# 查看完整版本 : ive化學 唔該哂

yuenhang2 2012-12-19 11:24 PM

## ive化學 唔該哂

jmlo 2012-12-20 02:59 AM

## 回覆 1# 的帖子

Q6. I got the same results as yours.

Q10. Note that
Fe3+ + e-  ---->  Fe2+   E = 0.77V
Cr2O7 2- + 14H+ + 6e-  ----->  2Cr3+ + 7H2O   E = 1.33V
It implies that in the reaction mixture, Cr2O7 2- is preferentially reduced to Cr3+ while Fe2+ is oxidized to Fe3+.

Starting from
Fe2+  ---->  Fe3+ + e-                                              (1)
Cr2O7 2- + 14H+ + 6e-  ----->  2Cr3+ + 7H2O            (2)

We multiply (1) by 6 in order to balance the electrons transferred. It gives
Cr2O7 2- + 6Fe2+ + 14H+  ---->  6Fe3+ + 2Cr3+ + 7H2O        6*(1) + (2)

Now input the counter ions into the equation based on the given information:
K2Cr2O7 + 6FeSO4 + 14H+  ---->  3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O

Note that H+ is not balanced yet. Since all the species are dissolved in H2SO4, we can use SO4 2- as counter ions to balance H+:
K2Cr2O7 + 6FeSO4 + 7H2SO4  ---->  3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O

We notice also that K+ is not balanced. By the same token, we can add one SO4 2- on the RHS to give
K2Cr2O7 + 6FeSO4 + 7H2SO4  ----> 3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O + K2SO4

Finally, we check all ions on both sides again to make sure they are properly balanced.

[[i] 本帖最後由 jmlo 於 2012-12-20 03:00 AM 編輯 [/i]]

yuenhang2 2012-12-20 09:36 AM

[quote]原帖由 [i]jmlo[/i] 於 2012-12-20 02:59 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=349984241&ptid=21275395][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
Q6. I got the same results as yours.

Q10. Note that
Fe3+ + e-  ---->  Fe2+   E = 0.77V
Cr2O7 2- + 14H+ + 6e-  ----->  2Cr3+ + 7H2O   E = 1.33V
It implies that in the reaction mixture, Cr2O7 2- i ... [/quote]

Fe3+ + e-  ---->  Fe2+   E = 0.77V
Cr2O7 2- + 14H+ + 6e-  ----->  2Cr3+ + 7H2O   E = 1.33V

jmlo 2012-12-20 10:38 AM

## 回覆 3# 的帖子

[quote] 想問清楚d

Fe3+ + e-  ---->  Fe2+   E = 0.77V
Cr2O7 2- + 14H+ + 6e-  ----->  2Cr3+ + 7H2O   E = 1.33V [/quote]
The Fe2+/Fe3+ and Cr2O7 2-/Cr3+ are redox pairs that will participate in the actual oxidation and reduction processes. All other species such as K+ and SO4 2- are irrelevant.

[quote] 好像cr3+  +  e- >>>> cr2+    E= -0.41

The notation of standard reduction potentials is that positive value means favorable reduction. Since the process
Cr3+ + e-  ----->  Cr2+
has a negative reduction potential, it is not favorable.

yuenhang2 2012-12-20 11:22 AM

[quote]原帖由 [i]jmlo[/i] 於 2012-12-20 10:38 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=349994808&ptid=21275395][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

The Fe2+/Fe3+ and Cr2O7 2-/Cr3+ are redox pairs that will participate in the actual oxidation and reduction processes. All other species such as K+ and SO4 2- are irrelevant.

The notation of st ... [/quote]

thx
:victory:

yuenhang2 2012-12-20 12:14 PM

stock solution=10ppm
10ppm=10mg/dm3 = 0.01g/dm3
=0.01/55.847(fe 的 molecular weight)
=1.79x10^-4 mol/dm3

:smile_41:

[[i] 本帖最後由 yuenhang2 於 2012-12-20 12:26 PM 編輯 [/i]]

jmlo 2012-12-20 12:58 PM

[quote] 呢條個conc點計

stock solution=10ppm
10ppm=10mg/dm3 = 0.01g/dm3
=0.01/55.847(fe 的 molecular weight)
=1.79x10^-4 mol/dm3

Yes, something similar. Once you find the concentration of stock Fe solution, you can work out the concentrations of the standard solutions by dilution. For example,
Solution 1:
[Fe] = (1.79 x 10^-4 M)*(10/250) = 7.16 x 10^-6 M
Using the calculated concentrations, you can plot absorbance against the standard concentrations according to the Beer's law:
Absorbance = [concentration]*(e*l) = m*[concentration]
assuming extinction coefficient e and cell length l are constant in the measurements.

From the calibration plot, you know the slope m. Then in part (b) you can obtain the concentration by plugging in A = 0.327 into the calibration. One more step is needed to convert the concentration to the actual concentration of Fe in steam water sample; recall that the sample is diluted 10 times before experiment. Therefore, you need to multiply the concentration by 10 to get the final answer.