6L.5 b

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Gabriella Bates 2L
Posts: 113
Joined: Thu Jul 11, 2019 12:15 am

6L.5 b

Postby Gabriella Bates 2L » Sun Mar 01, 2020 12:30 pm

b. Ce4+(aq) + I-(aq) --> I2(s) + Ce3+(aq)

When writing the cell diagram, the solution writes Pt on the iodine anode side of the cell and says, "An inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solution." Why do do they assume that I is in solution? The I2 in the products is a solid, so initially it doesn't seem to be in solution. Thanks!

Sue Bin Park 2I
Posts: 52
Joined: Mon Jun 17, 2019 7:24 am

Re: 6L.5 b

Postby Sue Bin Park 2I » Sun Mar 01, 2020 1:15 pm

in the process of oxidation in the anode, you will be going from I-(aq) to I2(s). the half rxn is 2I-(aq) -> I2(s) + 2e-

before the redox begins, you will have, as an anode, only an aqueous solution of I-(no I2 present yet). therefore you will need some inert conducting metal to carry the current in order to begin producing I2(s), which would probably start forming on the surface of the Pt electrode.


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