题目
14.已知电池 |Zn(Cl)_(2)(b=0.01molcdot (Kg)^-1)|AgCl(s)|Ag 在25℃时的电动势 =1.1566V,-|||-电动势的温度系数 (dfrac (partial E)(partial T))=4.02times (10)^-4Vcdot (K)^-1, 电极电势 ({E)_(2n)}^2+/(n)_(n)=-0.763V, _(AgC{H)_(Ag)}=0.222V-|||-(1)写出电池的电极反应和电池反应;-|||-(2)计算25℃时电池反应的 Delta (G)_(m) (Delta )_(1)(H)_(m) (Delta )_(1)(S)_(m) 及标准平衡常数 ^theta -|||-(3)计算25℃时 .01molcdot (kg)^-1Zn(Cl)_(2) 溶液的平均活度及平均活度因子。
题目解答
答案
解析
步骤 1：写出电池的电极反应和电池反应
负极反应：$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}$
正极反应：$2AgCl(s) + 2e^{-} \rightarrow 2Ag(s) + 2Cl^{-}(aq)$
电池反应：$Zn(s) + 2AgCl(s) \rightarrow 2Ag(s) + ZnCl_{2}(aq)$
步骤 2：计算25℃时电池反应的 $\Delta G_{m}$, $\Delta H_{m}$, $\Delta S_{m}$ 及标准平衡常数 $K^{\theta}$
$\Delta G_{m} = -nFE = -2 \times 96485 \times 1.1566 = -223.225 kJ/mol$
$\Delta H_{m} = \Delta G_{m} + T\Delta S_{m}$
$\Delta S_{m} = \frac{\partial E}{\partial T} \times nF = 4.02 \times 10^{-4} \times 2 \times 96485 = 72.386 J/(mol\cdot K)$
$\Delta H_{m} = -223.225 \times 10^{3} + 298.15 \times 72.386 = -208.99 kJ/mol$
$K^{\theta} = e^{\frac{-\Delta G_{m}^{\theta}}{RT}} = e^{\frac{223.225 \times 10^{3}}{8.314 \times 298.15}} = 4.93 \times 10^{-34}$
步骤 3：计算25℃时 $0.01mol\cdot kg^{-1}ZnCl_{2}$ 溶液的平均活度及平均活度因子
由Nernst方程：$E = E^{\theta} - \frac{RT}{zF} \ln \frac{a_{Zn^{2+}}}{a_{Cl^{-}}^{2}}$
$E^{\theta} = E_{AgCl/Ag} - E_{Zn^{2+}/Zn} = 0.985 V$
$a_{Zn^{2+}} = \gamma_{Zn^{2+}} \frac{b_{Zn^{2+}}}{b^{\theta}}$
$a_{Cl^{-}} = \gamma_{Cl^{-}} \frac{b_{Cl^{-}}}{b^{\theta}}$
$b_{Zn^{2+}} = 0.01 mol/kg$
$b_{Cl^{-}} = 0.02 mol/kg$
$b^{\theta} = 1 mol/kg$
代入Nernst方程，解得：$\gamma_{Zn^{2+}} = 0.73$
$a_{Zn^{2+}} = 0.73 \times \frac{0.01}{1} = 0.0073$
负极反应：$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}$
正极反应：$2AgCl(s) + 2e^{-} \rightarrow 2Ag(s) + 2Cl^{-}(aq)$
电池反应：$Zn(s) + 2AgCl(s) \rightarrow 2Ag(s) + ZnCl_{2}(aq)$
步骤 2：计算25℃时电池反应的 $\Delta G_{m}$, $\Delta H_{m}$, $\Delta S_{m}$ 及标准平衡常数 $K^{\theta}$
$\Delta G_{m} = -nFE = -2 \times 96485 \times 1.1566 = -223.225 kJ/mol$
$\Delta H_{m} = \Delta G_{m} + T\Delta S_{m}$
$\Delta S_{m} = \frac{\partial E}{\partial T} \times nF = 4.02 \times 10^{-4} \times 2 \times 96485 = 72.386 J/(mol\cdot K)$
$\Delta H_{m} = -223.225 \times 10^{3} + 298.15 \times 72.386 = -208.99 kJ/mol$
$K^{\theta} = e^{\frac{-\Delta G_{m}^{\theta}}{RT}} = e^{\frac{223.225 \times 10^{3}}{8.314 \times 298.15}} = 4.93 \times 10^{-34}$
步骤 3：计算25℃时 $0.01mol\cdot kg^{-1}ZnCl_{2}$ 溶液的平均活度及平均活度因子
由Nernst方程：$E = E^{\theta} - \frac{RT}{zF} \ln \frac{a_{Zn^{2+}}}{a_{Cl^{-}}^{2}}$
$E^{\theta} = E_{AgCl/Ag} - E_{Zn^{2+}/Zn} = 0.985 V$
$a_{Zn^{2+}} = \gamma_{Zn^{2+}} \frac{b_{Zn^{2+}}}{b^{\theta}}$
$a_{Cl^{-}} = \gamma_{Cl^{-}} \frac{b_{Cl^{-}}}{b^{\theta}}$
$b_{Zn^{2+}} = 0.01 mol/kg$
$b_{Cl^{-}} = 0.02 mol/kg$
$b^{\theta} = 1 mol/kg$
代入Nernst方程，解得：$\gamma_{Zn^{2+}} = 0.73$
$a_{Zn^{2+}} = 0.73 \times \frac{0.01}{1} = 0.0073$