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aaronq Group Title

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  • 2 years ago
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    • 2 years ago
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    @Carl_Pham hey man you seem like you know a lot, can you take a look at this?

    • 2 years ago
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    To disproportionate, you'd need from the same species a reduction (following the arrow) and an oxidation (going against the arrow), such that subtracting the potential above the oxidation from the potential above the reduction still left you with a positive number. That would tell you the overall cell potential (which is always Ered - Eox) is positive, and the reaction is spontaneous. In this case, HOI qualifies: \[{\rm HOI}(aq) + {\rm H}^{+}(aq) + e^- \rightarrow \frac{1}{2} {\rm I}_2(s) + {\rm H}_2{\rm O}(l)\] \[{\rm HOI}(aq) + 2 {\rm H}_2{\rm O}(l) \rightarrow {\rm IO_3}^{-}(aq) + 5 {\rm H}^{+}(aq) + 4 e^-\] The potential for the reduction (top) is 1.44 V, and the potential for the oxidation (bottom) is -1.14 V, so the overall cell potential is +0.30 V, and the redox reaction will be spontaneous. You generally expect reduction potentials to fall with oxidation state, so that the most highly oxidized species has the biggest reduction potential. When that trend is interrupted -- when some species in the middle has an unusually low reduction potential -- then you can get disproportionation. The only place where that happens here is IO3- to HOI, so I don't seen any other candidates.

    • 2 years ago
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    thanks man, i didn't know these diagrams represented cell potentials. it's funny it's in the midterm practice test yet we never talked about it in class.

    • 2 years ago
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    They don't. They represent standard reduction potentials, which are, if you will "half" cell potentials. You need to add a reduction and then reverse one of them to be the oxidation, to find the cell potetial.

    • 2 years ago
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    i worded my statement incorrectly, std reduction potentials. thanks for the input (y)

    • 2 years ago
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    Another way to think about them is as the free energy changes for each step. So a standard reduction potential of +1.0 V means dG for that step is -97 kJ/mol per electron. Going backward, oxidizing, you reverse the sign, so going backward along the arrow represents dG = +97 kJ/mol. You want to pick a combination of going foward and backward so that the dG's add up to a negative number.

    • 2 years ago
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    sweet thanks dude, i know dG= nFE(std).

    • 2 years ago
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