Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    20
    ReputationRep:
    Hi,
    I'm just self learning aqa transition metals: chromium chemistry
    What's the point of it?
    Why do we need to reduce Cr2O72- by zinc?


    Posted from TSR Mobile
    Offline

    17
    ReputationRep:
    (Original post by Bloom77)
    Hi,
    I'm just self learning aqa transition metals: chromium chemistry
    What's the point of it?
    Why do we need to reduce Cr2O72- by zinc?


    Posted from TSR Mobile
    What do you mean? Are you asking why you need to learn about transition metals?
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by langlitz)
    What do you mean? Are you asking why you need to learn about transition metals?
    No, just what's the use of oxidising and then reducing chromium ions continuously?


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Bloom77)
    Hi,
    I'm just self learning aqa transition metals: chromium chemistry
    What's the point of it?
    Why do we need to reduce Cr2O72- by zinc?


    Posted from TSR Mobile
    Because dichromate (VII) Cr2O7 2- is a widely used oxidizing agent that you should remember from AS Organic and CHEM 4 for the test tube reaction's of primary/secondary alcohols.

    Cr2O7 2- is reduced to Cr 3+, here is the half equation:

    Cr2O7 2- + 14H+ + 6 e- -> 2Cr3+ + 7H2O

    So the dichromate needs to gain 6 e- to be reduced, it can do that by oxidising zinc.

    Zn -> Zn (2+) + 2 e-

    If you combine the 1/2 equations by multiplying Zn by 3 to balance the electrons you get the eq:

    Cr2O7(2-) + 14H+ + 3 Zn -> 2Cr(3+) + 7H2O + 3 Zn (2+)

    As the reaction progresses you'll see the dichromate solution go from orange to green. (If the reaction is done in a o2 free environment the chromium ion will reduce further to Cr (2+) which will be a blue solution). Also, if you wanted to you can oxidise the Cr(3+) to CrO4- by using H2O2 and OH-.
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by RME11)
    Because dichromate (VII) Cr2O7 2- is a widely used oxidizing agent that you should remember from AS Organic and CHEM 4 for the test tube reaction's of primary/secondary alcohols.

    Cr2O7 2- is reduced to Cr 3+, here is the half equation:

    Cr2O7 2- + 14H+ + 6 e- -> 2Cr3+ + 7H2O

    So the dichromate needs to gain 6 e- to be reduced, it can do that by oxidising zinc.

    Zn -> Zn (2+) + 2 e-

    If you combine the 1/2 equations by multiplying Zn by 3 to balance the electrons you get the eq:

    Cr2O7(2-) + 14H+ + 3 Zn -> 2Cr(3+) + 7H2O + 3 Zn (2+)

    As the reaction progresses you'll see the dichromate solution go from orange to green. (If the reaction is done in a o2 free environment the chromium ion will reduce further to Cr (2+) which will be a blue solution). Also, if you wanted to you can oxidise the Cr(3+) to CrO4- by using H2O2 and OH-.
    Oh yeah! Thank you!


    Posted from TSR Mobile
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.