Tarnish on iron is the compound FeO. A tarnished iron plate is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears.

Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge.

Calculate the standard cell potential given the following standard reduction potentials:
Al3++3e−→Al;E∘=−1.66 V
Fe2++2e−→Fe;E∘=−0.440 V

I thought it was 2.1 V but i am wrong...

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asked by John
  1. The two half reactions are
    2Al ==> 2Al^3+ + 6e Eo = 1.66
    3Fe^2+ + 6e ==> 3Fe Eo = -0.44
    Add half cells to obtain rxn.
    Add Eo to obtain Eocell.

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  2. The reduction potential that is the highest is where reduction is taken place, and reduction takes place at the cathode.

    Fe2++2e−→Fe;E∘=−0.440 V <==Cathode

    Al3++3e−→Al;E∘=−1.66 V <==Anode


    Ecell=−0.440 V-(−1.66 V)


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    posted by Devron
  3. The initial equation should read Ecell=Ecat-Eano


    I just want to add one thing: The way that Dr. Bob222 went about balancing the equation is correct, but one thing to keep in mind is that although you must balance the electrons which requires increasing moles or multiplying compounds by a number to accomplish this, reduction potential are left alone; this is a common mistake that I see done a lot.

    ***The way that I remember which goes, which is with a little rhyme:

    Electrons are negatively charge and are light, and like helium, they like to float from the ground, low potential, to the sky, high potential. Protons are positively charge and are heavy. They like to fall from the sky from the sky, high potential, to the ground, low potential.

    Oxidation loses electrons and reduction gains them.

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    posted by Devron

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