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aaronqBest ResponseYou've already chosen the best response.0
@Carl_Pham hey man you seem like you know a lot, can you take a look at this?
 one year ago

Carl_PhamBest ResponseYou've already chosen the best response.1
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.
 one year ago

aaronqBest ResponseYou've already chosen the best response.0
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.
 one year ago

Carl_PhamBest ResponseYou've already chosen the best response.1
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.
 one year ago

aaronqBest ResponseYou've already chosen the best response.0
i worded my statement incorrectly, std reduction potentials. thanks for the input (y)
 one year ago

Carl_PhamBest ResponseYou've already chosen the best response.1
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.
 one year ago

aaronqBest ResponseYou've already chosen the best response.0
sweet thanks dude, i know dG= nFE(std).
 one year ago
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