Electrochemistry is a fascinating branch of chemistry, where some substances undergo interesting reactions under the influence of electricity. For example, when electrolyzing acetic acid or sodium acetate solutions in different systems, the electrolysis products will vary depending on the most easily electron-losing part of acetic acid (sodium). Now there are the following statements:

1. When electrolyzing aqueous sodium acetate solution at normal temperature and pressure, 1.5 mol of gas will be produced at the anode for every 1 mol of electrons passed.

2. When electrolyzing aqueous sodium acetate solution at normal temperature and pressure, 0.25 mol of gas will be produced at the anode for every 1 mol of electrons passed.

3. When electrolyzing acetic acid in liquid $\mathrm{HF}$ solution at normal temperature and pressure, if the main product at the anode does not contain hydrogen and the mass fraction of $\mathrm{C}$ is $22.9\%$, then 0.167 mol of acetic acid is consumed for every 1 mol of electrons passed.

4. When electrolyzing acetic acid in liquid $\mathrm{HF}$ solution at normal temperature and pressure, if the oxidation product at the anode does not contain hydrogen and the mass fraction of $\mathrm{C}$ is $22.9\%$, then 0.125 mol of acetic acid is consumed for every 1 mol of electrons passed.

5. When electrolyzing acetic acid in liquid $\mathrm{HF}$ solution at normal temperature and pressure, if the oxidation product at the anode does not contain hydrogen and the mass fraction of $\mathrm{C}$ is $22.9\%$, then under this condition, acetic acid has two or more oxidation products, including $\mathrm{CO_2}$.

Which of the above statements are correct?
电化学是极富化学魅力的一个分支，一些物质在通电条件下会发生有趣的反应。例如在不同的体系下电解醋酸溶液或醋酸钠溶液时，电解产物会因乙酸(钠)最易失电子的部分的不同而不同。现在有以下说法：

1.常温常压电解醋酸钠水溶液时，每通过1 mol电子，阳极会产生1.5 mol气体

2.常温常压电解醋酸钠水溶液时，每通过1 mol电子，阳极会产生0.25 mol气体

3.常温常压电解醋酸的液态$\mathrm{HF}$溶液时，若阳极的主产物不含氢，$\mathrm{C}$的质量分数为$22.9\%$，则每通过1 mol电子，消耗0.167 mol乙酸

4.常温常压电解醋酸的液态$\mathrm{HF}$溶液时，若阳极的氧化产物不含氢，$\mathrm{C}$的质量分数为$22.9\%$，则每通过1 mol电子，消耗0.125 mol乙酸

5.常温常压电解醋酸的液态$\mathrm{HF}$溶液时，若阳极的氧化产物不含氢，$\mathrm{C}$的质量分数为$22.9\%$，则该条件下乙酸有包括$\mathrm{CO_2}$在内的两种或两种以上氧化产物

上述说法中正确的有
All statements are incorrect.
1
2
3
4
5
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4, 5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
1,3,4,5
2,3,4,5
所有说法均不正确
1
2
3
4
5
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
1,3,4,5
2,3,4,5
The reactions occurring during the electrolysis of aqueous sodium acetate solution are as follows:

Anode: $\mathrm{2CH_3COO^-\rightarrow C_2H_6+2CO_2+2e^-}$

<Checkpoint>[$\mathrm{2CH_3COO^-\rightarrow C_2H_6+2CO_2+2e^-}$](1)</Checkpoint>

Cathode: $\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}$

<Checkpoint>[Cathode: $\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}$](1)</Checkpoint>

Therefore, for every 1 mol of electrons passed, the anode produces 0.5 mol of $\mathrm{C_2H_6}$ and 1 mol of $\mathrm{CO_2}$, totaling 1.5 mol of gas.

Statement 1 is correct, statement 2 is incorrect.

<Checkpoint>[For every 1 mol of electrons passed, the anode produces 0.5 mol of $\mathrm{C_2H_6}$ and 1 mol of $\mathrm{CO_2}$](1)</Checkpoint>

During the electrolysis of acetic acid in liquid HF solution, it is known that the anodic oxidation product does not contain hydrogen, so the possible elements are only $\mathrm{C、O、F}$. Let the chemical formula be $\mathrm{C_xO_yF_z}$. Based on the mass fraction, we can write:

$\mathrm{\frac{12.01x}{12.01x+16y+19z}=22.9\%}$

Increase $\mathrm{x}$ starting from 1. An integer solution appears starting at $\mathrm{x=3}$, and we calculate $x:y:z=3:4:3$, i.e., $\mathrm{C_3O_4F_3}$. At this point, a reasonable chemical formula cannot be written, so it is discarded.

An integer solution appears at $\mathrm{x=4}$, and we calculate $x:y:z=4:3:6$, i.e., $\mathrm{C_4O_3F_6}$. The only reasonable chemical formula that can be written is $\mathrm{(CF_3CO)_2O}$.

Since the substrate is acetic acid, we temporarily do not consider complex products with a larger number of carbons. Consider the rationality of this product (trifluoroacetic anhydride):

In this system, acetic acid exists in molecular form and is not easily decarboxylated, but instead oxidizes the methyl side, replacing hydrogen with fluorine to become trifluoromethyl, "having different reactivity from aqueous sodium acetate solution", which is consistent with the conditions of the question. Therefore, this product is indeed the anodic oxidation product of acetic acid.

<Checkpoint>[The molecular formula of the anode product is $\mathrm{C_4O_3F_6}$](2)</Checkpoint>

<Checkpoint>[The chemical formula of the anode product is $\mathrm{(CF_3CO)_2O}$](1)</Checkpoint>

Therefore, the electrolysis equation can be written as:

$\mathrm{2CH_3COOH+6HF\rightarrow (CF_3CO)_2O+6H_2+H_2O}$

<Checkpoint>[$\mathrm{2CH_3COOH+6HF\rightarrow (CF_3CO)_2O+6H_2+H_2O}$](1)</Checkpoint>

Or express $\mathrm{H_2O}$ as $\mathrm{H_3O^+·HF_2^-}$

Then, for every 1 mol of electrons passed, $\frac{1}{6}$ mol of acetic acid is consumed. Statement 3 is correct, statement 4 is incorrect.

Acetic acid has only one oxidation product, and no $\mathrm{CO_2}$ is produced. Statement 5 is incorrect.

Select G
电解醋酸钠水溶液时，发生的反应如下：

阳极：$\mathrm{2CH_3COO^-\rightarrow C_2H_6+2CO_2+2e^-}$

<Checkpoint>[$\mathrm{2CH_3COO^-\rightarrow C_2H_6+2CO_2+2e^-}$](1)</Checkpoint>

阴极：$\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}$

<Checkpoint>[阴极：$\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}$](1)</Checkpoint>

则每通过1 mol电子，阳极产生0.5 mol$\mathrm{C_2H_6}$和1 mol$\mathrm{CO_2}$，共1.5 mol气体

说法1正确，说法2错误

<Checkpoint>[每通过1 mol电子，阳极产生0.5 mol$\mathrm{C_2H_6}$和1 mol$\mathrm{CO_2}$](1)</Checkpoint>

电解醋酸的液态HF溶液时，已知阳极氧化产物不含氢，则可能的元素只有$\mathrm{C、O、F}$，设化学式为$\mathrm{C_xO_yF_z}$，根据质量分数可列出：

$\mathrm{\frac{12.01x}{12.01x+16y+19z}=22.9\%}$

使$\mathrm{x}$从1开始增大，在$\mathrm{x=3}$开始有一个整数解，算得$x:y:z=3:4:3$，即$\mathrm{C_3O_4F_3}$，此时无法写出合理的化学式，舍去；

在$\mathrm{x=4}$有一个整数解，算得$x:y:z=4:3:6$，即$\mathrm{C_4O_3F_6}$，可以写出唯一合理的化学式为$\mathrm{(CF_3CO)_2O}$

由于底物为乙酸，暂不考虑碳数较大的复杂产物，考虑此产物(三氟乙酸酐)的合理性：

此体系下乙酸以分子形式存在，不易脱羧，转而氧化甲基侧，用氟取代氢使之变成三氟甲基，“具有与乙酸钠水溶液不同的反应活性”，与题干条件相符，故此产物确实为乙酸的阳极氧化产物。

<Checkpoint>[阳极产物的分子式为$\mathrm{C_4O_3F_6}$](2)</Checkpoint>

<Checkpoint>[阳极产物的化学式为$\mathrm{(CF_3CO)_2O}$](1)</Checkpoint>

因此可以写出电解的方程式：

$\mathrm{2CH_3COOH+6HF\rightarrow (CF_3CO)_2O+6H_2+H_2O}$

<Checkpoint>[$\mathrm{2CH_3COOH+6HF\rightarrow (CF_3CO)_2O+6H_2+H_2O}$](1)</Checkpoint>

或者将$\mathrm{H_2O}$表示为$\mathrm{H_3O^+·HF_2^-}$

则每通过1 mol电子，消耗$\frac{1}{6}$ mol乙酸，说法3正确，说法4错误

乙酸仅有一种氧化产物，且没有$\mathrm{CO_2}$产生，说法5错误

选G