题目

某HCl溶液中c_(mathrm{Fe)^3+}=c_(mathrm{Fe)^2+}=1,mathrm(mol)cdotmathrm(L)^-1，此时溶液中铁电对的条件电位E_(mathrm{Fe)^3+,mathrm(Fe)^2+}^theta'为（）A. E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta'=E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta+0.059lg(I_(mathrm{Fe)^3+})/(I_(mathrm{Fe)^2+)}（I为离子强度）B. E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta'=E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta+0.059lg(alpha_(mathrm{Fe)^2+})/(alpha_(mathrm{Fe)^3+)}C. E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta'=E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^thetaD. E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta'=E_(mathrm{Fe)^3+/mathrm(Fe)^2+}^theta+0.059lg(gamma_(mathrm{Fe)^3+}alpha_(mathrm{Fe)^2+})/(gamma_(mathrm{Fe)^2+)alpha_(mathrm{Fe)^3+}}

某HCl溶液中$c_{\mathrm{Fe}^{3+}}=c_{\mathrm{Fe}^{2+}}=1\,\mathrm{mol}\cdot\mathrm{L}^{-1}$，此时溶液中铁电对的条件电位$E_{\mathrm{Fe}^{3+},\mathrm{Fe}^{2+}}^{\theta'}$为（）

A. $E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta'}=E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta}+0.059\lg\frac{I_{\mathrm{Fe}^{3+}}}{I_{\mathrm{Fe}^{2+}}}$（$I$为离子强度）

B. $E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta'}=E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta}+0.059\lg\frac{\alpha_{\mathrm{Fe}^{2+}}}{\alpha_{\mathrm{Fe}^{3+}}}$

C. $E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta'}=E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta}$

D. $E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta'}=E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\theta}+0.059\lg\frac{\gamma_{\mathrm{Fe}^{3+}}\alpha_{\mathrm{Fe}^{2+}}}{\gamma_{\mathrm{Fe}^{2+}}\alpha_{\mathrm{Fe}^{3+}}}$

题目解答

答案

根据条件电极电位公式： \[ E^{\theta'} = E^{\theta} + 0.059 \lg \frac{a_{\text{Fe}^{3+}}}{a_{\text{Fe}^{2+}}} = E^{\theta} + 0.059 \lg \frac{\gamma_{\text{Fe}^{3+}} \alpha_{\text{Fe}^{2+}}}{\gamma_{\text{Fe}^{2+}} \alpha_{\text{Fe}^{3+}}} \] 在HCl溶液中，$ \text{Fe}^{3+} $ 和 $ \text{Fe}^{2+} $ 的活度系数 $ \gamma $ 及副反应系数 $ \alpha $ 均需考虑。A项未考虑 $ \alpha $，B项未考虑 $ \gamma $，C项未考虑任何影响，均不正确。D项全面符合条件电极电位的计算。答案：D. $ E^{\theta'}_{\text{Fe}^{3+}/\text{Fe}^{2+}} = E^{\theta}_{\text{Fe}^{3+}/\text{Fe}^{2+}} + 0.059 \lg \frac{\gamma_{\text{Fe}^{3+}} \alpha_{\text{Fe}^{2+}}}{\gamma_{\text{Fe}^{2+}} \alpha_{\text{Fe}^{3+}}} $