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    hello here's a question im stuck on:

    "When a solution of copper (I) sulfate is left to stand a red solid appears and the solution changes from colourless to blue" explain these observations:
    identify the products -
    use electrode potentials to explain the reaction
    give a balanced ionic equation

    the products are copper and copper (II) sulfate im guessing???
    does this mean that half of the copper ions in copper (I) sulfate form Cu(s) and the other form Copper (II) sulfate???

    Electrode potentials:
    Cu+ [+] e- ---> Cu E=+0.52 V
    Cu2+ [+] e- ----> Cu+ E=+0.15 V

    any help would be great
    thanks
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    Sounds right with the products - one Cu(I) reduced to Cu(0), one oxidised to Cu(II) sulphate.

    I've never heard of Cu(I) sulphate - I would imagine it would have to be Cu2SO4? But you're right in suggesting that for each ion oxidised, another must be reduced. And if the redox is being done solely by Cu and not by air then you'll have equal amounts of Cu and CuSO4.

    Can help you with the electrode potentials I'm afraid - long time since I did A-Level electrochemistry! But your half-equations look OK.
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    (Original post by matt^)
    hello here's a question im stuck on:

    "When a solution of copper (I) sulfate is left to stand a red solid appears and the solution changes from colourless to blue" explain these observations:
    identify the products -
    use electrode potentials to explain the reaction
    give a balanced ionic equation

    the products are copper and copper (II) sulfate im guessing???
    does this mean that half of the copper ions in copper (I) sulfate form Cu(s) and the other form Copper (II) sulfate???

    Electrode potentials:
    Cu+ [+] e- ---> Cu E=+0.52 V
    Cu2+ [+] e- ----> Cu+ E=+0.15 V

    any help would be great
    thanks
    Well this is a typical disproportionation example.
    1) write balanced eqn
    2Cu(I) -----> Cu + Cu(II)

    2) Find E = RHE - LHE = What is being reduced - What is being oxidised = +0.52 - (+0.15) = +0.37 V

    3) Use delta G = - nFE
    As E is positive, F is a positive constant, n is no of electrons transferred, no worries, it is always positive, hence
    +ve E ---> negative delta G -----> reaction is spontaneous
    [your syllabus might not cover gibbs energy G, but thought you'd like to know]
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    (Original post by Preasure)
    Sounds right with the products - one Cu(I) reduced to Cu(0), one oxidised to Cu(II) sulphate.

    I've never heard of Cu(I) sulphate - I would imagine it would have to be Cu2SO4? But you're right in suggesting that for each ion oxidised, another must be reduced. And if the redox is being done solely by Cu and not by air then you'll have equal amounts of Cu and CuSO4.

    Can help you with the electrode potentials I'm afraid - long time since I did A-Level electrochemistry! But your half-equations look OK.
    (Original post by shengoc)
    Well this is a typical disproportionation example.
    1) write balanced eqn
    2Cu(I) -----> Cu + Cu(II)

    2) Find E = RHE - LHE = What is being reduced - What is being oxidised = +0.52 - (+0.15) = +0.37 V

    3) Use delta G = - nFE
    As E is positive, F is a positive constant, n is no of electrons transferred, no worries, it is always positive, hence
    +ve E ---> negative delta G -----> reaction is spontaneous
    [your syllabus might not cover gibbs energy G, but thought you'd like to know]

    Thanks to both of you
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    (Original post by shengoc)
    Well this is a typical disproportionation example.
    1) write balanced eqn
    2Cu(I) -----> Cu + Cu(II)

    2) Find E = RHE - LHE = What is being reduced - What is being oxidised = +0.52 - (+0.15) = +0.37 V

    3) Use delta G = - nFE
    As E is positive, F is a positive constant, n is no of electrons transferred, no worries, it is always positive, hence
    +ve E ---> negative delta G -----> reaction is spontaneous
    [your syllabus might not cover gibbs energy G, but thought you'd like to know]

    Hi,
    Please can you help me with the electrode potentials i am getting confused and dont understand some parts of the book.

    1. What do the E^o/V values mean?
    2. How do i know which direction to draw the arrows from the electrodes as from looking at examples it does vary i am sure there is some logic to this but i don't seem to be understanding it. If you are unable to explain then please can you direct me to a good website.

    Thanks
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    (Original post by gcseangel)
    Hi,
    Please can you help me with the electrode potentials i am getting confused and dont understand some parts of the book.

    1. What do the E^o/V values mean?
    2. How do i know which direction to draw the arrows from the electrodes as from looking at examples it does vary i am sure there is some logic to this but i don't seem to be understanding it. If you are unable to explain then please can you direct me to a good website.

    Thanks
    Read up,
    http://www.chemguide.co.uk/physical/...roduction.html
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    (Original post by gcseangel)
    Hi,
    Please can you help me with the electrode potentials i am getting confused and dont understand some parts of the book.

    1. What do the E^o/V values mean?
    2. How do i know which direction to draw the arrows from the electrodes as from looking at examples it does vary i am sure there is some logic to this but i don't seem to be understanding it. If you are unable to explain then please can you direct me to a good website.

    Thanks
    gcseangel, can you please delete some of your private messages because they exceeded the limit

    ** Sorry I had to quote your reply, to get your attention! I didn't know how else I could do it ! **
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    (Original post by gcseangel)
    Hi,
    Please can you help me with the electrode potentials i am getting confused and dont understand some parts of the book.

    1. What do the E^o/V values mean?
    2. How do i know which direction to draw the arrows from the electrodes as from looking at examples it does vary i am sure there is some logic to this but i don't seem to be understanding it. If you are unable to explain then please can you direct me to a good website.

    Thanks
    If you start with simple systems:

    An Eº value in Volts is the tendency that a metal has to lose electrons when placed in contact with a solution of its own ions.

    There is no absolute value for this so it is measured relative to the tendency of hydrogen to lose electrons when in contact with a 1 M solution of H+ ions in the presence of a platinum black catalyst. This is the standard defined as 0 Volts.

    For example, zinc tends to lose electrons when compared to hydrogen to the tune of -0.76 V. That means if we complare the two processes:

    2H+(aq) + 2e <--> H2(g)
    Zn2+(aq) + 2e <--> Zn(s)

    the zinc half equation lies further to the LHS than the hydrogen half equation.

    Notice that the equations are written as reductions. This is only a convention and it doesn't mean that this is the direction of change.

    The fact that Zn(s) is better able to lose electrons than H2(g) means that zinc is a better reducing agent than hydrogen.

    Conversely hydrogen ions are a better oxidising agent than zinc ions.

    ---------------------------------------------------------------

    Now consider copper:

    Cu2+(aq) + 2e <--> Cu(s) Eº = +0.34 V

    so the copper half equation tends to go to the RHS more than the zinc half equation.

    Cu(s) is a poorer reducing agent than zinc, while copper ions are a better oxidising agent than zinc.

    ----------------------------------------------------------------

    Using Eº values
    ---------------

    We know that oxidising agents react wiht reducing agents so we can calculate whether a given oxidising agent (from the LHS) will react with a given reducing agent (RHS) by considereing their electrode potentials.

    If the reaction proceeds the reducing agent gets oxidised and the oxidising agent gets reduced.

    We can apply the equation E(prediction) = Eº(reduced state) - Eº(oxidised state)

    And if the value of E(prediction) is positive and greater than +0.3 V the reaction is predicted to be spontaneous thermodynamically. If the value if between 0 V and +0.3 V an equilibrium is established.

    Example
    --------

    Will copper ions react with zinc metal?

    If this reaction were to proceed the copper ions would get reduced and the zinc metal oxidised.

    E(prediction) = Eº(Cu2+) - Eº(Zn)
    E(prediction) = +0.34 - -0.76 V = +1.10 V

    Therefore the reaction is spontaneous.
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    (Original post by charco)
    If you start with simple systems:

    An Eº value in Volts is the tendency that a metal has to lose electrons when placed in contact with a solution of its own ions.

    There is no absolute value for this so it is measured relative to the tendency of hydrogen to lose electrons when in contact with a 1 M solution of H+ ions in the presence of a platinum black catalyst. This is the standard defined as 0 Volts.

    For example, zinc tends to lose electrons when compared to hydrogen to the tune of -0.76 V. That means if we complare the two processes:

    2H+(aq) + 2e <--> H2(g)
    Zn2+(aq) + 2e <--> Zn(s)

    the zinc half equation lies further to the LHS than the hydrogen half equation.

    Notice that the equations are written as reductions. This is only a convention and it doesn't mean that this is the direction of change.

    The fact that Zn(s) is better able to lose electrons than H2(g) means that zinc is a better reducing agent than hydrogen.

    Conversely hydrogen ions are a better oxidising agent than zinc ions.

    ---------------------------------------------------------------

    Now consider copper:

    Cu2+(aq) + 2e <--> Cu(s) Eº = +0.34 V

    so the copper half equation tends to go to the RHS more than the zinc half equation.

    Cu(s) is a poorer reducing agent than zinc, while copper ions are a better oxidising agent than zinc.

    ----------------------------------------------------------------

    Using Eº values
    ---------------

    We know that oxidising agents react wiht reducing agents so we can calculate whether a given oxidising agent (from the LHS) will react with a given reducing agent (RHS) by considereing their electrode potentials.

    If the reaction proceeds the reducing agent gets oxidised and the oxidising agent gets reduced.

    We can apply the equation E(prediction) = Eº(reduced state) - Eº(oxidised state)

    And if the value of E(prediction) is positive and greater than +0.3 V the reaction is predicted to be spontaneous thermodynamically. If the value if between 0 V and +0.3 V an equilibrium is established.

    Example
    --------

    Will copper ions react with zinc metal?

    If this reaction were to proceed the copper ions would get reduced and the zinc metal oxidised.

    E(prediction) = Eº(Cu2+) - Eº(Zn)
    E(prediction) = +0.34 - -0.76 V = +1.10 V

    Therefore the reaction is spontaneous.
    Charco, that is an amazing explanation you have there. That is basically what i learnt over two to three weeks while at school.
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    (Original post by shengoc)
    Charco, that is an amazing explanation you have there. That is basically what i learnt over two to three weeks while at school.
    why thank eee Sir
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    (Original post by algérie_mon_amour)
    gcseangel, can you please delete some of your private messages because they exceeded the limit

    ** Sorry I had to quote your reply, to get your attention! I didn't know how else I could do it ! **
    Thank you for telling me it is now done. If you still have the pm saved then please send it to me.
 
 
 
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