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Titation between Manganate and iron(II) ions - Electrode Potentials

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    Question:


    Values from Data Booklet:


    What I don't understand is why the Permanganate ions will reduce to Mn(II) ions instead of MnO2, as stated in the mark scheme? Is it because of the fact that we don't get black/brown solid(MnO2) at the end of the titration?
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    (Original post by Zishi)
    Question:


    Values from Data Booklet:


    What I don't understand is why the Permanganate ions will reduce to Mn(II) ions instead of MnO2, as stated in the mark scheme? Is it because of the fact that we don't get black/brown solid(MnO2) at the end of the titration?
    I'm guessing your exam board uses the clockwise rule? Edexcel use the anti-clockwise rule (most negative on top of list, most positive at the bottom). Therefore, anything on the left will react with ANYTHING above it and to the right, I apologise if this is confusing.
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    (Original post by Zishi)
    Question:


    Values from Data Booklet:


    What I don't understand is why the Permanganate ions will reduce to Mn(II) ions instead of MnO2, as stated in the mark scheme? Is it because of the fact that we don't get black/brown solid(MnO2) at the end of the titration?
    The more positive a reduction potential value is, the better oxidising agent it is. Therefore, it will become reduced more easily, and thus gain electrons more easily.

    I think it's just because MnO4- is a very strong oxidising agent (compare to the likes of K2Cr2O7), and thus will become reduced very easily to Mn2+, instead of Mn4+ in MnO2.
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    Hmm, seems like Iron(II) is strong enough reducing agent to reduce manganate(VII) ions to manganate(II) ions. Thanks to both of you.

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