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...
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=Ecat-Ered
Ecell=−0.440 V-(−1.66 V)
Ecell=1.22V
The two half reactions are
2Al ==> 2Al^3+ + 6e Eo = 1.66
3Fe^2+ + 6e ==> 3Fe Eo = -0.44
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Add half cells to obtain rxn.
Add Eo to obtain Eocell.
The initial equation should read Ecell=Ecat-Eano
cat=cat
ano=anode
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.
To calculate the standard cell potential of a redox reaction, you need to subtract the standard reduction potential of the anode half-reaction from the standard reduction potential of the cathode half-reaction.
In this case, the anode half-reaction is the reduction of Fe2+ to Fe: Fe2+ + 2e- -> Fe with a standard reduction potential (E°) of -0.440 V.
The cathode half-reaction is the reduction of Al3+ to Al: Al3+ + 3e- -> Al with a standard reduction potential (E°) of -1.66 V.
When the two half-reactions are connected through a wire and a salt bridge, the anode half-reaction occurs at the anode (negative electrode), and the cathode half-reaction occurs at the cathode (positive electrode).
To calculate the standard cell potential (E°cell), you subtract the anode half-reaction potential from the cathode half-reaction potential:
E°cell = E°cathode - E°anode
E°cell = (-1.66 V) - (-0.440 V)
E°cell = -1.22 V
So, the correct standard cell potential for this reaction is -1.22 V, not 2.1 V.