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Half equations to ionic equations to full equations watch

    • Thread Starter

    Right i know how to combine 2 half equations to an ionic equation how do i get from that to a full equation? Any help please?

    Just pair up the ions to make salts?

    You need to know the E.M.F values to see which reaction is feasible. What is an example of the question you're trying to do?

    soz misread it, yeah, u pair up the ions


    Say for example they give you this:
    2C02 + 2e -----> C2O4 2- (Equation 1)
    MnO4- +8H+ +5e ---> Mn2+ +4H2O (Equation 2)

    Reverse the first equation to get C2O4 2- ----> 2CO2 + 2e (Equation 3)

    tHEN BALANCE THE e, this is done by multiplying equation 3 by 5 and equation 2 by 2

    To get 5C2O4 2- ---->10CO2 + 10E
    2MnO4- + 16H+ + 10e ----> 2Mn2+ + 8H20

    Then add both together, to get the overall equation:

    5c2042- + 2Mn04- +16 H+ -----> 10C02 + 2Mn2+ + 8H20

    Please note when adding the H+ may cancel out, just like this equation
    And do not add the electrons in the final equation
    • Thread Starter

    Why is the first equation reversed can someone explain please i understand the rest.
    • Community Assistant
    • Study Helper

    Community Assistant
    Study Helper
    you have to pair up a reduction with an oxidation to make an equal number of electrons on both sides when you add the equations together.

    What you are effectively saying is that equation 1 must provide the electrons that are going to be used in equation 2 (or vice versa)

    The way to find out which of the half equations is a reduction and which is an oxidation requires use ofelectrode potentials.

    Put together a pair of half equations - one reduction (electrons on the left hand side)and one oxidation (electrons on the right hand side)

    now check out E= E(red) - E(ox)
    ...where E(red) is the electrode potential of the half equation undergoing reduction and E(ox) is the electrode potential of the half equation undergoing oxidation.

    If the value for Eº given is positive and greater than +0.3 (this value differs in different courses some say +0.4) then the reaction is spontanous when the two half equations are added together.

    If you are happy with the equations this way round (ie Eº is > +0.3) then multiply each equation to make the number of electrons on the left in one half equation equal to the number of electrons on the RHS of the other equation and simply add the two equations together, cancelling out the electrons in the process.

    To make a full equation WITH balancing ions you have to add in all of the original balancing ions from the species that you originally used - this process is fundamentally worthless as all the other ions are spectator ions and not involved in the reaction.

    But, if you are desperate to do so for example:

    after constructing the ionic equation you may get...

    MnO4(-) + 5Fe2+ + 8H+ --> Mn2+ + 5Fe3+ + 4H2O

    the MnO4(-) was added using KMnO4 so that makes one K+ ion
    the Fe2+ was added using FeSO4 so that makes 5SO4(2-) ions
    the H+ weas added using H2SO4 so that makes 4 more SO4(2-) ions

    so now our equation looks like:
    KMnO4 + 5FeSO4 + 4H2SO4 --> MnSO4 + 5/2Fe2(SO4)3 + 4H2O + 1/2K2SO4

    multiply through by two to remove fractions

    2KMnO4 +10FeSO4 +8H2SO4 --> 2MnSO4 + 5Fe2(SO4)3 + 8H2O + K2SO4

    pretty useless but correct!
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