Iodometry is a very commonly used analytical method in analytical chemistry. Based on the type of titration, it can be divided into direct iodometry and indirect iodometry. Among the following ten substances, some are suitable for direct iodometry, while others are suitable for indirect iodometry.  

1. $\mathrm{Cl_2}$  
2. $\mathrm{N_2H_4}$  
3. $\mathrm{O_3}$  
4. $\mathrm{NO_2^-}$  
5. $\mathrm{HCN}$  
6. $\mathrm{H_3AsO_3}$  
7. $\mathrm{H_3AsO_4}$  
8. $\mathrm{SO_2}$  
9. $\mathrm{Ce}^{4+}$  
10. $\mathrm{C_2H_5SH}$  

Regarding the type of iodometry suitable for these ten substances and the pH conditions during the titration process, the following statements are correct:
碘量法是分析化学中非常常用的分析方法。其根据滴定类型可分为直接碘量法和间接碘量法两种方式，以下十种物质有些适用于直接碘量法，有些适用于间接碘量法。

1. $\mathrm{Cl_2}$ 
2. $\mathrm{N_2H_4}$
3. $\mathrm{O_3}$
4. $\mathrm{NO_2^-}$ 
5. $\mathrm{HCN}$
6. $\mathrm{H_3AsO_3}$
7. $\mathrm{H_3AsO_4}$ 
8. $\mathrm{SO_2}$  
9. $\mathrm{Ce}^{4+}$ 
10. $\mathrm{C_2H_5SH}$

关于这十种物质所适用的碘量法类型，以及滴定过程中的酸碱度，下列说法正确的是：
Substances suitable for direct iodometry have even serial numbers.
Substances suitable for indirect iodometry have two sequence numbers that are not odd.
There are six types of substances suitable for indirect iodometry.
Substance No. 2 needs to be titrated in an acidic environment.
Substance No. 9 can be titrated under weakly alkaline conditions.
Substance No. 4 can be titrated under weakly alkaline conditions.
None of the above statements are correct
适用于直接碘量法的物质，其序号均为偶数
适用于间接碘量法的物质，有两个序号不为奇数
适用于间接碘量法的物质有6种
序号2的物质需要在酸性环境下滴定
序号9的物质可以在弱碱性条件下滴定
序号4的物质滴定可以在弱碱性条件下进行。
以上说法均不正确
The difference between direct iodometry and indirect iodometry is that direct iodometry uses a standardized $\mathrm{I_2}$ standard solution to directly titrate the substance, using excess iodine to indicate the endpoint; while indirect iodometry uses excess $\mathrm{I^-}$ to react with oxidizing substances to generate $\mathrm{I_2}$, and then uses sodium thiosulfate solution to titrate $\mathrm{I_2}$. Thus, the various substances are analyzed as follows:

<Checkpoint>[Direct iodometry uses a standardized $\mathrm{I_2}$ standard solution to directly titrate](1)</Checkpoint>

<Checkpoint>[Indirect iodometry uses excess $\mathrm{I^-}$ to react with oxidizing substances to generate $\mathrm{I_2}$, and then uses sodium thiosulfate solution to titrate $\mathrm{I_2}$](1)</Checkpoint>

1. $\mathrm{Cl_2}$: It cannot react directly with elemental iodine, but it has strong oxidizing properties and can oxidize $\mathrm{I^-}$, thus it uses indirect iodometry.

<Checkpoint>[$\mathrm{Cl_2}$ has strong oxidizing properties and can oxidize $\mathrm{I^-}$, thus it uses indirect iodometry](1)</Checkpoint>

2. $\mathrm{N_2H_4}$: It has strong reducing properties and can be oxidized by $\mathrm{I_2}$, thus it uses direct iodometry. Since the titration of $\mathrm{N_2H_4}$ will produce hydrogen ions:

$\mathrm{N_2H_4 + 2I_2 = N_2 + 4H^+ + 4I^-}$

The reaction produces hydrogen ions, so it should not be titrated in an acidic environment, but should be titrated under weakly alkaline buffer conditions, so option D is incorrect.

<Checkpoint>[$\mathrm{N_2H_4}$ has strong reducing properties and can be reduced by $\mathrm{I_2}$, thus it uses direct iodometry](1)</Checkpoint>

<Checkpoint>[$\mathrm{N_2H_4 + 2I_2 = N_2 + 4H^+ + 4I^-}$ produces hydrogen ions, and should be titrated under weakly alkaline buffer conditions](1)</Checkpoint>

3. $\mathrm{O_3}$ has strong oxidizing properties and can oxidize $\mathrm{I^-}$, thus it uses indirect iodometry.

<Checkpoint>[$\mathrm{O_3}$ has strong oxidizing properties, thus it uses indirect iodometry](1)</Checkpoint>

4. $\mathrm{NO_2^-}$ has oxidizing properties and can oxidize $\mathrm{I^-}$, thus it uses indirect iodometry. This reaction cannot be carried out under alkaline conditions, because the oxidation products of $\mathrm{NO_2^-}$ under neutral/weakly alkaline conditions are not fixed, and may be ammonia or nitrogen gas; but under acidic conditions, the oxidation product is fixed as nitric oxide, so option F is incorrect.

<Checkpoint>[$\mathrm{NO_2^-}$ has oxidizing properties, thus it uses indirect iodometry](1)</Checkpoint>

<Checkpoint>[The oxidation products of $\mathrm{NO_2^-}$ under neutral/weakly alkaline conditions are not fixed, and titration should be performed under acidic conditions](1)</Checkpoint>

5. $\mathrm{HCN}$ can be oxidized by elemental iodine to produce the pseudohalogen $\mathrm{ICN}$, so it can be titrated by direct iodometry.

<Checkpoint>[$\mathrm{HCN}$ can be oxidized by elemental iodine to produce the pseudohalogen $\mathrm{ICN}$, thus it uses direct iodometry](1)</Checkpoint>

6/7. $\mathrm{H_3AsO_3}$ has reducing properties, and arsenic in the +3 valence state will be oxidized to +5 by elemental iodine. Similarly, arsenic in the +5 valence state, $\mathrm{H_3AsO_4}$, can oxidize $\mathrm{I^-}$ and be reduced to +3. Therefore, $\mathrm{H_3AsO_3}$ uses direct iodometry, and $\mathrm{H_3AsO_4}$ uses indirect iodometry.

<Checkpoint>[$\mathrm{H_3AsO_3}$ in the +3 valence state will be oxidized to +5 by elemental iodine. Similarly, arsenic in the +5 valence state, $\mathrm{H_3AsO_4}$, can oxidize $\mathrm{I^-}$ and be reduced to +3](1)</Checkpoint>

8. $\mathrm{SO_2}$ can be oxidized to sulfate by $\mathrm{I_2}$, so direct iodometry is used.

<Checkpoint>[$\mathrm{SO_2}$ can be oxidized to sulfate by $\mathrm{I_2}$, thus it uses direct iodometry](1)</Checkpoint>

9. $\mathrm{Ce}^{4+}$ is a strong oxidant, and uses indirect iodometry to oxidize $\mathrm{I^-}$. $\mathrm{Ce}^{4+}$ will turn into a precipitate under alkaline conditions, so it cannot be titrated under weakly alkaline conditions, so option E is incorrect.

<Checkpoint>[$\mathrm{Ce}^{4+}$ is a strong oxidant, thus it uses indirect iodometry](1)</Checkpoint>

<Checkpoint>[$\mathrm{Ce}^{4+}$ will turn into a precipitate under alkaline conditions, and cannot be titrated under weakly alkaline conditions](1)</Checkpoint>

10. Thiols have reducing properties and can form disulfide bonds, so $\mathrm{I_2}$ oxidation is used, which is direct iodometry.

<Checkpoint>[Thiols have reducing properties and can form disulfide bonds, thus $\mathrm{I_2}$ oxidation is used, which is direct iodometry](1)</Checkpoint>

In summary, direct iodometry includes 2, 5, 6, 8, 10; indirect iodometry includes 1, 3, 4, 7, 9, so options A, B, and C are all incorrect.

<Checkpoint>[Direct iodometry includes 2, 5, 6, 8, 10; indirect iodometry includes 1, 3, 4, 7, 9](1)</Checkpoint>

In summary, options A-F are all incorrect, and option G is correct.
直接碘量法和间接碘量法的区别在于，直接碘量法利用标定好的$\mathrm{I_2}$标准溶液直接滴定物质，利用过量的碘指示终点；而间接碘量法是利用过量$\mathrm{I^-}$与氧化性物质生成$\mathrm{I_2}$，再利用硫代硫酸钠溶液滴定$\mathrm{I_2}$。由此对各个物质进行分析：

<Checkpoint>[直接碘量法利用标定好的$\mathrm{I_2}$标准溶液直接滴定](1)</Checkpoint>

<Checkpoint>[间接碘量法是利用过量$\mathrm{I^-}$与氧化性物质生成$\mathrm{I_2}$，再利用硫代硫酸钠溶液滴定$\mathrm{I_2}$](1)</Checkpoint>

1. $\mathrm{Cl_2}$：不能直接与碘单质反应，但具有强氧化性，可氧化$\mathrm{I^-}$，从而为间接碘量法。

<Checkpoint>[$\mathrm{Cl_2}$具有强氧化性，可氧化$\mathrm{I^-}$，为间接碘量法](1)</Checkpoint>

2. $\mathrm{N_2H_4}$：具有强还原性，可被$\mathrm{I_2}$氧化，为直接碘量法。由于$\mathrm{N_2H_4}$滴定会生成氢离子：

$\mathrm{N_2H_4 + 2I_2 = N_2 + 4H^+ + 4I^-}$

反应产生氢离子，故不应在酸性环境下滴定，应当在弱碱性缓冲条件下滴定，选项D错误。

<Checkpoint>[$\mathrm{N_2H_4}$具有强还原性，可被$\mathrm{I_2}$还原，为直接碘量法](1)</Checkpoint>

<Checkpoint>[$\mathrm{N_2H_4 + 2I_2 = N_2 + 4H^+ + 4I^-}$产生氢离子，应当在弱碱性缓冲条件下滴定](1)</Checkpoint>

3. $\mathrm{O_3}$具有强氧化性，可氧化$\mathrm{I^-}$，为间接碘量法。

<Checkpoint>[$\mathrm{O_3}$具有强氧化性，为间接碘量法](1)</Checkpoint>

4. $\mathrm{NO_2^-}$具有氧化性，可氧化$\mathrm{I^-}$，为间接碘量法。该反应不能在碱性条件下进行，因为$\mathrm{NO_2^-}$在中性/弱碱性条件下的氧化产物不固定，可能为氨气或氮气；但酸性条件下氧化产物固定为一氧化氮，从而选项F错误。

<Checkpoint>[$\mathrm{NO_2^-}$具有氧化性，为间接碘量法](1)</Checkpoint>

<Checkpoint>[$\mathrm{NO_2^-}$在中性/弱碱性条件下的氧化产物不固定，需在酸性条件滴定](1)</Checkpoint>

5. $\mathrm{HCN}$可以被碘单质氧化生成拟卤素$\mathrm{ICN}$，从而可用直接碘量法滴定。

<Checkpoint>[$\mathrm{HCN}$可以被碘单质氧化生成拟卤素$\mathrm{ICN}$，为直接碘量法](1)</Checkpoint>

6/7. $\mathrm{H_3AsO_3}$具有还原性，+3价态的砷会被碘单质氧化到+5价，同理+5价的砷$\mathrm{H_3AsO_4}$可氧化$\mathrm{I^-}$被还原为+3价。因此$\mathrm{H_3AsO_3}$为直接碘量法，$\mathrm{H_3AsO_4}$为间接碘量法。

<Checkpoint>[+3价态的$\mathrm{H_3AsO_3}$会被碘单质氧化到+5价，同理+5价的砷$\mathrm{H_3AsO_4}$可氧化$\mathrm{I^-}$被还原为+3价](1)</Checkpoint>

8. $\mathrm{SO_2}$ 可被$\mathrm{I_2}$氧化为硫酸根，从而采用直接碘量法。

<Checkpoint>[$\mathrm{SO_2}$ 可被$\mathrm{I_2}$氧化为硫酸根，为直接碘量法](1)</Checkpoint>

9. $\mathrm{Ce}^{4+}$ 为强氧化剂，采用间接碘量法氧化$\mathrm{I^-}$。$\mathrm{Ce}^{4+}$碱性条件下会变为沉淀，从而不能在弱碱性条件下滴定，选项E错误。

<Checkpoint>[$\mathrm{Ce}^{4+}$为强氧化剂，为间接碘量法](1)</Checkpoint>

<Checkpoint>[$\mathrm{Ce}^{4+}$碱性条件下会变为沉淀，不能在弱碱性条件下滴定](1)</Checkpoint>

10. 硫醇具有还原性，可生成二硫键，因此采用$\mathrm{I_2}$氧化，为直接碘量法。

<Checkpoint>[硫醇具有还原性，可生成二硫键，为直接碘量法](1)</Checkpoint>

综上，直接碘量法为2，5，6，8，10；间接碘量法为1，3，4，7，9，从而选项A,B,C均错误。

<Checkpoint>[直接碘量法为2，5，6，8，10；间接碘量法为1，3，4，7，9](1)</Checkpoint>

综上，选项A-F均不正确，选项G正确。