bootcamp Redox question..

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HdK

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I thought I had a full understanding of redox and how to do the questions, but for some reaosn this question is completely confusing..and the answer confuses me even more.

2. In an electrochemical cell, an overall negative cell potential means that:
(yes I know the overall is nonspontaneous)

the answer is

the oxidation half reaction potential is greater than that of the reduction half reaction potential

The correct answer is D. This is a pretty hard question, and you have to keep in mind which reactions produce what. The oxidation reaction results in oxidation of a metal and release of electrons, whereas the reduction reaction results in gain of electrons by the receiving metal. Thus, a cell with a negative potential means that the oxidation half reaction potential is greater than that of the reduction half reaction potential, aka the reaction does not proceed forward. Remember, for a reaction to move forward spontaneously, E°cell (electrode potential) must be positive.

Huh? What? I feel like everyhting I have known about redox is a lie haha

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what it says its true...think it about it terms of an example with reductional potentials

the oxidation half reaction has +3.4V as E^o red potential (thus -3.4V as E^o ox potential)
the reduction half reaction has < +3.4V as E^o red potential

thus the overall E is always negative when the oxidation half reaction potential is greater than that of the reduction half reaction potential. Keep in mind the rxns would be opposite if looking for a spontaneous cell (E>0).

Easier way is the E(overall) = E(red) - E(ox) so the E(ox) has to be greater than E(red) for E(overall) to be negative
 
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Oh yeah! I meant to say E(overall) = E(cat) - E(an). thus E(an) has to be bigger and since oxidation occurs at the anode, the oxidation potential has to be bigger.


Regarding E = E(red) + E(ox), I guess the E(ox) has to be "bigger" in terms of absolute value even though its a negative number?
 
Thanks for the help but still confuzzled T.T hahah ill take look at it again later
 
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