}}^0\text{ Cell - }\frac{{\displaystyle 0.026}}{6}\text{ln}\frac{{\displaystyle [0.1{]}^2}}{{\displaystyle [0.01{]}^3}} \\ \begin{array}{r}0.56=E^0\text{ Cell }-0.004\times \ln {10}^4\\ \\ \ln {10}^4=2.303\log {10}^4\\ \\ \ln {10}^4=9.212\\ \\ 0.56=E^0\text{ Cell }-0.0368\\ \\ E^0\text{ Cell }=0.596\mathrm{V}\end{array} \end{gathered}$$