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Electrode Potentials and Redox reactions 
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#1
Feb213, 03:09 AM

P: 247

What is the relation between Electrode Potentials and Redox reactions ?
Can we determine the relative strength of reduction of a metal by looking just at the electrode potentials ? Here is a particular thing that confuses me : We know that the standard electrode potential of Lithium is 3.05V I know that this implies that the bulk of the following reaction is to the right . (ie Lithium is a very good reducing agent. Li [itex]\longleftrightarrow[/itex] [itex]Li^{+}[/itex] + [itex]e^{}[/itex] So here is the confusion : 1. What is the electrode potential of this reaction ? Li [itex]\rightarrow[/itex] [itex]Li^{+}[/itex] + [itex]e^{}[/itex] 2.What is the electrode potential of this reaction ? Li [itex]\leftarrow[/itex] [itex]Li^{+}[/itex] + [itex]e^{}[/itex] are these going to be exactly the same as 3.05V or would we see a sign flip ? Also, what exactly is a reduction potential and an oxidation potential ? (is there a difference between the two) ? 


#2
Feb213, 08:57 AM

P: 59

The reduction half reaction would be the negative one, and the oxidation half reaction would have the same magnitude as the standard reduction potential of lithium, but opposite in sign (so in short, positive). The values for the standard potential are usually listed in a table called standard reduction potential table, which I'm assuming you know. Those values are for the reduction reaction. The reverse of the reaction (the oxidation) has the cell potential opposite to the sign of the reaction cell potential of the other reaction.
The standard reduction potential values are from an experiment made with a voltaic cell, with the other electrode being SHE (Standard Hydrogen Electrode). The value for H+ is assigned with a value of 0 V arbitrarily. The reduction potential, as its name suggests, gives you an idea how that particular element is more likely to be reduced. And the oxidation potential gives you an idea how that particular element is more likely to be oxidized. If the value for the reduction potential is positive, then it is spontaneous (meaning this occurs in nature). If it is negative, then it isn't likely to occur. As you may see from your equation, oxidation and reduction is seen by the forward and reverse equations. This means that one will have the positive value, and the other one having a negative value. Only one direction will be favored and be spontaneous. 


#3
Feb713, 10:15 AM

Admin
P: 23,579

This is a misleading approach.
Reaction goes both ways at exactly the same potential, doesn't matter if it is oxidation or reduction. When we want to calculate cell potential difference we should subtract EOx from ERed  and the sign will "flip itself" whenever necessary, this is simple math. When calculating temperature difference between two objects of which one is at 10°C and the other at 10°C, we subtract 10(10)=20. Sure, we can say one has a "positive temperature" of 10°C and the other has "negative temperature" of 10°C, and the difference is sum of these  but it is just making things confusing when they are not. 


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