1. When you want to identify an oxidising agent using sodium thiosulphate, is it a MUST to do so under acidic conditions?
2. and what does it mean when it is said that sodium thiosulphate can be used to analyse oxidising agent which liberate iodine from KI (iodometric titration)
does it mean that Na2S2O3 reacts with KI to give iodine, which as a reducing agent can then reduce the Oxidising agent to prove the O.A presence?
Oh and another question.
A mixture of Na2O and Na2O2, is dissolved in 100cm3 od distilled water to give solution H. Their reaction with water is as follows
Na2O + H2O -> 2NaOH
Na2O2 + 2H2O -> 2NaOH + H2O2
10.0cm3 of 0.0200mol dm-3 KMnO4 was required to titrate with 25.0cm3 of solution H. Calculate the total amount in moles of H2O2 in solution H.
Ok, so no. of moles of KMnO4 needed to react with 25.0cm of H is 0.0002mol so to titrate 100cm3 of H, it will be 0.0008mol.
The next step is weird...
Since 2MnO4- + 5H2O2 + 6H+ -> 2Mn2+ + 5O2 + 8H2O, (UNDER ACIDIC CONDITION!)
nH2O2 = 0.000800 X 2.5 = 0.00200 mol
I thought that equation is meant for the titration under acidic conditions?With NaOH, wouldn't it be considered alkaline and thus the equation
2MnO4 - + 3H2O2 -> 2MnO2 + 2OH- + 2H2O + 3O2
even if one is saying the NaOH is the product of the reaction and thus is not present initially in the experiment, there's still nothing to say this is done under acidic conditions. Can someone advise? Thanks :)
Originally posted by Audi:1. When you want to identify an oxidising agent using sodium thiosulphate, is it a MUST to do so under acidic conditions?
2. and what does it mean when it is said that sodium thiosulphate can be used to analyse oxidising agent which liberate iodine from KI (iodometric titration)
does it mean that Na2S2O3 reacts with KI to give iodine, which as a reducing agent can then reduce the Oxidising agent to prove the O.A presence?
Oh and another question.
A mixture of Na2O and Na2O2, is dissolved in 100cm3 od distilled water to give solution H. Their reaction with water is as follows
Na2O + H2O -> 2NaOH
Na2O2 + 2H2O -> 2NaOH + H2O2
10.0cm3 of 0.0200mol dm-3 KMnO4 was required to titrate with 25.0cm3 of solution H. Calculate the total amount in moles of H2O2 in solution H.
Ok, so no. of moles of KMnO4 needed to react with 25.0cm of H is 0.0002mol so to titrate 100cm3 of H, it will be 0.0008mol.
The next step is weird...
Since 2MnO4- + 5H2O2 + 6H+ -> 2Mn2+ + 5O2 + 8H2O, (UNDER ACIDIC CONDITION!)
nH2O2 = 0.000800 X 2.5 = 0.00200 mol
I thought that equation is meant for the titration under acidic conditions?With NaOH, wouldn't it be considered alkaline and thus the equation
2MnO4 - + 3H2O2 -> 2MnO2 + 2OH- + 2H2O + 3O2
even if one is saying the NaOH is the product of the reaction and thus is not present initially in the experiment, there's still nothing to say this is done under acidic conditions. Can someone advise? Thanks :)
1) Depends on the oxidizing agent. Not always necessary. Many oxidizing agents work best in acidic conditions, though.
2) It means that after your mystery oxidizing agent has oxidized I- to I2, you can use thiosulfate ions to reduce the I2 back to I-, using starch as an indicator for the presence of I2. How much thiosulfate ions you used (until the starch indicator shows you've completely reduce all I2 to I-), can tell you (by stoichiometry) how much I2 was present, which can tell you (by stoichiometry) how much of the oxidizing agent was present (which can help in identifying it, if you don't already know exactly what it was).
3) Assuming (as the question does) you want MnO4- to be reduced to Mn2+ (instead of MnO2), then you need acidic conditions, which you can provide by using excess of a non-redox-reactive acid, as part of the acidified KMnO4 solution used. This will not only neutralize all the OH- present, but will supply enough H+ such that MnO4- can be reduced to Mn2+.
thanks UltimaOnline!
I have got 2 questions which I do not know how to solve. (URGENT!)
1. A solution of FA2 was prepared by dissolving 9.60g of 'iron tablet' in dilute H2SO4. THe solution was then made up to 250cm3 with distilled water. 42.00cm3 of FA3 containing 0.100mol dm-3 KMnO4 was put into a 250cm3 graduated flask and made up to the mark with distlled water. The solution was then labelled FA4 and put into a second burette. 25.0cm3 of FA2 was pipetted into a conical flask and acidified with dilute H2SO4. The resultant solution requires 21.65cm3 of FA4 for reaction.
a) calculate amt of KMnO4 in 250cm3 if FA4 = 0.100 X (250/1000) = 0.0250mol
b) calculate amt of KMnO4 in vol. of FA4 needed for oxidation of 25.0cm3 of FA2
0.100 X (21.65/1000) = 2.17 X 10^-3 mol
c) calculate amt of Fe2+ ions reacted with the KMnO4 from part b
d) calculate amt of Fe2+ in 250cm3 of FA2
e) calculate %tage by mass of Fe2+ ions in the tablets used to make FA2 (Ar of Fe =55.8) answer is 10.6%
cant do question c and d. by the way, i'm not sure why question c comes before question d because if i were to do the question, i thought i had to solve question d first. Or did I cauclated wrongly for the first 2 questions?
Pls give me detailed guidelines on how to solve the last 3 parts (ok, maybe just for part C)
2. when a 1cm length of iron wire is dissolved in acid and titrated with 0.025mol dm-3 of K2Cr2O7, 10.50cm3 of oxidant solution is required to reach the end point. Calculate the mass of iron in the wire.
All I did was find the no. of moles of K2Cr2O7 needed to reach end-point and I do not know how to go on from there.
Thanks for helping!
c) apply stoichiometry of balanced redox between Fe2+ and MnO4-. This will give u amt of Fe2+ in 25cm3.
d) multiply ur answer from (c) by 10, to give u amt of Fe2+ in 250cm3.
e) From amt of Fe2+, find sample mass of Fe2+. Then take % by mass out of 9.60g
Q2) apply stoichiometry of balanced redox btwn Fe2+ and Cr2O7 2-.
Originally posted by UltimaOnline:c) apply stoichiometry of balanced redox between Fe2+ and MnO4-. This will give u amt of Fe2+ in 25cm3.
d) multiply ur answer from (c) by 10, to give u amt of Fe2+ in 250cm3.
e) From amt of Fe2+, find sample mass of Fe2+. Then take % by mass out of 9.60g
Q2) apply stoichiometry of balanced redox btwn Fe2+ and Cr2O7 2-.
Thanks for the valuable advice!
Hello to fellow forumers again. Have a couple of questions which I need help, in screening the answers, etc.
1. A piece of iron wire weighs 0.2756 g. When dissolved in acid it is oxidised to Fe 2+ and titrated with 40.80 cm3 of a 0.0200mol dm-3 KMnO4 solution. What is the percentage purity of the iron wire?
I wrote out the overall eqn which is
5Fe + 2MnO4- + 16H+ -> 5 Fe2+ + 2Mn2+ + 8H2O
Amt of KMnO4 needed for titration is 8.16 X 10^-4 mol
comparing mole ratio amt of pure Fe should be 2.04 X 10^-3 mol and the mass, after multiplying by molar mass is given by 0.1138g
so %tage purity i got is 0.1138/0.2756 X 100 = 41.3%, however answer given is 82.6%. Did I miss out anything?
2. 25.0 cm3 of a solution of iron(II) ethandioate FeC2O4 was titrated against a 0.0200mol dm-3 solution of acidified potassium manganate (VII) solution. It was found that 33.75 cm3 of the potassium manganate (VII) solution was needed for complete reaction. What was the concentration of the iron (II) ethandioate solution?
Reminder : Both Fe2+ and C2O4 2- are oxidized by MnO4-
Got the oxidation reactions for both Fe 2_ and C2O4 2-.
MnO4- + 8H+ + 5 Fe 2+ -> Mn2+ + 4H2O +5Fe3+
2MnO4 - + 16H+ + 5C2O4 2 -> 2Mn2+ 8H2O + 10CO2
Overall eqn being
4MnO4- + 32H+ +10Fe2+ + 5C2O4 2- -> 4Mn2+ + 16H2O + 10Fe3+ + 10CO2
amt of KMnO4 needed for complete rxn is 6.75 X 10^-4 mol.
so from the overall eqn, I can find the total amt of Fe2+ and C2O4 2-
which is
(6.75 X 10^-4) / 4 X 15 = 2.53125 X 10^-3 mol
to find contrn,
I used 2.53125 X 10^-3 / (25/1000) = 0.10125mol dm-3.
answer given is 0.0450mol dm-3. been staring at my workings for so long and yet cant deduce my mistake. LOL.
Need advice urgently. Thanks!
Originally posted by Audi:
Thanks for the valuable advice!Hello to fellow forumers again. Have a couple of questions which I need help, in screening the answers, etc.
1. A piece of iron wire weighs 0.2756 g. When dissolved in acid it is oxidised to Fe 2+ and titrated with 40.80 cm3 of a 0.0200mol dm-3 KMnO4 solution. What is the percentage purity of the iron wire?
I wrote out the overall eqn which is
5Fe + 2MnO4- + 16H+ -> 5 Fe2+ + 2Mn2+ + 8H2O
Amt of KMnO4 needed for titration is 8.16 X 10^-4 mol
comparing mole ratio amt of pure Fe should be 2.04 X 10^-3 mol and the mass, after multiplying by molar mass is given by 0.1138g
so %tage purity i got is 0.1138/0.2756 X 100 = 41.3%, however answer given is 82.6%. Did I miss out anything?
2. 25.0 cm3 of a solution of iron(II) ethandioate FeC2O4 was titrated against a 0.0200mol dm-3 solution of acidified potassium manganate (VII) solution. It was found that 33.75 cm3 of the potassium manganate (VII) solution was needed for complete reaction. What was the concentration of the iron (II) ethandioate solution?
Reminder : Both Fe2+ and C2O4 2- are oxidized by MnO4-
Got the oxidation reactions for both Fe 2_ and C2O4 2-.
MnO4- + 8H+ + 5 Fe 2+ -> Mn2+ + 4H2O +5Fe3+
2MnO4 - + 16H+ + 5C2O4 2 -> 2Mn2+ 8H2O + 10CO2
Overall eqn being
4MnO4- + 32H+ +10Fe2+ + 5C2O4 2- -> 4Mn2+ + 16H2O + 10Fe3+ + 10CO2
amt of KMnO4 needed for complete rxn is 6.75 X 10^-4 mol.
so from the overall eqn, I can find the total amt of Fe2+ and C2O4 2-
which is
(6.75 X 10^-4) / 4 X 15 = 2.53125 X 10^-3 mol
to find contrn,
I used 2.53125 X 10^-3 / (25/1000) = 0.10125mol dm-3.
answer given is 0.0450mol dm-3. been staring at my workings for so long and yet cant deduce my mistake. LOL.
Need advice urgently. Thanks!
Q1. Your error is that MnO4- oxidizes Fe2+ to Fe3+, not Fe to Fe2+
Q2. See this other thread here : http://www.sgforums.com/forums/2297/topics/393535
Originally posted by UltimaOnline:c) apply stoichiometry of balanced redox between Fe2+ and MnO4-. This will give u amt of Fe2+ in 25cm3.
d) multiply ur answer from (c) by 10, to give u amt of Fe2+ in 250cm3.
e) From amt of Fe2+, find sample mass of Fe2+. Then take % by mass out of 9.60g
hi UltimaOnline,
I did this question as you had specified and the overall equation is
(c)MnO4- + 8H+ + 5Fe 2+ -> Mn2+ + 4H2O +5Fe3+
From the earlier part we know that amt of KMnO4 needed for oxidation of 25.0cm3 of FA2 is 2.165 X 10^-3 mol.
therefore amt of Fe2+ = 2.165 X 10^-3 X 5 = 0.0108 mol
(d) ans : 0.108 mol of Fe 2+
(e) %tage by mass = (0.108 X 55.8) / 9.60 X 100 = 62.775% not 10.2%
Dont really see any mistakes, did i go wrong somewhere?
'O' & 'A' Level Qn.
A solution of FA2 was prepared by dissolving 9.60g of 'iron tablet' in dilute H2SO4. THe solution was then made up to 250cm3 with distilled water. 42.00cm3 of FA3 containing 0.100mol dm-3 KMnO4 was put into a 250cm3 graduated flask and made up to the mark with distlled water. The solution was then labelled FA4 and put into a second burette. 25.0cm3 of FA2 was pipetted into a conical flask and acidified with dilute H2SO4. The resultant solution requires 21.65cm3 of FA4 for reaction. Find the % by mass of iron in the tablets.
Solution (in white) :
Let x be the moles of Fe in 9.6g of tablet.
This implies amt of Fe in 250cm3 solution = x mol.
This implies amt of Fe in 25cm3 solution = 0.1x mol.
Amt of KMnO4 in 250cm3 solution = 4.2x10^-3 mol.
This implies amt of KMnO4 in 21.65cm3 solution = 3.6372x10^-4 mol.
Balanced on stoichiometry of balanced redox, 0.1x mol of Fe2+ requires (1/5)*(0.1x) = 0.02x mol of MnO4-.
Since we know 3.6372x10^-4 mol of MnO4- was used, this implies :
0.02x = 3.6372x10^-4
x = 0.01816 mol.
Hence sample mass of Fe in tablet = 1.0148g.
Hence % by mass = 1.0148g / 9.6g = 10.57% = 10.6% (3 s.f.)
Hi everyone! I do have questions on redox too, can someone help me out with this?
1. The metallic ion X^n+ is oxidized to XO3- by MnO4- ion in acidic solution. If 1.93 X 10^-3 mol of X^n+ required 1.16 X 10^-3 mols of MnO4- for oxidation, what is the value of n?
2. An aq. solution contains 1 mol of S2O3 2- ions and this reduces 4 mol of Cl2 molecules. What is the sulfur containing product of this reaction? Present your workings clearly.
Not very sure of this, been writing down equations for hours on end but still cant solve. My guess is S4O6 2-, well, i'm not sure. haha.
3. If NH2OH is oxidized by Fe3+, can someone explain to me what you will get.
thanks for all the help
Originally posted by anpanman:Hi everyone! I do have questions on redox too, can someone help me out with this?
1. The metallic ion X^n+ is oxidized to XO3- by MnO4- ion in acidic solution. If 1.93 X 10^-3 mol of X^n+ required 1.16 X 10^-3 mols of MnO4- for oxidation, what is the value of n?
2. An aq. solution contains 1 mol of S2O3 2- ions and this reduces 4 mol of Cl2 molecules. What is the sulfur containing product of this reaction? Present your workings clearly.
Not very sure of this, been writing down equations for hours on end but still cant solve. My guess is S4O6 2-, well, i'm not sure. haha.
3. If NH2OH is oxidized by Fe3+, can someone explain to me what you will get.
thanks for all the help
1.
Is your XO3- uninegative or trinegative? Be careful.
Based on stoichiometry of the reduction half-equation of MnO4- to Mn2+, work out moles of e- transferred per mole of XO3-.
Based on final O.S. of X (which depends on whether XO3- is uninegative or trinegative), and moles of e- removed (by MnO4-), you can work out original O.S. of X.
2.
(edited : see next 2 posts below)
3.
NH2OH (O.S. -1) can be oxidized to a variety of compounds (depending on the nature, strength and mechanism of the oxidation process) in which N has a higher O.S., eg. N2O (O.S. +1), NO2- (O.S. +3), NO2 (O.S. +4), NO3- (O.S. +5), etc.
You will have to determine which compound NH2OH is oxidized to, based on information given in the question (eg. based on moles of e- transferred, work out new O.S. in the oxidation product).
Originally posted by UltimaOnline:4 moles of Cl2 will donate 8 moles of e-.
Hence final O.S. will be original O.S. + e- accepted, ie. +2 + -8 = -6. Hence final sulfur containing product will be sulfate(VI) ion, SO4 2-.
One question. Because Cl2 will be reduced, shouldnt it gain electrons? Why does it "donate" e- away in this case? thanks
Yes, you're right there. My bad for the error.
The relevant redox equation is :
4Cl2 + S2O3 2- + 5H2O --> 2SO4 2- + 10H+ + 8Cl-
4 moles of Cl2 will ACCEPT 8 moles of e-. Hence final O.S. of S in the oxidized product, will be original O.S. + e- donated/accepted, ie. (+2) + 0.5(+8) = (+2) + (+4) = +6. Hence final sulfur containing product will be sulfate(VI) ion, SO4 2-.
The 0.5(+8) is due to the stoichiometry : two sulfate(VI) ions are generated per four moles of chlorine reduced. Even if you didn't know the equations (though for H2/H1 Chem you should), you could still obtain the answer because there are two S atoms in thiosulfate ion, so EACH S atom would donate 4e- to provide the 8e- required to reduce 4 chlorine molecules into 8 chloride ions.
(Actually, this does not show the entire picture. To gain a better understanding, you need to draw the structure of thiosulfate, tetrathionate and sulfate(VI) ions, and determine the OS of the individual atoms. You'd realize that the two S atoms in thiosulfate each have a different O.S.: +0 and +4, which gives an average of +2. To understand *exactly* what goes on in any redox reaction, ie. the flow of electrons, you'll need to study the mechanism (redox mechanisms are more complex than the mechanisms covered in H2 Org Chem), which is beyond 'A' levels.)
Originally posted by UltimaOnline:Yes, you're right there. My bad for the error.
The relevant redox equation is :
4Cl2 + S2O3 2- + 5H2O --> 2SO4 2- + 10H+ + 8Cl-
4 moles of Cl2 will ACCEPT 8 moles of e-. Hence final O.S. of S in the oxidized product, will be original O.S. + e- donated/accepted, ie. (+2) + 0.5(+8) = (+2) + (+4) = +6. Hence final sulfur containing product will be sulfate(VI) ion, SO4 2-.
The 0.5(+8) is due to the stoichiometry : two sulfate(VI) ions are generated per four moles of chlorine reduced. Even if you didn't know the equations (though for H2/H1 Chem you should), you could still obtain the answer because there are two S atoms in thiosulfate ion, so EACH S atom would donate 4e- to provide the 8e- required to reduce 4 chlorine molecules into 8 chloride ions.
(Actually, this does not show the entire picture. To gain a better understanding, you need to draw the structure of thiosulfate, tetrathionate and sulfate(VI) ions, and determine the OS of the individual atoms. You'd realize that the two S atoms in thiosulfate each have a different O.S.: +0 and +4, which gives an average of +2. To understand *exactly* what goes on in any redox reaction, ie. the flow of electrons, you'll need to study the mechanism (redox mechanisms are more complex than the mechanisms covered in H2 Org Chem), which is beyond 'A' levels.)
But if I present the workings you have written, wouldnt I be like "working backwards"? I do understand the part where you say each S atom has to contribute 4 e-, but how do I derive the answer from that part?
Some other questions I have probs with.
By writing half equations, give the full balanced equation of the following rxns described:
1. When potassium iodide is added to acidified H2O2, a brown solution of iodine appears. No oxygen is obtained.
2. IO3- + I- + H+ + Cl- -> ICl2- (there are no other Cl containing compounds)
Yeah, not very sure how to write the half equations for this 2.
thanks a lot
EDITED: I understand the reasoning that UltimaOnline gave regarding the suplhur question already... but still dont get why your explanation only takes into account the S atoms giving away e- but not the O atoms. Is it because the question only gives focus on the sulphur product, so we need only focus on the S atoms?
The reason is because only the O.S. of the S atoms changes, while the O.S. of the O atoms remain as -2 (this is because O is more electronegative than S; oxidation state = formal charge + electronegativity consideration, meaning O.S. is the charge of an atom after taking electronegativity into account).
To reiterate, drawing out the structure of the thiosulfate S2O3 2- ion, you'll see that the two S atoms in thiosulfate S2O3 2- each has a different O.S. : +0 and +4.
In terms of O.S., 1 of the S atoms (with initial O.S. of +4) loses 2 electrons, while the other S atom (with initial O.S. of +0) loses 6 electrons.
This makes a total of 8 electrons, used for reducing 4Cl2 to 8Cl-.
The thiosulfate S2O3 2- ion then rearranges itself (upon losing the 8 e-) into 2 sulfate(VI) SO4 2- ions.
-----------------------------------------
1. When potassium iodide is added to acidified H2O2, a brown solution of iodine appears. No oxygen is obtained.
[reduction] H2O2 --> H2O
[oxidation] I- --> I2
Balance the half equations above, then combine them to obtain the balanced redox equation.
(Note that the brown solution is the uninegative triiodide ion I3-(aq), but unless otherwise specified by the question, it's ok to write it as I2(aq). Iodine being non-polar, does not dissolve well in hydrogen-bonded water.)
-----------------------------------------------------
'A' Level Qn
Draw the uninegative iodine dichloride ion. State all formal charges and oxidation states within the ion. State its electron and ionic geometry.
Balance the following equation :
IO3(-) + I- + H+ + Cl- ---> ICl2(-) + H2O
Solution :
In the iodine dichloride ion, [Cl-I-Cl]-, the 2 Cl atoms have no formal charge, and an O.S. of -1. The I atom has a 1- formal charge, and an O.S. of +1. The sum of formal charges, as well as the sum of oxidation states, adds up to the uninegative ionic charge.
Electron geometry = trigonal bipyramidal.
Ionic geometry = linear.
What is happening during the redox reaction : the iodate(V) ion is reduced, and the iodide ion is oxidized, both becoming the iodine atom in the iodine dichloride ion (O.S. of +1). Notice that the O.S. of Cl does not change (-1 before redox, -1 after redox).
[reduction] 4e- + 6H+ + 2Cl- + IO3(-) ---> [Cl-I-Cl]- + 3H2O
[oxidation] 2Cl- + I- ---> [Cl-I-Cl]- + 2e-
[balanced redox] 6Cl- + 6H+ + 2I- + IO3(-) ---> 3[Cl-I-Cl]- + 3H2O