可以用来测定(AgCl(s))的标准摩尔生成Gibbs自由能Delta_(f) G_(m)^Theta的电池为（）A. (Ag(s)) mid (AgCl(s)) mid (HCl(aq)) mid (AgCl(s)) mid (Ag(s))B. (Ag(s)) mid (AgNO)_(3)((aq)) mid (HCl(aq)) mid (Cl)_(2)((p)_({Cl)_(2)}) mid (Pt(s))C. (Ag(s)) mid (AgNO)_(3)((aq)) mid mid (HCl(aq)) mid (AgCl(s)) mid (Ag(s))D. (Ag(s)) mid (AgCl(s)) mid (HCl(aq)) mid (Cl)_(2)((p)_({Cl)_(2)}) mid (Pt(s))
A. $\text{Ag(s)} \mid \text{AgCl(s)} \mid \text{HCl(aq)} \mid \text{AgCl(s)} \mid \text{Ag(s)}$
B. $\text{Ag(s)} \mid \text{AgNO}_{3}(\text{aq}) \mid \text{HCl(aq)} \mid \text{Cl}_{2}(\text{p}_{\text{Cl}_{2}}) \mid \text{Pt(s)}$
C. $\text{Ag(s)} \mid \text{AgNO}_{3}(\text{aq}) \mid \mid \text{HCl(aq)} \mid \text{AgCl(s)} \mid \text{Ag(s)}$
D. $\text{Ag(s)} \mid \text{AgCl(s)} \mid \text{HCl(aq)} \mid \text{Cl}_{2}(\text{p}_{\text{Cl}_{2}}) \mid \text{Pt(s)}$
题目解答
答案
解析
本题考查利用原电池测定物质的标准摩尔生成Gibbs自由能的知识，解题的关键在于找到一个电池反应，其电池反应的标准电动势$E^{\Theta}$与$\text{AgCl(s)}$的标准摩尔生成Gibbs自由能$\Delta_{f} G_{m}^{\Theta}$存在关联，即通过电池反应的$\Delta_{r}G_{m}^{\Theta}=-zFE^{\Theta}$，且该电池反应能对应$\text{AgCl(s)}$的生成反应。
选项A
电池$\text{Ag(s)} \mid \text{AgCl(s)} \mid \text{HCl(aq)} \mid \text{AgCl(s)} \mid \text{Ag(s)}$，其负极反应为$\text{Ag(s)}+\text{Cl}^{-}(aq)\rightarrow\text{AgCl(s)}+e^{-}$，正极反应为$\text{AgCl(s)}+e^{-}\rightarrow\text{Ag(s)}+\text{Cl}^{-}(aq)$，电池总反应为$0$，不能用于测定$\text{AgCl(s)}$的标准摩尔生成Gibbs自由能。
选项B
电池$\text{Ag(s)} \mid \text{AgNO}_{3}(\text{aq}) \mid \text{HCl(aq)} \mid \text{Cl}_{2}(\text{p}_{\text{Cl}_{2}}) \mid \text{Pt(s)}$，负极反应为$\text{Ag(s)}\rightarrow\text{Ag}^{+}(aq)+e^{-}$，正极反应为$\frac{1}{2}\text{Cl}_{2}(p_{\text{Cl}_{2}})+e^{-}\rightarrow\text{Cl}^{-}(aq)$，电池总反应为$\text{Ag(s)}+\frac{1}{2}\text{Cl}_{2}(p_{\text{Cl}_{2}})\rightarrow\text{Ag}^{+}(aq)+\text{Cl}^{-}(aq)$，该反应不是$\text{AgCl(s)}$的生成反应，不能用于测定$\text{AgCl(s)}$的标准摩尔生成Gibbs自由能。
选项C
电池$\text{Ag(s)} \mid \text{AgNO}_{3}(\text{aq}) \mid \mid \text{HCl(aq)} \mid \text{AgCl(s)} \mid \text{Ag(s)}$，负极反应为$\text{Ag(s)}\rightarrow\text{Ag}^{+}(aq)+e^{-}$，正极反应为$\text{AgCl(s)}+e^{-}\rightarrow\text{Ag(s)}+\text{Cl}^{-}(aq)$，电池总反应为$\text{AgCl(s)}\rightarrow\text{Ag}^{+}(aq)+\text{Cl}^{-}(aq)$，这是$\text{AgCl(s)}$的溶解反应，不是生成反应，不能用于测定$\text{AgCl(s)}$的标准摩尔生成Gibbs自由能。
选项D
电池$\text{Ag(s)} \mid \text{AgCl(s)} \mid \text{HCl(aq)} \mid \text{Cl}_{2}(\text{p}_{\text{Cl}_{2}}) \mid \text{Pt(s)}$，负极反应为$\text{Ag(s)}+\text{Cl}^{-}(aq)\rightarrow\text{AgCl(s)}+e^{-}$，正极反应为$\frac{1}{2}\text{Cl}_{2}(p_{\text{Cl}_{2}})+e^{-}\rightarrow\text{Cl}^{-}(aq)$，电池总反应为$\text{Ag(s)}+\frac{1}{2}\text{Cl}_{2}(p_{\text{Cl}_{2}})\rightarrow\text{AgCl(s)}$，此反应正是$\text{AgCl(s)}$的生成反应。根据$\Delta_{r}G_{m}^{\Theta}=-zFE^{\Theta}$，其中$z = 1$（反应转移电子数），$F$为法拉第常数，$E^{\Theta}$为该电池的标准电动势，通过测定$E^{\Theta}$就可以计算出$\text{AgCl(s)}$的标准摩尔生成Gibbs自由能$\Delta_{f} G_{m}^{\Theta}$。