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Balance the ewuation using thr oxidation number method

Here is what i did

Au+ + H20---> Au2O3 + Au + H+

Au+ losses 3e- to form Au2O3-->oxidation
Au+ gets reduced to for Au

therefore the balanced chemical equation is 3Au+ + 3H20---> Au2O3 +Au +6H+

Does it make sense?

  • Chemistry -

    Yes and no.
    The equation isn't balanced. (I know it looks balanced at first glance but it isn't.)
    To be balanced, an equation must
    1. have the same number of atoms on each side
    2. have the same charge on each side.
    3. electrons lost = electrons gained.
    Your equation has 1 ok.
    The charge doesn't balance (3+ on left and 6+ on right).
    Electron change doesn't balance.

    Here is the way I would do it.
    Au^+ ==> Au2O3
    First you make the Au atoms equal so we don't "split" charges.
    2Au^+ ==> Au2O3
    Oxidation state is +2 on left for 2 atoms and +6 on right for 2 atoms. We balance the change in state with electrons.
    2Au^+ ==> Au2O3 + 4e
    Count the charge on the left (+2) and right (-4) and balance with H^+.
    2Au^+ ==> Au2O3 + 4e + 6H^+
    Now add water to the other side to balance the H^+ added.
    3H2O + 2Au^+ ==> Au2O3 + 4e + 6H^+

    Next half cell.
    Au^+ + e ==> Au

    Now multiply the first equation by 1 and the second equation by 4 and add to obtain
    3H2O + 6Au^+ ==> Au2O3 + 4Au + 6H^+
    1. atoms balance 6H left and right. 3 O left and right. 6 Au left and right.
    2. charge balances. +6 on left and right.
    3.electron change balances. 2Au^+ on left move to Au2O3 where both Au atoms are +6. Loss of 4e. Other 4 Au^+ atoms move to 4Au on right. Gain of 4e.

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