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    Hi, there's two questions that I'm slightly confused on regarding their answers. For the first one, I don't understand why 12e- are on the products side, as although this balances the equation, 12e- were not lost in terms of oxidation numbers. Also, could anyone help on explaining an answer for the second question. Thank you
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    (Original post by candyhearts)
    Hi, there's two questions that I'm slightly confused on regarding their answers. For the first one, I don't understand why 12e- are on the products side, as although this balances the equation, 12e- were not lost in terms of oxidation numbers. Also, could anyone help on explaining an answer for the second question. Thank you
    The half equation for the first question is:

    C2H5OH + 3 H2O ----> 2 CO2 + 12 H+ + 12 e-.

    Start off by having:

    C2H5OH ----> CO2.

    You then balance the carbons:

    C2H5OH ----> 2 CO2.

    Then balance the oxygens with water molecules:

    C2H5OH + 3 H2O ----> 2 CO2.

    You then balance the hydrogen atoms with protons (H+ ions):

    C2H5OH + 3 H2O ----> 2 CO2 + 12 H+.

    You then balance the overall charge using electrons:

    C2H5OH + 3 H2O ----> 2 CO2 + 12 H+ + 12 e-.
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    (Original post by thegodofgod)
    The half equation for the first question is:

    C2H5OH + 3 H2O ----> 2 CO2 + 12 H+ + 12 e-.

    Start off by having:

    C2H5OH ----> CO2.

    You then balance the carbons:

    C2H5OH ----> 2 CO2.

    Then balance the oxygens with water molecules:

    C2H5OH + 3 H2O ----> 2 CO2.

    You then balance the hydrogen atoms with protons (H+ ions):

    C2H5OH + 3 H2O ----> 2 CO2 + 12 H+.

    You then balance the overall charge using electrons:

    C2H5OH + 3 H2O ----> 2 CO2 + 12 H+ + 12 e-.
    Usually I'd first consider the electrons, how come that doesn't work with this example?
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    (Original post by candyhearts)
    Usually I'd first consider the electrons, how come that doesn't work with this example?
    It does:

    The overall charge in C2H5OH is 0, as it is a neutral compound. Each H atom has an oxidation state of +1 and each O atom has an oxidation state of of -2.

    Therefore algebraically, you have the following equation:

    2C + 6(+1) + (-2) = 0

    2C + 6(+1) = 2

    2C = -4

    C = -2.

    This means that the oxidation state of each C atom in each C2H5OH molecule is -2. However, as there are 2 C atoms per molecule of C2H5OH, the total oxidation states of the C atoms is 2(-2) = -4.

    The overall charge in CO2 is 0, as it is a neutral compound. Each O atom has an oxidation state of -2.

    Therefore algebraically, you have the following equation:

    C + 2(-2) = 0

    C = +4.

    This means that the oxidation state of each C atom in each CO2 molecule is +4.

    However, bearing in mind that there are 2 molecules of CO2 in the reaction, the total oxidation state of both C atoms in 2 CO2 is 2(+4) = +8.

    Okay, so going from an oxidation state from -4 to +8 requires a loss of: 8 - (-4) = 12 electrons.

    Therefore, you add 12 electrons to the products side.

    I hope this is clear enough
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    (Original post by thegodofgod)
    It does:

    The overall charge in C2H5OH is 0, as it is a neutral compound. Each H atom has an oxidation state of +1 and each O atom has an oxidation state of of -2.

    Therefore algebraically, you have the following equation:

    2C + 6(+1) + (-2) = 0

    2C + 6(+1) = 2

    2C = -4

    C = -2.

    This means that the oxidation state of each C atom in each C2H5OH molecule is -2. However, as there are 2 C atoms per molecule of C2H5OH, the total oxidation states of the C atoms is 2(-2) = -4.

    The overall charge in CO2 is 0, as it is a neutral compound. Each O atom has an oxidation state of -2.

    Therefore algebraically, you have the following equation:

    C + 2(-2) = 0

    C = +4.

    This means that the oxidation state of each C atom in each CO2 molecule is +4.

    However, bearing in mind that there are 2 molecules of CO2 in the reaction, the total oxidation state of both C atoms in 2 CO2 is 2(+4) = +8.

    Okay, so going from an oxidation state from -4 to +8 requires a loss of: 8 - (-4) = 12 electrons.

    Therefore, you add 12 electrons to the products side.

    I hope this is clear enough
    Thank you so much Looking at it now, I'm not sure why I was even confused in the first place as they way you've explained it makes so much sense
 
 
 
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