1. 5 cm3 of 0.10 mol dm-3 of potassium manganate (VII) reacts with 12.0cm3 of nitrogen monoxide at r.t.p. A colourless solution containing a brown precipitate of manganese (IV) oxide, MnO2 was produced. Determine the final oxidation number of nitrogen in the compound and hence suggest a possible formula of the nitrogen-containing product of the reaction. [5]
2. Iodine and chlorine react together to form compound X (ICl n).
When 0.0010 mol of X was react with an excess of KI(aq), all of its iodine was converted to I2. The iodine liberated required 40.0 cm3 of 0.10 mol dm-3 sodium thiosulfate for complete reaction.
i)calculate the amount (in moles) of iodine produced. [2]
ii)hence calculate the value of n in ICl n. [2]
iii) Write a balanced equation for the reaction between Cl2 and I2. [1]
A level redox chemistry help!
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can anybody help?
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..now i just realised that these qns were in my homework too . i dunno how to do the 1st qn and probably the 2nd as well
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Originally posted by tay77:1. 5 cm3 of 0.10 mol dm-3 of potassium manganate (VII) reacts with 12.0cm3 of nitrogen monoxide at r.t.p. A colourless solution containing a brown precipitate of manganese (IV) oxide, MnO2 was produced. Determine the final oxidation number of nitrogen in the compound and hence suggest a possible formula of the nitrogen-containing product of the reaction. [5]
2. Iodine and chlorine react together to form compound X (ICl n).
When 0.0010 mol of X was react with an excess of KI(aq), all of its iodine was converted to I2. The iodine liberated required 40.0 cm3 of 0.10 mol dm-3 sodium thiosulfate for complete reaction.
i)calculate the amount (in moles) of iodine produced. [2]
ii)hence calculate the value of n in ICl n. [2]
iii) Write a balanced equation for the reaction between Cl2 and I2. [1]
Q1.Find O.S. of N in NO.
Find no. of e- accepted by so-much of MnO4- (when it's reduced to MnO2).
Find no. of e- accepted by so-much of MnO4- (when it's reduced to MnO2), per mole of NO.
Hence calculate new O.S. = initial O.S. + moles of e- removed.
Q2.
Write a balanced equation with coefficients in terms of n, for the reaction between X and KI to generate I2 and KCl.
[Ans : ICln + nKI --> ((n+1)/2)I2 + nKCl ]
From stoichiometry of balanced redox between thiosulfate and iodine, find amt of iodine present. [Ans = 0.02 mol]
Hence find amt of iodine per mole of X. [Ans = 2 mol]
Therefore, by inspection of the balanced equation (which you wrote at the beginning of this solution), you realize ((n+1)/2) = 2.
Finally solve for n, and hence write the balanced equation between I2 and Cl2 to generate ICln.
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oo is the answer N2O5 for qn 1?
I wasn't sure because it seems to be impossible to balance the half equation of the oxidation in NO:
NO -> N2O5 + 3e
Thanks again :)
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Originally posted by qdtimes2:
oo is the answer N2O5 for qn 1?
I wasn't sure because it seems to be impossible to balance the half equation of the oxidation in NO:
NO -> N2O5 + 3e
Thanks again :)
New O.S. = (+2) + (+3) = +5Hence, N2O5 (in which N has an O.S. of +5) is indeed a valid answer.
The oxidation half-equation can be balanced as follows :
2NO + 3H2O --> N2O5 + 6H+ + 6e-
You can double-check by writing the balanced redox, which does indeed give a 1:1 stoichiometry betwen NO and MnO4-.
The data in the question indeed corroborates this :
"5 cm3 of 0.10 mol dm-3 (ie. 5x10^-4 mol) of potassium manganate (VII) reacts with 12.0cm3 of nitrogen monoxide at r.t.p. (ie. 5x10^-4 mol)."
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hey, thanks for the replies, but i have one last qn to ask!
i)copper can reduce niric acid to NO2. Write a balanced equation for this reaction.
im not sure if its:
1) Cu +2HNO3 +2H+ --> Cu (2+) + 2NO2 + 2H20 OR
2) 4NO3 --> 2H20 +4NO2+O2 OR
3) Cu +4H(+) + 2N03(-) --> Cu(2+) +2NO2+ 2H20
can anybody explain to me which is the right ans?
in addition, i also does not know how to write a balanced equation for the reduction of nitric acid to NH4+ by aluminium.
suggest a reason why the reduction products of nitric acid are different for each of these two metals. ( i think it has something to do with aluminium protecting the metal from undergoing further reaction, but i do not know how to explain it in detail)
pls help!!
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ohh. for the second equation. I mis-typed. shld be:
4HNO3 --> 2H2O + 4NO2 +O2.
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'O' & 'A' Level Qn.
Copper can reduce nitric acid to NO2. Write a balanced equation for this reaction.
Is it :
1) Cu +2HNO3 +2H+ --> Cu (2+) + 2NO2 + 2H20 OR
2) Cu +4H(+) + 2NO3(-) --> Cu(2+) +2NO2+ 2H20
Write a balanced equation for the reduction of nitric acid to NH4+ by aluminium.
Suggest a reason why the reduction products of nitric acid are different for each of these two metals.
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Solution :
It depends on whether concentrated (ie. pure liquid, no water present) or aqueous acid is used. If concentrated, the HNO3 exists as a covalent, unionized molecule. If aqueous, it dissociates and ionizes to H+ and NO3-.
If concentrated, use equation #1. If aqueous, use equation #2. If the exam question is ambiguous, give both with qualifications (ie. explanations).
Regarding the QA test for nitrate(V) ions, the aluminium foil or dervada's alloy will reduce nitrate(V) to ammonium cation, which will undergo a Bronsted-Lowry acid-base proton transfer reaction with the hydroxide ion (from the alkali added), to generate aqueous ammonia, which when heated overcomes the hydrogen bonds with water and generates gaseous ammonia, which can be detected by the hydrolysis of ammonia on moist red litmus paper generating the hydroxide ion, turning the litmus paper blue.
Reduction half-equation :
NO3(-) + 10H(+) + 8e- ---> NH4(+) + 3H2O
Oxidation half-equation :
Al ---> Al(3+) + 3e-
Combine the two half-equations to get the balanced redox equation.
The reasons for the different redox products of nitric(V) acid with aluminium versus copper, has to do with the different oxidation potentials (ie. reducing strength) of Al versus Cu. Al is a lot more reactive, has a more positive oxidation potential, and is hence a more powerful reducing agent than Cu.
Hence Al is able to reduce the O.S. of N from +5 (in HNO3) all the way to -3 (in NH4+), while Cu is only able to reduce the O.S. of N from +5 (in HNO3) down slightly to +4 (in NO2).
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but if the qn is only worth 1 mark, do u also quote both equation?
Regarding the QA test for nitrate(V) ions, in reference to the reduction equation, why only N03(-) is involved in the reaction, but not HNO3?
and how do you determine which reactants is more reactive and oxidation potential?
In addition, for the first 2nd qn which i posted on the first message (Write a balanced equation with coefficients in terms of n, for the reaction between X and KI to generate I2 and KCl.) i have another method for finding n, though it gives a weird answer, but im wondering why its wrong.0.00200 mole of I2 reacts with 0.00100 mole of X.mole ratio: 2 I2= X= 4esince 2 mole of I2 gained 4 mole of electrons,1 mole of x lost 4 mole of electrons.-1+n(-1)= 4n= -5pls pardon me if i ask too many qns! :s -
Originally posted by tay77:
but if the qn is only worth 1 mark, do u also quote both equation?
Regarding the QA test for nitrate(V) ions, in reference to the reduction equation, why only N03(-) is involved in the reaction, but not HNO3?
and how do you determine which reactants is more reactive and oxidation potential?
In addition, for the first 2nd qn which i posted on the first message (Write a balanced equation with coefficients in terms of n, for the reaction between X and KI to generate I2 and KCl.) i have another method for finding n, though it gives a weird answer, but im wondering why its wrong.0.00200 mole of I2 reacts with 0.00100 mole of X.mole ratio: 2 I2= X= 4esince 2 mole of I2 gained 4 mole of electrons,1 mole of x lost 4 mole of electrons.-1+n(-1)= 4n= -5pls pardon me if i ask too many qns! :s>>> but if the qn is only worth 1 mark, do u also quote both equation? <<<
If you're exam smart and value every single mark, then yes give both alternative answers, with qualification (ie. explanation). It's the question's fault for being ambiguous, so the candidate must compensate for the flaw in the question.
Such ambiguous question can appear throughout any topic, eg. a hydrocarbon is combusted, and a contraction of so-much cm3 was observed when cooled to rtp. Is the question referring to contraction in reactant gases or residual product gases? Depending on your interpretation, you get two different answers, so an exam-smart candidate must give both alternative workings and answers.
>>> Regarding the QA test for nitrate(V) ions, in reference to the reduction equation, why only N03(-) is involved in the reaction, but not HNO3? <<<
Because this is a test for the NO3- ion, not HNO3. Eg. the test solution given is AgNO3, and you need to confirm the presence of NO3- ions. Obviously you're testing for NO3- and not HNO3 specifically.
>>> i have another method for finding n, though it gives a weird answer, but im wondering why its wrong. 0.00200 mole of I2 reacts with 0.00100 mole of X. mole ratio: 2 I2= X= 4e since 2 mole of I2 gained 4 mole of electrons, 1 mole of x lost 4 mole of electrons. -1+n(-1)= 4; n= -5 <<<
It's wrong because your fundamental assumptions are wrong. When iodine reacts with chlorine to form ICl3, iodine is oxidized and chlorine is reduced. Subsequently when reacted with KI, iodine is simultaneously oxidized (from O.S. -1 in KI) and reduced (from O.S. +3 in ICl3), to generate I2 (O.S. 0). Your errorneous method disregarded the fact that iodine is both simultaneously oxidized and reduced at the same time.>>> how do you determine which reactants is more reactive and oxidation potential? <<<
By referring to the data booklet.
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wow! thanks for the treasure data booklet ^
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thanks alot!
but why there is two different equation for the reaction btw nitric acid and copper and only one btw nitric acid and aluminium? -
>>> but why there is two different equation for the reaction btw nitric acid and copper and only one btw nitric acid and aluminium? <<<
For copper, which is an unreactive metal, usually concentrated nitric(V) acid is used. But with high enough molarity and activation energy, it's possible for aqueous nitric(V) acid to work too. So both equations are more relevant.
In contrast, for reactive aluminium, which is often used on salt solutions containing nitrate(V) ions, as a QA test, only the NO3- equation is usually relevant. If you used concentrated nitric(V) acid, then the other equation would then be relevant, but usually that is not used.
In other words, to oxidize stubborn copper, no choice but to use HNO3. But for willing-partner aluminium, use NO3- can liao.
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ok! thanks!!