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F325 -balancing

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    hi can someone please show me how to balance this

    Zn gives Zn(2+) + e-
    Cr2o7(2-) + H+ + e- gives Cr(2+) + H20
    the brackets are the charges

    ok i know how to balance the first one
    you do
    Zn gives Zn(+2) + 2e-

    then second part i balanced everything except the electron how do u balance the electron they got 8 ? how?
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    and should there have to be a 2 in front of cr2o7(2-)?
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    Ok, this is how you form and balance half equations:

    First off, learn the basic first step. Dichromate ions are reduced to Chromium ions:

    \displaystyle (Cr_2O_7)^2^-  \to \ 2(Cr)^3^+

    Oxygen always comes out as water:

    \displaystyle (Cr_2O_7)^2^-  \to \ 2(Cr)^3^+ +7H_2O

    We need more hydrogens on the LHS to balance the equation so add H+:

    \displaystyle (Cr_2O_7)^2^- +14H^+ \to \ 2(Cr)^3^+ +7H_2O

    Now balance the electrons:

    \displaystyle (Cr_2O_7)^2^- +14H^++6e^- \to \ 2(Cr)^3^+ +7H_2O
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    (Original post by Dreamweaver)
    Ok, this is how you form and balance half equations:

    First off, learn the basic first step. Dichromate ions are reduced to Chromium ions:

    \displaystyle (Cr_2O_7)^2^-  \to \ 2(Cr)^3^+

    Oxygen always comes out as water:

    \displaystyle (Cr_2O_7)^2^-  \to \ 2(Cr)^3^+ +7H_2O

    We need more hydrogens on the LHS to balance the equation so add H+:

    \displaystyle (Cr_2O_7)^2^- +14H^+ \to \ 2(Cr)^3^+ +7H_2O

    Now balance the electrons:

    \displaystyle (Cr_2O_7)^2^- +14H^++6e^- \to \ 2(Cr)^3^+ +7H_2O
    its wrong the electrons has to be 8?
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    (Original post by otrivine)
    its wrong the electrons has to be 8?
    No it's not wrong. If you add up the charges you will find it is:

    +6 -> +6

    Therefore the charges are balanced.
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    (Original post by Dreamweaver)
    No it's not wrong. If you add up the charges you will find it is:

    +6 -> +6

    Therefore the charges are balanced.
    http://pdf.ocr.org.uk/download/ms_10..._gce_f325.pdf?

    mark scheme question 6)b) check
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    (Original post by otrivine)
    http://pdf.ocr.org.uk/download/ms_10..._gce_f325.pdf?

    mark scheme question 6)b) check
    na what he is saying is right you have 8 electrons for the second equation then u multiply the zinc equation by 4 hence u get 8 electrons i dunno where u got 6, i understand the over equation needs 6 electrons plus 2- from the cr2072- you get 4+ on both sides hence an equal and balanced equation.

    check it if u dont believe me.
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    (Original post by otrivine)
    http://pdf.ocr.org.uk/download/ms_10..._gce_f325.pdf?

    mark scheme question 6)b) check

    (Original post by kurdishboy94)
    na what he is saying is right you have 8 electrons for the second equation then u multiply the zinc equation by 4 hence u get 8 electrons i dunno where u got 6, i understand the over equation needs 6 electrons plus 2- from the cr2072- you get 4+ on both sides hence an equal and balanced equation.

    check it if u dont believe me.
    Both equations are correct. On the MS, the Cr ion has a 2+ charge but in the book you will find examples where Cr is 3+. This is because Cr can have different oxidation states. You need to be careful and always look at the question.

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