A student conducted a titration analysis on a sample of $K_3Fe(C_2O_4)_3\cdot3H_2O$. The method involved: weighing an appropriate amount of the sample, dissolving it in dilute sulfuric acid, titrating it with $KMnO_4$ under heating until the endpoint was reached, then reducing the above titrated solution with $Zn$ until no color reaction with $KSCN$ was observed. After filtering out the excess $Zn$, the solution was titrated with $K_2Cr_2O_7$ until the endpoint was reached. Based on the results of these two titrations, the molar ratio of $Fe$ to $C_2O_4^{2-}$ in the sample could be calculated.

The student might have committed certain improper operations during the experiment:

1. The sample was left in the air for too long before titration, leading to efflorescence.
2. The sample was exposed to sunlight before titration, causing a color change.
3. A small amount of the solution splashed out while dissolving the sample.
4. A transparent burette was used instead of a brown one for the $KMnO_4$ titration.
5. Inadequate washing after filtering the $Zn$.

Assuming no other errors beyond the issues mentioned above, how many of these improper operations could potentially result in an elevated molar ratio value of $n(Fe)/n(C_2O_4^{2-})$?
某学生对一份$K_3Fe(C_2O_4)_3\cdot3H_2O$样品进行滴定分析。方法是：称量适量样品溶于稀硫酸，加热下用$KMnO_4$滴定至终点，接着用$Zn$将上述滴定液还原至对$KSCN$不显色后，过滤掉过量$Zn$，再用$K_2Cr_2O_7$滴定至终点。根据上述两次滴定的结果，可以计算出样品中$Fe$和$C_2O_4^{2-}$的摩尔比。

该生在实验时可能存在某些不当操作：

1. 滴定前样品放在空气中过久导致风化。
2. 滴定前样品暴露于阳光下，颜色发生变化。
3. 溶解样品时有少量溶液溅出。
4. 用$KMnO_4$滴定时没有用棕色滴定管，而是用透明滴定管。
5. 过滤$Zn$后洗涤不充分。

假定除上述问题外不存在其他过失，则以上不当操作中有几种可能导致$Fe$与$C_2O_4^{2-}$的摩尔比值$n(Fe)/n(C_2O_4^{2-})$偏高？
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Titration of $C_2O_4^{2-}$ content with $KMnO_4$.

<Checkpoint>[$KMnO_4$ titration of $C_2O_4^{2-}$ content](1)</Checkpoint>

Titration of $Fe^{2+}$ (i.e., Fe content) with $K_2Cr_2O_7$.

<Checkpoint>[$K_2Cr_2O_7$ titration of Fe content](1)</Checkpoint>

If the ratio $n(Fe)/n(C_2O_4^{2-})$ is high, it means the actual amount of $KMnO_4$ used is high or the actual amount of $K_2Cr_2O_7$ used is low.

<Checkpoint>[Actual amount of $KMnO_4$ used is high or actual amount of $K_2Cr_2O_7$ used is low](1)</Checkpoint>

Operation 1 does not affect the ratio. Weathering only affects the water of crystallization and does not influence the redox titration part.

<Checkpoint>[Operation 1 does not affect the ratio](1)</Checkpoint>

Operation 2 does not affect the ratio.

The photolysis reaction is  

$2K_3Fe(C_2O_4)_3\cdot3H_2O = 3K_2C_2O_4 + 2FeC_2O_4 + 2CO_2 +6H_2O$  

Although it superficially reduces the amount of $C_2O_4^{2-}$, since $Fe^{2+}$ is generated, the total electron transfer number remains unchanged during $KMnO_4$ titration, and thus the calculated $C_2O_4^{2-}$ content is unaffected.

<Checkpoint>[Operation 2 does not affect the ratio](5)</Checkpoint>

Operation 3 does not affect the ratio. If the solution splashes out before titration, the values for both titrations decrease proportionally, leaving the ratio unaffected.

<Checkpoint>[Operation 3 does not affect the ratio](1)</Checkpoint>

Operation 4 causes the ratio to decrease. Photodecomposition of $KMnO_4$ reduces its concentration, leading to higher titration usage.

<Checkpoint>[Operation 4 causes the ratio to decrease](1)</Checkpoint>

Operation 5 causes the ratio to decrease. If the residue is not fully washed, $Fe^{2+}$ does not completely enter the next titration step, resulting in a lower measured Fe content.

<Checkpoint>[Operation 5 causes the ratio to decrease](1)</Checkpoint>

In conclusion, no operation causes the ratio to increase.

<Checkpoint>[0 operations cause the ratio to increase](1)</Checkpoint>
$KMnO_4$滴定$C_2O_4^{2-}$含量。

<Checkpoint>[$KMnO_4$滴定$C_2O_4^{2-}$含量](1)</Checkpoint>

$K_2Cr_2O_7$滴定$Fe^{2+}$即$Fe$含量。

<Checkpoint>[$K_2Cr_2O_7$滴定$Fe$含量](1)</Checkpoint>

若$n(Fe)/n(C_2O_4^{2-})$偏高，意味着$KMnO_4$的实际用量偏高或$K_2Cr_2O_7$的实际用量偏低。

<Checkpoint>[$KMnO_4$的实际用量偏高或$K_2Cr_2O_7$的实际用量偏低](1)</Checkpoint>

操作1不影响比值。风化只影响结晶水，不影响氧化还原滴定的部分。

<Checkpoint>[操作1不影响比值](1)</Checkpoint>

操作2不影响比值。

光解反应为

$2K_3Fe(C_2O_4)_3\cdot3H_2O = 3K_2C_2O_4 + 2FeC_2O_4 + 2CO_2 +6H_2O$

虽然表面上减少了$C_2O_4^{2-}$的量，但由于生成了$Fe^{2+}$，在$KMnO_4$滴定时总电子得失数不变，不会影响由此算得的$C_2O_4^{2-}$含量。

<Checkpoint>[操作2不影响比值](5)</Checkpoint>

操作3不影响比值。滴定前溶液溅出，两部分滴定的数值同比例减少，不影响。

<Checkpoint>[操作3不影响比值](1)</Checkpoint>

操作4使得比值下降。$KMnO_4$光分解，使得浓度降低，从而滴定用量偏高。

<Checkpoint>[操作4使得比值下降](1)</Checkpoint>

操作5使得比值下降。滤渣未充分洗涤，则$Fe^{2+}$没有完全进入下一步滴定，测得$Fe$含量偏低。

<Checkpoint>[操作5使得比值下降](1)</Checkpoint>


综上，没有操作使比值升高。

<Checkpoint>[0种操作使比值升高](1)</Checkpoint>