On the Standard Reduction Potentials of Half-Cells table, why does the Oxidizing agent have to be higher than Reducing agent for a redox reaction to spontaneously occur?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ I thought that the only requirement was: if $E^{\circ}$ is positive = the redox is spontaneous. I thought that it did not matter which one is more positive. $\endgroup$– CoffeeIsLifeJan 15, 2017 at 20:46
-
$\begingroup$ You're right. It has to be positive and I get that. However, that's the mathematical explanation. But how can you explain it scientifically? $\endgroup$– A.AKJan 16, 2017 at 3:22
-
$\begingroup$ Need some clarification here based on your comment. Are you interested in why a positive cell potential means a negative deltaG or are you interested in why a negative deltaG means spontaneous reaction? $\endgroup$– Burak UlgutJan 16, 2017 at 5:48
-
$\begingroup$ The sign convention may be considered to be 'back to front' since $\Delta G=-nF\Delta E$ with $\Delta E$ as the difference in standard redox potentials for the reaction. $\Delta G$ is negative for a spontaneous reaction. Oxidising agents have more positive redox $E^0$ in tables of redox potentials than reducing agents do. It is the difference in redox that determines if a reaction is spontaneous. 'Scientifically' a reducing agent is an electron donor and so an oxidising agent is an electron acceptor. $\endgroup$– porphyrinJan 16, 2017 at 10:24
Add a comment
|