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    Hi,

    When working out the e.m.f value I believe you do the

    cathode (more positive number) - anode (more negative number)

    so for cr3+ +3e- --> cr ( -o.74)
    and cd2+ + 2e- --> Cd (-0.40)

    I'd do -0.40 - (-0.74)
    and get 0.34 ( a positive value so the reaction is feasible)

    This appears to be the right answer, but in this book I have there is a question...

    Fe3+ Ag --> fe2+ Ag+

    Fe3+ e- --> fe2+ is 0.77v
    and
    Ag+ --> Ag + e- is 0.88v

    when I do cathode -anode I get 0.88 -(o.77) to get 0.11v so its feasible, but the answer is that it is not feasible.

    Why? If I write then the other way round I get a negative answer, but I thought it was cathode-anode and the more negative being at the anode.
    If its because of the direction of eq being left for the anode, how come it isn't in the question I did above and got right?


    Thanks in advance! rep for you if you can help
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    (Original post by AishaTara)
    Hi,

    When working out the e.m.f value I believe you do the

    cathode (more positive number) - anode (more negative number)

    so for cr3+ +3e- --> cr ( -o.74)
    and cd2+ + 2e- --> Cd (-0.40)

    I'd do -0.40 - (-0.74)
    and get 0.34 ( a positive value so the reaction is feasible)

    This appears to be the right answer, but in this book I have there is a question...

    Fe3+ Ag --> fe2+ Ag+

    Fe3+ e- --> fe2+ is 0.77v
    and
    Ag+ --> Ag + e- is 0.88v

    when I do cathode -anode I get 0.88 -(o.77) to get 0.11v so its feasible, but the answer is that it is not feasible.

    Why? If I write then the other way round I get a negative answer, but I thought it was cathode-anode and the more negative being at the anode.
    If its because of the direction of eq being left for the anode, how come it isn't in the question I did above and got right?


    Thanks in advance! rep for you if you can help
    To decide on feasibility you do E(red) - E(ox)

    In the proposed reaction the iron(III) gets reduced and the silver gets oxidised, so the sum is 0.77 - 0.88 = -0.11V
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    (Original post by charco)
    To decide on feasibility you do E(red) - E(ox)

    In the proposed reaction the iron(III) gets reduced and the silver gets oxidised, so the sum is 0.77 - 0.88 = -0.11V
    my teacher was saying the lower value (more negative) one is at the anode.:O

    your explanation makes sense
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    (Original post by AishaTara)
    Hi,

    When working out the e.m.f value I believe you do the

    cathode (more positive number) - anode (more negative number)

    so for cr3+ +3e- --> cr ( -o.74)
    and cd2+ + 2e- --> Cd (-0.40)

    I'd do -0.40 - (-0.74)
    and get 0.34 ( a positive value so the reaction is feasible)

    This appears to be the right answer, but in this book I have there is a question...

    Fe3+ Ag --> fe2+ Ag+

    Fe3+ e- --> fe2+ is 0.77v
    and
    Ag+ --> Ag + e- is 0.88v

    when I do cathode -anode I get 0.88 -(o.77) to get 0.11v so its feasible, but the answer is that it is not feasible.

    Why? If I write then the other way round I get a negative answer, but I thought it was cathode-anode and the more negative being at the anode.
    If its because of the direction of eq being left for the anode, how come it isn't in the question I did above and got right?


    Thanks in advance! rep for you if you can help
    If you look at the given equation Fe 3 + Ag-- > Fe 3+ Ag +
    You can see the forward reaction (hence the cathode) is Fe 3 +e --->fe 2+ which is 0.77
    So cathode- anode =0.77-0.88=-0.11

    Hope that clears it up.
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    (Original post by eggfriedrice)
    If you look at the given equation Fe 3 + Ag-- > Fe 3+ Ag +
    You can see the forward reaction (hence the cathode) is Fe 3 +e --->fe 2+ which is 0.77
    So cathode- anode =0.77-0.88=-0.11

    Hope that clears it up.
    thank you! it makes sense, using OIL RIG and then the cathode'anode

    I'm not sure why my teacher said the anode is always the negative value out of the too (so in this case 0.77)
    if that was the case, all reactions would have a positive value and would be feasible.

    Nevermind, atleast its clear to me.
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    (Original post by AishaTara)
    thank you! it makes sense, using OIL RIG and then the cathode'anode

    I'm not sure why my teacher said the anode is always the negative value out of the too (so in this case 0.77)
    if that was the case, all reactions would have a positive value and would be feasible.

    Nevermind, atleast its clear to me.
    She probably meant in a feasible electrochemical cell the anode is always the positive value (: (since anode is the positive electrode, and cathode is the negative electrode ).
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    (Original post by eggfriedrice)
    She probably meant in a feasible electrochemical cell the anode is always the positive value (: (since anode is the positive electrode, and cathode is the negative electrode ).
    In an electrochemical cell the anode is the NEGATIVE half cell. For example in a zinc/copper electrochemical cell the Zn2+|Zn half cell is the negative anode in which Zn --> Zn2+ + 2e

    This is the opposite of an electrolytic cell. Oxidation always occurs at the anode and reduction at the cathode in both types of cell...
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    (Original post by charco)
    In an electrochemical cell the anode is the NEGATIVE half cell. For example in a zinc/copper electrochemical cell the Zn2+|Zn half cell is the negative anode in which Zn --> Zn2+ + 2e

    This is the opposite of an electrolytic cell. Oxidation always occurs at the anode and reduction at the cathode in both types of cell...
    Ah my bad, I meant oxidation occurs in the more negative electrode, I didn't realise they switched the terms of anode and cathode around ^^; that just confuses things. I'm more of a physicist so I assumed the more negative electrode would be the cathode.
 
 
 
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