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Posted by on Thursday, December 13, 2012 at 11:00am.

For the cell shown below, which will increase the cell voltage the most?

Fe2+ | Fe3+ || Cu2+ | Cu

A) Halve[Cu2+]
B) Double[Cu2+]
C) Double[Fe2+]
D) Halve[Fe2+]
E) Cut Cu electrode in half


...
Ecell = E0 cell + 0.059/n * log([Cu+]/[Fe2+])

Ecell is directly proportional [Cu+] to and inversely proportional to [Fe2+]..

so i would think the answer would be, B and D both


but there can only be one answer , so i don't really know how else to work it out :/

pls help!

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 1:54pm

    I'm not 100% sure, but I think its like this.
    -------

    The half rxns are:

    Fe^2+ --> Fe3+ + e [Reduction]
    Cu^2+ + 2e --> Cu [Oxidation]

    Reduction is increase of electrons while oxidation is decrease.

    Reduction=cathode
    Oxidation=anode

    The equation is Ecell=Ecathode-Eanode.

    So I think the cathode should be higher than anode to gain maximum Ecell and the cathode is Fe so that should be doubled.
    -------

    Can you help me with question #10 and #25?

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 2:22pm

    10 is b)

    its a 2:1 ratio

    so you multipy the rate by 1/2

    and 25 is 110.6 kj.

    first you find delta H , then delta S , then you put those values into :

    delta G = delta H - T*deltaS

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 3:03pm

    thanks your helping me alot do you know 13?

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 3:11pm

    I know that the rxn is:

    Ag^+ + e --> Ag

    n=M/V

    n=0.250/0.6=0.42

    to convert from mols to mols of electrons you take into account the ratio. The ratio is 1

    so mols of e's=0.42

    1 mol e = 96500 C

    96500*0.42 = 40530C
    ===================

    I don't know where I went wrong from this point. There are no answers on the multiple choice like this.

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 3:17pm

    np. number 13 is like this :

    first find moles of Ag which is volume (in L) times molarity. should get 0.15 moles Ag.

    to convert to coulombs you multiply it by faraday's constant. (approx 96500 C/mol) . the units will cancel out the moles and you should be left with 1.45 * 10^4 C

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 3:18pm

    if you have your text on you, you think you can help me out with number 77 on pg 859 ? i have no clue how to do it :S

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 3:22pm

    wow I messed up the mol calculation. I made it n=c/V when its n=c*v

    Thanks

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 4:56pm

    For 77 pg 859

    I think first you write up the half rxns.

    Zn --> Zn^2+ + 2e
    Ni --> Ni^2+ + 2e

    Then I think the E values are in the texbook. The E value that's is greater is the cathode and the lesser one is the anode. From there its:

    Ecell=Ecathode-Eanode

    I'm not sure about b) and c)

    Even what I just said im not 100%.

    ================

    Do you know how to do 18 and 20 are on the old exam?

  • chemistry - exam practice help!! - , Thursday, December 13, 2012 at 4:57pm

    sorry for the late reply by the way.

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