Using copper phthalocyanine and 2-methylimidazole as precursors, student Xiaoming directly synthesized a catalyst $\mathrm{Cu@ZIF-8}$ via a hydrothermal reaction. Experimental studies have shown that this catalyst can reduce $\mathrm{CO_2}$ to products such as ethanol and acetone when a reduction potential is applied, but it also produces many reduction by-products, such as $\mathrm{H_2}$ and $\mathrm{CO}$. To further investigate the performance of the catalyst, Xiaoming dropped $\mathrm{0.4 mg}$ of the catalyst onto a $\mathrm{1 cm^2}$ carbon paper electrode and placed it in an electrolytic cell with a volume of $\mathrm{31 mL}$. First, Xiaoming added $\mathrm{15 mL}$ of $\mathrm{0.5 mol/L}$ $\mathrm{KOH}$ solution to the electrolytic cell and purged it with $\mathrm{Ar}$ gas for $\mathrm{20min}$. After chronoamperometry electrolysis under sealed conditions, the mole fraction of $\mathrm{H_2}$ in the electrolytic cell was measured to be $\mathrm{1.584 \%}$, and the integral of the $\mathrm{I-t}$ curve of the chronoamperometry yielded a charge of $\mathrm{2.203 C}$. Then, student Xiaoming purged $\mathrm{CO_2}$ gas for $\mathrm{60min}$ under stirring conditions, and then switched to a closed system and performed chronoamperometry electrolysis. The mole fraction of $\mathrm{CO}$ in the electrolytic cell was measured to be $\mathrm{0.2110 \%}$, and the integral of the $\mathrm{I-t}$ curve of the chronoamperometry yielded a charge of $\mathrm{1.677 C}$. Assuming the experimental conditions are room temperature and normal atmospheric pressure, what is the amount of substance of $\mathrm{CO}$ obtained during the test? Considering the gas tightness of the device, what should be the theoretical Faradaic efficiency of the $\mathrm{CO}$ by-product?
小明同学使用酞菁铜和2-甲基咪唑作为前驱体，直接水热反应合成出了一种催化剂$\mathrm{Cu@ZIF-8}$。经过实验研究表明，该催化剂可以在施加还原电位时将 $\mathrm{CO_2}$ 还原为乙醇、丙酮等产物，但是也产生了很多还原副产物，例如 $\mathrm{H_2}$、$\mathrm{CO}$ 等。为了深入研究催化剂的性能，小明将 $\mathrm{0.4 mg}$ 催化剂滴在 $\mathrm{1 cm^2}$ 的碳纸电极上，放入体积为 $\mathrm{31 mL}$ 的电解池中。首先，小明向电解池加入 $\mathrm{15 mL}$ 的 $\mathrm{0.5 mol/L}$ $\mathrm{KOH}$ 溶液，通入 $\mathrm{20min}$ 的 $\mathrm{Ar}$ 气体。密闭下经过计时电流电解后，测量得到电解池内 $\mathrm{H_2}$ 的摩尔分数为$\mathrm{1.584 \%}$，计时电流的 $\mathrm{I-t}$ 图像经过积分后得到电荷量为 $\mathrm{2.203 C}$。之后小明同学在搅拌条件下通入了 $\mathrm{60min}$ 的 $\mathrm{CO_2}$ 气体，之后改为密闭体系、经过计时电流电解，测量得到电解池内 $\mathrm{CO}$ 的摩尔分数为 $\mathrm{0.2110  \%}$ ，计时电流的 $\mathrm{I-t}$ 图像经过积分后得到电荷量为 $\mathrm{1.677 C}$。假设实验条件为室温、正常大气压，请问测试过程中得到的 $\mathrm{CO}$ 的物质的量为多少？考虑装置气密性后，理论的 $\mathrm{CO}$ 副产物的法拉第效率应为多少？
$$1.380 \times 10^{-6} \text{ mol}$$ ; 15.88 %
$$1.380 \times 10^{-6} \text{ mol}$$ ; 17.50 %
$$1.380\times 10^{-6} \text{ mol}$$ ; 31.76 %
$$2.761\times 10^{-6} \text{ mol}$$ ; 15.88 %
$$2.761\times 10^{-6} \text{ mol}$$ ; 17.50 %
$$2.761\times 10^{-6} \text{ mol}$$ ; 31.76 %
Insufficient device airtightness $90 \%$, unable to calculate accurately
$$1.380\times 10^{-6} \text{ mol}$$ ; 15.88 %
$$1.380\times 10^{-6} \text{ mol}$$ ; 17.50 %
$$1.380\times 10^{-6} \text{ mol}$$ ; 31.76 %
$$2.761\times 10^{-6} \text{ mol}$$ ; 15.88 %
$$2.761\times 10^{-6} \text{ mol}$$ ; 17.50 %
$$2.761\times 10^{-6} \text{ mol}$$ ; 31.76 %
装置气密性不足 $90 \%$，无法准确计算
Faradaic efficiency calculation formula:

$$ FE =\frac{n_{CO}\times z_{CO}\times F}{Q}\times 100 \% $$

<Checkpoint>[$$ FE =\frac{n_{CO}\times z_{CO}\times F}{Q}\times 100 \% $$](1)</Checkpoint>

In this problem, the amount of substance of the gas needs to be calculated using the ideal gas law:
$$ n_{CO} =\frac{p\times V}{R\times T} $$

<Checkpoint>[$$ n_{CO} =\frac{p\times V}{R\times T} $$](1)</Checkpoint>

During the electrochemical reduction process, the reactions producing $\mathrm{1 mol }$ of $\mathrm{H_2}$ and $\mathrm{CO}$ both involve the transfer of $\mathrm{2 mol }$ of electrons:

$$ \mathrm{CO_2+2e+2H^+\rightarrow CO+H_2O} $$

$$ \mathrm{2H^++2e\rightarrow H_2} $$

<Checkpoint>[ $\mathrm{ z_{CO} = z_{H_2} = 2 }$](1)</Checkpoint>

Analyzing the question, the volume of the gas container is $\mathrm{31-15=16mL}$, the mole fraction of $\mathrm{CO}$ is $\mathrm{0.211 \% }$, the temperature is $\mathrm{298.15 K }$, the pressure is $\mathrm{101.325 kPa }$, R is $\mathrm{ 8.314 J/mol \cdot K }$, F is $\mathrm{ 96485 C/mol }$, and the measured Q is $\mathrm{ 1.677 C }$.

Therefore:

$$ n_{CO} =\frac{101.325\times 16\times 0.2110/100000}{8.314\times 298.15} =1.380\times 10^{-6}$$ mol

<Checkpoint>[$$ n_{CO} =1.380\times 10^{-6} \text{ mol}$$](1)</Checkpoint>

However, the question emphasizes the importance of the airtightness of the device, and the previous experiment with Ar atmosphere in KOH was to test this. Since only $\mathrm{H_2}$ is produced under these conditions, the theoretical FE should be 100% if there is no leakage, which is a common method for calibrating airtightness in electrochemistry.

<Checkpoint>[To be precise, under Ar protection and KOH electrolyte conditions, the electroreduction reaction can only produce hydrogen, so the Faradaic efficiency of hydrogen can be used to test the airtightness of the device](3)</Checkpoint>

$$ n_{H_2} =\frac{101.325\times 16\times 1.584/100000}{8.314\times 298.15} =1.036\times 10^{-5} mol$$

$$ {FE}_{H_2} =\frac{1.036\times 10^{-5}\times 2\times 96485}{2.203}=90.74 \%$$

The airtightness of the device is greater than 90%, and the overall airtightness is good.

<Checkpoint>[$$ {FE}_{H_2}=90.74 \% > 90 \% $$，good airtightness](2)</Checkpoint>

Therefore, the correction factor is obtained:
$\eta = 1/0.9074=1.102$

<Checkpoint>[Correction factor $\eta =1.102$](2)</Checkpoint>

The corrected ideal $CO$ amount:

$$ n_{CO} =1.380\times 10^{-6} \times 1.102 = 1.521 \times 10^{-6} $$ mol

<Checkpoint>[Corrected $$ n_{CO}  = 1.521\times 10^{-6} \text{ mol}$$](2)</Checkpoint>

The final true Faradaic efficiency:
$$ FE =\frac{1.521\times 10^{-6} \times 2\times 96485}{1.677}=17.50 \%$$

<Checkpoint>[Corrected $$ {FE}_{CO} =17.50 \%$$](2)</Checkpoint>
法拉第效率计算公式：

$$ FE =\frac{n_{CO}\times z_{CO}\times F}{Q}\times 100 \% $$

<Checkpoint>[$$ FE =\frac{n_{CO}\times z_{CO}\times F}{Q}\times 100 \% $$](1)</Checkpoint>

本题中，气体的物质的量需要用理想气体状态方程计算：
$$ n_{CO} =\frac{p\times V}{R\times T} $$

<Checkpoint>[$$ n_{CO} =\frac{p\times V}{R\times T} $$](1)</Checkpoint>

电化学还原过程中，产生$\mathrm{1 mol }$ $\mathrm{H_2}$、$\mathrm{CO}$的反应都是转移 $\mathrm{2 mol }$ 电子：

$$ \mathrm{CO_2+2e+2H^+\rightarrow CO+H_2O} $$

$$ \mathrm{2H^++2e\rightarrow H_2} $$

<Checkpoint>[ $\mathrm{ z_{CO} = z_{H_2} = 2 }$](1)</Checkpoint>

分析题干，气体容器体积为 $\mathrm{31-15=16mL}$，$\mathrm{CO}$ 摩尔分数 $\mathrm{0.211 \% }$，温度为 $\mathrm{298.15 K }$，压强 $\mathrm{101.325 kPa }$，R为 $\mathrm{ 8.314 J/mol \cdot K }$，F为 $\mathrm{ 96485 C/mol }$，测量得到 Q 为 $\mathrm{ 1.677 C }$。

因此：

$$ n_{CO} =\frac{101.325\times 16\times 0.2110/100000}{8.314\times 298.15} =1.380\times 10^{-6} mol$$

<Checkpoint>[$$ n_{CO} =1.380\times 10^{-6} \text{ mol}$$](1)</Checkpoint>

但是题干中强调注意装置气密性，前面 Ar 气氛下在 KOH 的实验就是为了检验这一点。由于该条件下只会还原产生$\mathrm{H_2}$，因此不漏气的情况下，理论FE应为100 %，这是电化学中校正气密性的常用方法。

<Checkpoint>[准确说出在Ar保护、KOH电解质的条件下，电还原反应只能产生氢气，因此可以用氢气的法拉第效率检验装置气密性](3)</Checkpoint>

$$ n_{H_2} =\frac{101.325\times 16\times 1.584/100000}{8.314\times 298.15} =1.036\times 10^{-5} mol$$

$$ {FE}_{H_2} =\frac{1.036\times 10^{-5}\times 2\times 96485}{2.203}=90.74 \%$$

装置气密性大于90%，整体气密性良好

<Checkpoint>[$$ {FE}_{H_2}=90.74 \% > 90 \% $$，气密性良好](2)</Checkpoint>

因此得到校正系数：
$\eta = 1/0.9074=1.102$

<Checkpoint>[校正系数 $\eta =1.102$](2)</Checkpoint>

校正后的理想 $CO$ 物质的量：

$$ n_{CO} =1.380\times 10^{-6} \times 1.102 = 1.521\times 10^{-6}\text{mol}$$

<Checkpoint>[校正后 $$ n_{CO}  = 1.521\times 10^{-6} \text{mol}$$ ](2)</Checkpoint>

最终真实的法拉第效率：
$$ FE =\frac{1.521\times 10^{-6} \times 2\times 96485}{1.677}=17.50 \%$$

<Checkpoint>[校正后 $$ {FE}_{CO} =17.50 \%$$](2)</Checkpoint>