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Chemistry
OpenStudy (anonymous):

http://screencast.com/t/Iw4lq8fdI help pplease

OpenStudy (anonymous):

@Kryten

OpenStudy (anonymous):

dont know why but loading of your link takes long time, probably my isp, can you put it on dropbox or some other image site

OpenStudy (anonymous):

ok i got it so question is which contained oxidizing agent right?

OpenStudy (anonymous):

yeah?

OpenStudy (anonymous):

just a minute i have a problem with translation, i need some time to dissolve it in my head... :)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

im not sure if i got this right but oxidising agent is reduced which means that oxidising agent recieves electrons, so something must release them, as we have KI and KMnO4 we can conclude that change in color can be only in KI cause Mn is in oxidising state VII so he cant release more electrons but it says also its potassum (VII) manganate where K2MnO4 where Mn has valence VI (potassum manganate) and K2MnO4 is potassum permanganate where Mn has valence VII so this confuses me, if it is case that there is K2MnO4 then answer is: only 1! but if it is KMnO4 then answer should be: only 2!

OpenStudy (anonymous):

sory KMnO4 is permanganate lapsus calami

OpenStudy (anonymous):

thats what even i thought but the mrking scheme says B (1 and 2 only) and i m wondern why

OpenStudy (anonymous):

you see my doubt heena....

OpenStudy (vincent-lyon.fr):

Usually, English texts say "manganate (VII)" according to UIPAC rules. In continental Europe, we usually say "permanganate" instead. So in any of these simple problems, "manganate" always means MnO4-

OpenStudy (anonymous):

i didnt know that

OpenStudy (anonymous):

then i would say B) 1 and 2 only but im currently little bit confuzzeld.. :)

OpenStudy (vincent-lyon.fr):

Solution No1 contains an oxidising agent and a reducing agent; Solution No2 contains an oxidising agent only Solution No3 contains a reducing agent only Solution No4 contains neither So there are two solution which contain an oxidising agent, namely 1 and 2. So answer must be B.

OpenStudy (anonymous):

ok now my brain hurts from overthinking but i think that you are right...

OpenStudy (anonymous):

@Vincent-Lyon.Fr can u explain me the method on how you got it :)

OpenStudy (anonymous):

we know MNO4- i strong oxdising agent if we ee its result we observe that 1 and 3 has change now coming to seee KI effect which show result only in 1 and 2 only so by this we conclude 1 option i right now the option must contain 1 solution

OpenStudy (anonymous):

now looking other solution we found in 3 solution change occur so i ll go with c)

OpenStudy (anonymous):

1: potassium iodine is colorless and when oxidised KI -> K+ + I2 color change KMnO4 purple reduced so color change (unknown other chemical so it cant be known which compound can be created) 2: again KI so again I2 color change and KMnO4 no change because it cant be oxidised if i got it right

OpenStudy (vincent-lyon.fr):

iodide ions (colourless) are reducing agents that can be oxidised into iodine I2 (brown/yellow in aqueous solution). So any change in colour is the proof of an oxidising agent in teh solution. Permanganate ions (purple) are oxidising agents and will be reduced to colourless Mn2+ ions (at low pH) by a reducing agent. So if those few added drops turn colourless, it is proof of a reducing agent in the solution.

OpenStudy (anonymous):

wow..u all are awesome

OpenStudy (anonymous):

yeah and just smart to know that I2 cant be diluted in water that is it dilutes very very slow and you dilute it so that you make KI solution and then add I2 it is called iodine tincture i know that heena and vincent probably know that but zaphod that is a "fun" fact :D

OpenStudy (anonymous):

ohhh thnx i forgot abu I2 cant be diluted in water @Kryten :)

OpenStudy (anonymous):

:D

OpenStudy (anonymous):

thanks alot once again :)

OpenStudy (anonymous):

well i learned something also so tnx for question zaphod

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