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    i dont understand why there are 8 electrons in the half equation so4 2- ----> h2s

    when doing the first few steps, i have to balance the electrons before adding h20 to balance o and h+ etc.

    how would i know how to add 8 e- in the first step because from this i would add 2e- because of the 2- charge?
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    half equation so4 2- ----> h2s

    i dont understand why there are 8 electrons

    when doing the first few steps and balancing the electrons i would put 2 e- because of the 2-

    this is before i have added h2o to balance o and h to balance h+ etc.

    so how would i know when to add 8 e-?
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    Don't try to balance the charge until you have all the other species present and balanced. It needs to be the last step.

    If you want to know already how many electrons you could work out the oxidation state of sulfur on each side.
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    I have. I work out the oxidation state of the two elements and balance the element being oxidised and then the electrons.

    How does that help. The oxidation state of so4 2- is +6 and the other side is -2 . How does that help. I thought that adding electrons was to balance the overall charge like the one on so4 2-
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    (Original post by esmeralda123)
    half equation so4 2- ----> h2s

    i dont understand why there are 8 electrons

    when doing the first few steps and balancing the electrons i would put 2 e- because of the 2-

    this is before i have added h2o to balance o and h to balance h+ etc.

    so how would i know when to add 8 e-?
    First you balance the key element- sulfur, it’s already balanced.

    Then you add h20 to balance the oxygens, so add 4h20 on the right hand side

    Then you balance the Hs, so you add 10H+ ions on the left hand side

    Then the balance the charges so you add 8e- on the left hand side.

    Always been taught to do it in this order and it always works

    S042- + 10H+ + 8e- -> h2s + 4h20
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