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What's the point of chromium chemistry? Transition metals watch

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    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?


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    (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?


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    What do you mean? Are you asking why you need to learn about transition metals?
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    (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?


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    (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?


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    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-.
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    (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!


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