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Disagree with some multiple choice answers watch

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    1) The standard electrode potentials for two half-cells involving iron are given below.
    Fe2+(aq) + 2e– → Fe(s) Eο = –0.44 V
    Fe3+(aq) + e–→ Fe2+(aq) Eο = +0.77 V
    What is the equation and the cell potential for the spontaneous reaction that occurs when the two half-cells are connected?
    A. 3Fe2+(aq) → Fe(s) + 2Fe3+(aq) Eο = +1.21 V
    B. Fe2+(aq) + Fe3+(aq) → 2Fe(s) Eο = +0.33 V
    C. Fe(s) + 2Fe3+(aq) → 3Fe2+(aq) Eο = +0.33 V
    D. Fe(s) + 2Fe3+(aq) → 3Fe2+(aq) Eο = +1.21 V

    In 1, I thought the correct answer is A, but it shows D?

    2) Consider the standard electrode potentials of the following reactions.
    Cr3+(aq) + 3e– →Cr(s) –0.75 V
    Cd2+(aq) + 2e– → Cd(s) –0.40 V

    What is the value of the cell potential (in V) for the following reaction?

    2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)

    A. –0.35
    B. –1.15
    C. +0.30
    D. +0.35

    And in 2, the answer is D, but why would not it be C?

    Many thanks in advance for your help!
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    1) D is correct because Fe(s) is being oxidised and Fe3+ is being reduced when the two cells are joined. In A, the arrow is the wrong way round.

    2) D is correct: -0.40 - (-0.75) = +0.35v
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    (Original post by Samsoonie)
    1) The standard electrode potentials for two half-cells involving iron are given below.
    Fe2+(aq) + 2e– → Fe(s) Eο = –0.44 V
    Fe3+(aq) + e–→ Fe2+(aq) Eο = +0.77 V
    What is the equation and the cell potential for the spontaneous reaction that occurs when the two half-cells are connected?
    A. 3Fe2+(aq) → Fe(s) + 2Fe3+(aq) Eο = +1.21 V
    B. Fe2+(aq) + Fe3+(aq) → 2Fe(s) Eο = +0.33 V
    C. Fe(s) + 2Fe3+(aq) → 3Fe2+(aq) Eο = +0.33 V
    D. Fe(s) + 2Fe3+(aq) → 3Fe2+(aq) Eο = +1.21 V

    In 1, I thought the correct answer is A, but it shows D?
    Many thanks in advance for your help!
    Perhaps a little explanation might help...

    The electrode potential shows the (relative to hydrogen) tendency to release electrons. The easier they release electrons the more negative the value of Eº.

    Metals such as alkali metals are very keen to lose electrons, and so they have high negative values of Eº. They are good reducing agents.

    Reducing agents have to give their electrons to a speicies capable of receiving these electrons - i.e. better oxidising agent (worse reducing agents)

    Hence, a species high in the electrochemical series (Electrode potential series) on the right hand side can react with a species lower down on the left hand side.

    Perhaps at this point it should be mentioned that the list is traditionally shown as reductions, ie

    Li+(aq) + 1e --> Li(s) Eº = -3.04

    If it helps, you could think of the negative sign here also meaning that the equilibrium has a tendency to go backwards (your choice), but in reality it means that the half cell would be the negative side of a full cell when connected to the standard hydrogen electrode.

    So, to decide whether two species react together:

    1. you must have one species from each side of the series list (one to release electrons and one to absorb electrons)
    2. The right hand side species MUST be higher in the series than the left hand side species (ie the oxidising agent MUST have a lower Eº (less negative) value than the reducing agent)
    3. The difference in Eº values MUST be greater than 0.4V for complete reaction and form 0 to 0.4V for equilibrium

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

    A simple mathematical way to rememeber all of the above is to apply the formula:

    E = E(red) - E(ox)

    where E(red) is the Eº of the species that gets reduced in the hypothetical equation.
    and E(ox) is the Eº of the species that gets oxidised in the hypothetical equation.

    Once again if the answer is positive and greater than 0.4V there is complete reaction.

    -------------------------------------------------
    Applying this to your question:

    Fe(s) will react with Fe3+(aq)

    because in the proposed reaction:

    Fe(s) + 2Fe3+(aq) --> 3Fe2+(aq)

    the Fe(s) is getting oxidised and the Fe3+(aq) is getting reduced.

    Hence E= Eº(Fe3+(aq)) - Eº(Fe(s)) = 0.77 - - 0.44 = + 1.21V

    The answer is positive and greater than 0.4V :yes:
 
 
 
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